orbital mechanics

earn Orbital ie Mechanics Engineering Students Orbital Mechanics for Engineering Students T o my parents, Rondo and Geraldine, and my wife, Connie Dee Orbital Mechanics for Engineering Students Howard D. Curtis Embry-Riddle Aeronautical University Daytona Beach, Florida AMSTERDAM •BOSTON •HEIDELBERG •LONDON •NEW YORK •OXFORD PARIS •SAN DIEGO •SAN FRANCISCO •SINGAPORE •SYDNEY •TOKYO Elsevier Butterworth-Heinemann Linacre House, Jordan Hill, Oxford OX2 8DP30 Corporate Drive, Burlington, MA 01803 First published 2005Copyright © 2005, Howard D. Curtis. All rights reservedThe right of Howard D. Curtis to be identified as the author of this work has been asserted in accordance with the Copyright, Design andPatents Act 1988 No part of this publication may be reproduced in any material form (including photocopying or storing in any medium by electronic means and whether ornot transiently or incidentally to some other use of this publication) without the written permission of the copyright holder except in accordance with theprovisions of the Copyright, Designs and Patents Act 1988 or under the terms of alicence issued by the Copyright Licensing Agency Ltd, 90 T ottenham Court Road,London, England W1T 4LP . Applications for the copyright holder’s writtenpermission to reproduce any part of this publication should be addressed tothe publisher Permissions may be sought directly from Elsevier’s Science & T echnology Rights Department in Oxford, UK: phone ( +44) 1865 843830, fax: (+44) 1865 853333, e-mail: [email protected]. Y ou may also complete your request on-line via the Elsevier homepage (http://www.elsevier.com), by selecting ‘Customer Support’ and then ‘Obtaining Permissions’ British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloguing in Publication Data A catalogue record for this book is available from the Library of Congress ISBN 0 7506 6169 0 For information on all Elsevier Butterworth-Heinemann publications visit our website at http://books.elsevier.com Typeset by Charon T ec Pvt. Ltd, Chennai, India www.charontec.comPrinted and bound in Great Britain by Biddles Ltd, King’s Lynn, Norfolk Contents Preface xi Supplements to the text xv Chapter1 Dynamics of point masses 1 1.1 Introduction 1 1.2 Kinematics 2 1.3 Mass, force and Newton’s law of gravitation 7 1.4 Newton’s law of motion 10 1.5 Time derivatives of moving vectors 15 1.6 Relative motion 20 Problems 29 Chapter2 The two-body problem 33 2.1 Introduction 33 2.2 Equations of motion in an inertial frame 34 2.3 Equations of relative motion 37 2.4 Angular momentum and the orbit formulas 42 2.5 The energy law 50 2.6 Circular orbits (e=0) 51 2.7 Elliptical orbits (0<e<1) 55 2.8 Parabolic trajectories (e=1) 65 2.9 Hyperbolic trajectories (e>1) 69 2.10 Perifocal frame 76 2.11 The Lagrange coefficients 78 2.12 Restricted three-body problem 89 2.12.1 Lagrange points 92 2.12.2 Jacobi constant 96 Problems 101 Chapter3 Orbital position as a function of time 107 3.1 Introduction 107 3.2 Time since periapsis 108 v vi Contents 3.3 Circular orbits 108 3.4 Elliptical orbits 109 3.5 Parabolic trajectories 124 3.6 Hyperbolic trajectories 125 3.7 Universal variables 134 Problems 145 Chapter4 Orbits in three dimensions 149 4.1 Introduction 149 4.2 Geocentric right ascension–declination frame 150 4.3 State vector and the geocentric equatorial frame 154 4.4 Orbital elements and the state vector 158 4.5 Coordinate transformation 164 4.6 Transformation between geocentric equatorial and perifocal frames 172 4.7 Effects of the earth’s oblateness 177 Problems 187 Chapter5 Preliminary orbit determination 193 5.1 Introduction 193 5.2 Gibbs’ method of orbit determination from threeposition vectors 194 5.3 Lambert’s problem 202 5.4 Sidereal time 213 5.5 Topocentric coordinate system 218 5.6 Topocentric equatorial coordinate system 221 5.7 Topocentric horizon coordinate system 223 5.8 Orbit determination from angle and rangemeasurements 228 5.9 Angles-only preliminary orbit determination 235 5.10 Gauss’s method of preliminary orbit determination 236 Problems 250 Chapter6 Orbital maneuvers 255 6.1 Introduction 255 6.2 Impulsive maneuvers 256 6.3 Hohmann transfer 257 Contents vii 6.4 Bi-elliptic Hohmann transfer 264 6.5 Phasing maneuvers 268 6.6 Non-Hohmann transfers with a common apse line 273 6.7 Apse line rotation 279 6.8 Chase maneuvers 285 6.9 Plane change maneuvers 290 Problems 304 Chapter7 Relative motion and rendezvous 315 7.1 Introduction 315 7.2 Relative motion in orbit 316 7.3 Linearization of the equations of relative motion in orbit 322 7.4 Clohessy–Wiltshire equations 324 7.5 Two-impulse rendezvous maneuvers 330 7.6 Relative motion in close-proximity circular orbits 338 Problems 340 Chapter8 Interplanetary trajectories 347 8.1 Introduction 347 8.2 Interplanetary Hohmann transfers 348 8.3 Rendezvous opportunities 349 8.4 Sphere of influence 354 8.5 Method of patched conics 359 8.6 Planetary departure 360 8.7 Sensitivity analysis 366 8.8 Planetary rendezvous 368 8.9 Planetary flyby 375 8.10 Planetary ephemeris 387 8.11 Non-Hohmann interplanetary trajectories 391 Problems 398 Chapter9 Rigid-body dynamics 399 9.1 Introduction 399 9.2 Kinematics 400 9.3 Equations of translational motion 408 9.4 Equations of rotational motion 410 viii Contents 9.5 Moments of inertia 414 9.5.1 Parallel axis theorem 428 9.6 Euler’s equations 435 9.7 Kinetic energy 441 9.8 The spinning top 443 9.9 Euler angles 448 9.10 Yaw, pitch and roll angles 459 Problems 463 Chapter10 Satellite attitude dynamics 475 10.1 Introduction 475 10.2 Torque-free motion 476 10.3 Stability of torque-free motion 486 10.4 Dual-spin spacecraft 491 10.5 Nutation damper 495 10.6 Coning maneuver 503 10.7 Attitude control thrusters 506 10.8 Yo-yo despin mechanism 509 10.9 Gyroscopic attitude control 516 10.10 Gravity-gradient stabilization 530 Problems 543 Chapter11 Rocket vehicle dynamics 551 11.1 Introduction 551 11.2 Equations of motion 552 11.3 The thrust equation 555 11.4 Rocket performance 557 11.5 Restricted staging in field-free space 560 11.6 Optimal staging 570 11.6.1 Lagrange multiplier 570 Problems 578 References and further reading 581 AppendixA Physical data 583 AppendixB A road map 585 Contents ix AppendixC Numerical integration of the n-body equations of motion 587 C.1 Function file accel_3body.m 590 C.2 Script file threebody.m 592AppendixD MATLAB algorithms 595 D.1 Introduction 596 D.2 Algorithm 3.1: solution of Kepler’s equation by Newton’s method 596 D.3 Algorithm 3.2: solution of Kepler’s equation for thehyperbola using Newton’s method 598 D.4 Calculation of the Stumpff functions S(z)and C(z) 600 D.5 Algorithm 3.3: solution of the universal Kepler’s equation using Newton’s method 601 D.6 Calculation of the Lagrange coefficients fand gand their time derivatives 603 D.7 Algorithm 3.4: calculation of the state vector ( r,v) given the initial state vector ( r0,v0) and the time lapse /Delta1t 604 D.8 Algorithm 4.1: calculation of the orbital elements from the state vector 606 D.9 Algorithm 4.2: calculation of the state vector fromthe orbital elements 610 D.10 Algorithm 5.1: Gibbs’ method of preliminary orbitdetermination 613 D.11 Algorithm 5.2: solution of Lambert’s problem 616 D.12 Calculation of Julian day number at 0 hr UT 621 D.13 Algorithm 5.3: calculation of local sidereal time 623 D.14 Algorithm 5.4: calculation of the state vectorfrom measurements of range, angular position andtheir rates 626 D.15 Algorithms 5.5 and 5.6: Gauss’s method of preliminaryorbit determination with iterative improvement 631 D.16 Converting the numerical designation of a month ora planet into its name 640 D.17 Algorithm 8.1: calculation of the state vector ofa planet at a given epoch 641 D.18 Algorithm 8.2: calculation of the spacecraft trajectoryfrom planet 1 to planet 2 648AppendixE Gravitational potential energy of a sphere 657 Index 661 This page intentionally left blank Preface This textbook evolved from a formal set of notes developed over nearly ten years of teaching an introductory course in orbital mechanics for aerospace engineeringstudents. These undergraduate students had no prior formal experience in the subject,but had completed courses in physics, dynamics and mathematics through differentialequations and applied linear algebra. That is the background I have presumed forreaders of this book. This is by no means a grand, descriptive survey of the entire subject of astronautics. It is a foundations text, a springboard to advanced study of the subject. I focus on thephysical phenomena and analytical procedures required to understand and predict, tofirst order, the behavior of orbiting spacecraft. I have tried to make the book readablefor undergraduates, and in so doing I do not shy away from rigor where it is neededfor understanding. Spacecraft operations that take place in earth orbit are consideredas are interplanetary missions. The important topic of spacecraft control systems isomitted. However, the material in this book and a course in control theory providethe basis for the study of spacecraft attitude control. A brief perusal of the Contents shows that there are more than enough topics to cover in a single semester or term. Chapter 1 is a review of vector kinematics inthree dimensions and of Newton’s laws of motion and gravitation. It also focuses onthe issue of relative motion, crucial to the topics of rendezvous and satellite attitudedynamics. Chapter 2 presents the vector-base d solution of the classical two-body problem, coming up with a host of practical formulas for orbit and trajectory analy-sis. The restricted three-body problem is covered in order to introduce the notion ofLagrange points. Chapter 3 derives Kepler’ s equations, which relate position to time for the different kinds of orbits. The concept of ‘universal variables’ is introduced.Chapter 4 is devoted to describing orbits in three dimensions and accounting for themajor effects of the earth’s oblate, non-sphe rical shape. Chapter 5 is an introduction to preliminary orbit determination, including Gibbs’ and Gauss’s methods and the solution of Lambert’s problem. Auxiliary topics include topocentric coordinate sys-tems, Julian day numbering and sidereal time. Chapter 6 presents the common meansof transferring from one orbit to another by impulsive delta-v maneuvers, includingHohmann transfers, phasing orbits and plane changes. Chapter 7 derives and employsthe equations of relative motion required to understand and design two-impulse ren-dezvous maneuvers. Chapter 8 explores the basics of interplanetary mission analysis.Chapter 9 presents those elements of rigid-body dynamics required to characterizethe attitude of an orbiting satellite. Chapte r 10 describes the methods of controlling, changing and stabilizing the attitude of spacecraft by means of thrusters, gyros andother devices. Finally, Chapter 11 is a brief introduction to the characteristics anddesign of multi-stage launch vehicles. Chapters 1 through 4 form the core of a first orbital mechanics course. The time devoted to Chapter 1 depends on the background of the student. It might be surveyed xi xii Preface briefly and used thereafter simply as a reference. What follows Chapter 4 depends on the objectives of the course. Chapters 5 through 8 carry on with the subject of orbital mechanics. Chapter 6 on orbital maneuvers should be included in any case. Coverage of Chapters 5, 7 and8 is optional. However, if all of Chapter 8 on interplanetary missions is to form a partof the course, then the solution of Lambert’s problem (Section 5.3) must be studiedbeforehand. Chapters 9 and 10 must be covered if the course objectives include an introduction to satellite dynamics. In that case Chapters 5, 7 and 8 would probably not be studiedin depth. Chapter 11 is optional if the engineering curriculum requires a separate course in propulsion, including rocket dynamics. T o understand the material and to solve problems requires using a lot of under- graduate mathematics. Mathematics, of course, is the language of engineering.Students must not forget that Sir Isaac Newton had to invent calculus so he could solveorbital mechanics problems precisely. Newton (1642–1727) was an English physi-cist and mathematician, whose 1687 publication Mathematical Principles of Natural Philosophy (‘the Principia ’) is one of the most influential scientific works of all time. It must be noted that the German mathematician Gottfried Wilhelm von Leibniz (1646–1716) is credited with inventing infinitesimal calculus independently of Newton inthe 1670s. In addition to honing their math skills, students are urged to take advantage of computers (which, incidentally, use the binary numeral system developed byLeibniz). There are many commercially available mathematics software packages forpersonal computers. Wherever possible they should be used to relieve the burden ofrepetitive and tedious calculations. Computer programming skills can and should beput to good use in the study of orbital mechanics. Elementary MATLAB® programs(M-files) appear at the end of this book to illustrate how some of the procedures devel-oped in the text can be implemented in software. All of the scripts were developedusing MATLAB version 5.0 and were successfully tested using version 6.5 (release 13).Information about MATLAB, which is a registered trademark of The MathWorks,Inc., may be obtained from: The MathWorks, Inc. 3 Apple Hill DriveNatick, MA, 01760-2098 USAT el: 508-647-7000Fax: 508-647-7101E-mail: [email protected] Web: www.mathworks.com The text contains many detailed explanations and worked-out examples. Their purpose is not to overwhelm but to elucidate. It is always assumed that the material isbeing seen for the first time and, wherever possible, solution details are provided so asto leave little to the reader’s imagination. There are some exceptions to this objective,deemed necessary to maintain the focus and control the size of the book. For example,in Chapter 6, the notion of specific impulse is laid on the table as a means of ratingrocket motor performance and to show precisely how delta-v is related to propellantexpenditure. In Chapter 10 Routh–Hurwitz stability criteria are used without proof to Preface xiii show quantitatively that a particular satellit e configuration is, indeed, stable. Specific impulse is covered in more detail in Chapter 11, and the stability of linear systems istreated in depth in books on control theory. See, for example, Nise (2003) and Ogata(2001). Supplementary material appears in the appendices at the end of the book. Appendix A lists physical data for use throughout the text. Appendix B is a ‘road map’ to guide the reader through Chapters 1, 2 and 3. Appendix C shows how to setup the n-body equations of motion and program them in MATLAB. Appendix D lists the MATLAB implementations of algorithms presented in several of the chapters. Appendix E shows that the gravitational field of a spherically symmetric body is the same as if the mass were concentrated at its center. The field of astronautics is rich and vast. References cited throughout this text are listed at the end of the book. Also listed are other books on the subject that might beof interest to those seeking additional insights. I wish to thank colleagues who provided helpful criticism and advice during the development of this book. Y echiel Crispin and Charles Eastlake were sources forideas about what should appear in the summary chapter on rocket dynamics. HabibEslami, Lakshmanan Narayanaswami, Mahmut Reyhanoglu and Axel Rohde all usedthe evolving manuscript as either a text or a reference in their space mechanics courses.Based on their classroom experiences, they gave me valuable feedback in the formof corrections, recommendations and much-needed encouragement. T ony Hagarvoluntarily and thoroughly reviewed the entire manuscript and made a number ofsuggestions, nearly all of which were incorporated into the final version of the text. I am indebted to those who reviewed the manuscript for the publisher for their many suggestions on how the book could be improved and what additional topicsmight be included. Finally, let me acknowledge how especially grateful I am to the students who, throughout the evolution of the book, reported they found it to be a helpful andunderstandable introduction to space mechanics. Howard D. Curtis Embry-Riddle Aeronautical University Daytona Beach, Florida This page int entionally left blank Supplements to the text For the student: •Copies of the MATLAB programs (M-files) that appear in Appendix D can be downloaded from the companion website accompanying this book. T o access these please visit http://books.elsevier.com/companions and follow theinstructions on screen. For the instructor: •A full Instructor’s Solutions Manual is available for adopting tutors, which pro- vides complete worked-out solutions to the problems set at the end of eachchapter. T o access these please visit http://books.elsevier.com/manuals and followthe instructions on screen. xv This page int entionally left blank 1Chapter Dynamics of point masses Chapter outline 1.1 Introduction 1 1.2 Kinematics 2 1.3 Mass, force and Newton’s law of gravitation 7 1.4 Newton’s law of motion 10 1.5 Time derivatives of moving vectors 15 1.6 Relative motion 20 Problems 29 1.1 Introduction This chapter serves as a self-contained reference on the kinematics and dynamics of point masses as well as some basic vector operations. The notation and concepts summarized here will be used in the following chapters. Those familiar withthe vector-based dynamics of particles can simply page through the chapter and thenrefer back to it later as necessary. Those who need a bit more in the way of reviewwill find the chapter contains all of the material they need in order to follow thedevelopment of orbital mechanics topics in the upcoming chapters. We begin with the problem of describing the curvilinear motion of particles in three dimensions. The concepts of force and mass are considered next, alongwith Newton’s inverse-square law of gravitation. This is followed by a presentation 1 2Chapter 1 Dynamics of point masses of Newton’s second law of motion (‘force equals mass times acceleration’) and the important concept of angular momentum. As a prelude to describing motion relative to moving frames of reference, we develop formulas for calculating the time derivatives of moving vectors. These areapplied to the computation of relative velocity and acceleration. Example problemsillustrate the use of these results as does a detailed consideration of how the earth’srotation and curvature influence our measurements of velocity and acceleration. Thisbrings in the curious concept of Coriolis force. Embedded in exercises at the end ofthe chapter is practice in verifying several fundamental vector identities that will beemployed frequently throughout the book. 1.2 Kinematics T o track the motion of a particle Pthrough Euclidean space we need a frame of reference, consisting of a clock and a cartesian coordinate system. The clock keepstrack of time tand the xyzaxes of the cartesian coordinate system are used to locate the spatial position of the particle. In non-relativistic mechanics, a single ‘universal’clock serves for all possible cartesian coordinate systems. So when we refer to a frameof reference we need think only of the mutually orthogonal axes themselves. The unit of time used throughout this book is the second (s). The unit of length is the meter (m), but the kilometer (km) will be the length unit of choice when largedistances and velocities are involved. Conversion factors between kilometers, milesand nautical miles are listed in Table A.3. Given a frame of reference, the position of the particle Pat a time tis defined by the position vector r(t) extending from the origin Oof the frame out to Pitself, as illustrated in Figure 1.1. (Vectors will always be indicated by boldface type.) The xyz O/rho1v a P s oPath Figure 1.1 Position, velocity and acceleration vectors. 1.2 Kinematics 3 components of r(t) are just the x,yand zcoordinates, r(t)=x(t)ˆi+y(t)ˆj+z(t)ˆk ˆi,ˆjandˆkare the unit vectors which point in the positive direction of the x,yand z axes, respectively. Any vector written with the overhead hat (e.g., ˆa)i st ob ec o n s i d e r e d a vector of unit dimensionless magnitude. The distance of Pfrom the origin is the magnitude or length of r, denoted /bardblr/bardblor justr, /bardblr/bardbl=r =/radicalBig x2+y2+z2 The magnitude of r,o ra n yv e c t o r Afor that matter, can also be computed by means of the dot product operation, r=√r·r/bardblA/bardbl=√ A·A The velocity vand acceleration aof the particle are the first and second time derivatives of the position vector, v(t)=dx(t) dtˆi+dy(t) dtˆj+dz(t) dtˆk=vx(t)ˆi+vy(t)ˆj+vz(t)ˆk a(t)=dvx(t) dtˆi+dvy(t) dtˆj+dvz(t) dtˆk=ax(t)ˆi+ay(t)ˆj+az(t)ˆk It is convenient to represent the time derivative by means of an overhead dot. In this shorthand notation, if ( ) is any quantity, then (·)≡d() dt(··)≡d2() dt2(···)≡d3() dt3,e t c. Thus, for example, v=˙r a=˙v=¨r vx=˙x vy=˙y vz=˙z ax=˙vx=¨xa y=˙vy=¨ya z=˙vz=¨z The locus of points that a particle occupies as it moves through space is called its path or trajectory. If the path is a straight line, then the motion is rectilinear. Otherwise, thepath is curved, and the motion is called curvilinear. The velocity vector vis tangent to the path. If ˆu tis the unit vector tangent to the trajectory, then v=vˆut where v, the speed, is the magnitude of the velocity v. The distance dsthat Ptravels along its path in the time interval dtis obtained from the speed by ds=vdt 4Chapter 1 Dynamics of point masses In other words, v=˙s The distance s, measured along the path from some starting point, is what the odome- ters in our automobiles record. Of course, ˙s, our speed along the road, is indicated by the dial of the speedometer. Note carefully that v/negationslash=˙r, i.e., the magnitude of the derivative of rdoes not equal the derivative of the magnitude of r. Example 1.1The position vector in meters is given as a function of time in seconds as r=(8t2+7t+6)ˆi+(5t3+4)ˆj+(0.3t4+2t2+1)ˆk(m) (a) Att=10 seconds, calculate v(the magnitude of the derivative of r) and ˙r(the derivative of the magnitude of r). The velocity vis found by differentiating the give n position vector with respect to time, v=dr dt=(16t+7)ˆi+15t2ˆj+(1.2t3+4t)ˆk The magnitude of this vector is the square root of the sum of the squares of its components, /bardblv/bardbl= (1.44t6+234.6t4+272t2+224t+49)1 2 Evaluating this at t=10 s, we get v=1953.3m/s Calculating the magnitude of rin (a), leads to /bardblr/bardbl= (0.09t8+26.2t6+68.6t4+152t3+149t2+84t+53)1 2 Differentiating this expression with respect to time, ˙r=dr dt=0.36t7+78.6t5+137.2t3+228t2+149t+42 (0.09t8+26.2t6+68.6t4+152t3+149t2+84t+53)1 2 Substituting t=10 s, yields ˙r=1935.5m/s Ifvis given, then we can find the components of the unit tangent ˆutin the cartesian coordinate frame of reference ˆut=v /bardblv/bardbl=vx vˆi+vy vˆj+vz vˆk/parenleftBig v=/radicalBig v2x+v2y+v2z/parenrightBig 1.2 Kinematics 5 The acceleration may be written, a=atˆut+anˆun where atand anare the tangential and normal components of acceleration, given by at=˙v(=¨s) an=v2 /rho1(1.1) /rho1is the radius of curvature, which is the distance from the particle Pto the center of curvature of the path at that point. The unit principal normal ˆunis perpendicular to ˆutand points towards the center of curvature C, as shown in Figure 1.2. Therefore, the position of Crelative to P, denoted rC/P,i s rC/P=/rho1ˆun The orthogonal unit vectors ˆutandˆunform a plane called the osculating plane. The unit normal to the osculating plane is ˆub, the binormal, and it is obtained from ˆut andˆunby taking their cross product, ˆub=ˆut׈un The center of curvature lies in the osculating plane. When the particle Pmoves an incremental distance dsthe radial from the center of curvature to the path sweeps out a small angle dφ, measured in the osculating plane. The relationship between this angle and dsis ds=/rho1dφ so that ˙s=/rho1˙φ,o r ˙φ=v /rho1(1.2) xyz OP Cdfdsut unub rOsculating plane ˆˆ ˆ Figure 1.2 Orthogonal triad of unit vectors associated with the moving point P. 6Chapter 1 Dynamics of point masses Example 1.2Relative to a cartesian coordinate system, the position, velocity and acceleration of a particle relative at a given instant are r=250ˆi+630ˆj+430ˆk(m) v=90ˆi+125ˆj+170ˆk(m/s) a=16ˆi+125ˆj+30ˆk(m/s2) Find the coordinates of the center of curvature at that instant. First, we calculate the speed v, v=/bardbl v/bardbl=/radicalbig 902+1252+1702=229.4m/s The unit tangent is, therefore, ˆut=v v=90ˆi+125ˆj+170ˆk 797.4=0.3923ˆi+0.5449ˆj+0.7411ˆk We project the acceleration vector onto the direction of the tangent to get its tangential component at, at=a·ˆut=(16ˆi+125ˆj+30ˆk)·(0.3923ˆi+0.5449ˆj+0.7411ˆk)=96.62 m/s2 The magnitude of ais a=/radicalbig 162+1252+302=129.5m/s2 Since a=atˆut+anˆunandˆutandˆunare perpendicular to each other, it follows that a2=a2 t+a2 n, which means an=/radicalBig a2−a2 t=/radicalbig 129.52−96.622=86.29 m/s2 Hence, ˆun=1 an(a−atˆut) =1 86.29[(16ˆi+125ˆj+30ˆk)−96.62(0.3923ˆi+0.5449ˆj+0.7411ˆk)] =− 0.2539ˆi+0.8385ˆj−0.4821ˆk The equation an=v2//rho1can now be solved for /rho1to yield /rho1=v2 an=229.42 86.29=609.9m 1.3 Mass, force and Newton’s law of gravitation 7 LetrCbe the position vector of the center of curvature C. Then rC=r+rC/P =r+/rho1ˆun=250ˆi+630ˆj+430ˆk+609.9(−0.2539 ˆi+0.8385 ˆj−0.4821 ˆk) =95.16ˆi+1141ˆj+136.0ˆk(m) That is, the coordinates of Care x=95.16 m y=1141 m z=136.0m 1.3 Mass, force and Newton’s law of gravitation Mass, like length and time, is a primitive physical concept: it cannot be defined in terms of any other physical concept. Mass is simply the quantity of matter. Morepractically, mass is a measure of the inertia of a body. Inertia is an object’s resistanceto changing its state of motion. The larger its inertia (the greater its mass), the moredifficult it is to set a body into motion or bring it to rest. The unit of mass is thekilogram (kg). Force is the action of one physical body on another, either through direct contact or through a distance. Gravity is an example of force acting through a distance, as aremagnetism and the force between charged particles. The gravitational force betweentwo masses m 1and m2having a distance rbetween their centers is Fg=Gm1m2 r2(1.3) This is Newton’s law of gravity, in which G, the universal gravitational constant, has the value 6.6742 ×1011m3/kg·s2. Due to the inverse-square dependence on distance, the force of gravity rapidly diminishes with the amount of separation between the two masses. In any case, the force of gravity is minuscule unless at least one of themasses is extremely big. The force of a large mass (such as the earth) on a mass many orders of magnitude smaller (such as a person) is called weight, W. If the mass of the large object is Mand that of the relatively tiny one is m, then the weight of the small body is W=GMm r2=m/parenleftbiggGM r2/parenrightbigg or W=mg (1.4) where g=GM r2(1.5) 8Chapter 1 Dynamics of point masses ghas units of acceleration (m/s2) and is called the acceleration of gravity. If planetary gravity is the only force acting on a body, then the body is said to be in free fall. Theforce of gravity draws a freely falling object towards the center of attraction (e.g.,center of the earth) with an acceleration g. Under ordinary conditions, we sense our own weight by feeling contact forces acting on us in opposition to the force of gravity.In free fall there are, by definition, no contact forces, so there can be no sense of weight.Even though the weight is not zero, a person in free fall experiences weightlessness,or the absence of gravity. Let us evaluate Equation 1.5 at the surface of the earth, whose radius according to Table A.1 is 6378 km. Letting g 0represent the standard sea-level value of g,w eg e t g0=GM R2 E(1.6) In SI units, g0=9.807 m/s (1.7) Substituting Equation 1.6 into Equation 1.5 and letting zrepresent the distance above the earth’s surface, so that r=RE+z, we obtain g=g0R2 E (RE+z)2=g0 (1+z/RE)2(1.8) Commercial airliners cruise at altitudes on the order of 10 kilometers (six miles). At that height, Equation 1.8 reveals that g(and hence weight) is only three-tenths of a percent less than its sea-level value. Thus, under ordinary conditions, we ignore thevariation of gwith altitude. A plot of Equation 1.8 out to a height of 1000 km (the upper limit of low-earth orbit operations) is shown in Figure 1.3. The variation ofgover that range is significant. Even so, at space station altitude (300 km), weight is only about 10 percent less that it is on the earth’s surface. The astronauts experienceweightlessness, but they clearly are not weightless. 200 400 600 8000.70.80.91.0 1000 z, kmg/g0 00 Figure 1.3 Variation of the acceleration of gravity with altitude. 1.3 Mass, force and Newton’s law of gravitation 9 Example 1.3Show that in the absence of an atmosphere, the shape of a low altitude ballistic trajectory is a parabola. Assume the acceleration of gravity gis constant and neglect the earth’s curvature. xy (x0, y0)υ0 P gg0 Figure 1.4 Flight of a low altitude projectile in free fall (no atmosphere). Figure 1.4 shows a projectile launched at t=0w i t has p e e d v0at a flight path angle γ0from the point with coordinates ( x0,y0). Since the projectile is in free fall after launch, its only acceleration is that of gravity in the negative y-direction: ¨x=0 ¨y=− g Integrating with respect to time and applying the initial conditions leads to x=x0+(v0cosγ0)t (a) y=y0+(v0sinγ0)t−1 2gt2(b) Solving (a) for tand substituting the result into (b) yields y=y0+(x−x0) tanγ0−1 2g v0cosγ0(x−x0)2(c) This is the equation of a second-degree curve, a parabola, as sketched in Figure 1.4. Example 1.4An airplane flies a parabolic trajectory like that in Figure 1.4 so that the passengers will experience free fall (weightlessness). What is the required variation of the flightpath angle γwith speed v? Ignore the curvature of the earth. Figure 1.5 reveals that for a ‘flat’ earth, dγ=− dφ, i.e., ˙γ=−˙φ 10 Chapter 1 Dynamics of point masses (Example 1.4 continued)It follows from Equation 1.2 that /rho1˙γ=−v (1.9) The normal acceleration anis just the component of the gravitational acceleration g in the direction of the unit principal normal to the curve (from Ptowards C). From Figure 1.5, then, an=gcosγ (a) Substituting Equation 1.1 into (a) and solving for the radius of curvature yields /rho1=v2 gcosγ(b) Combining Equations 1.9 and (b), we find the time rate of change of the flight path angle, ˙γ=−gcosγ v dfgdg Cry xgP g Figure 1.5 Relationship between dγand dφfor a ‘flat’ earth. 1.4 Newton’s law of motion Force is not a primitive concept like mass because it is intimately connected with the concepts of motion and inertia. In fact, the only way to alter the motion of a body isto exert a force on it. The degree to which the motion is altered is a measure of theforce. This is quantified by Newton’s second law of motion. If the resultant or netforce on a body of mass misF net, then Fnet=ma (1.10) 1.4 Newton’s law of motion 11 xyz r Inertial frame ijk OmFneta v ˆ ˆ ˆ Figure 1.6 The absolute acceleration of a particle is in the direction of the net force. In this equation, ais the absolute acceleration of the center of mass. The absolute acceleration is measured in a frame of reference which itself has neither translationalnor rotational acceleration relative to the fixed stars. Such a reference is called anabsolute or inertial frame of reference. Force, then, is related to the primitive concepts of mass, length and time by Newton’s second law. The unit of force, appropriately, is the Newton, which is theforce required to impart an acceleration of 1 m/s 2to a mass of 1 kg. A mass of one kilogram therefore weighs 9.81 Newtons at the earth’s surface. The kilogram is not aunit of force. Confusion can arise when mass is expressed in units of force, as frequently occurs in US engineering practice. In common parlance either the pound or the ton (2000pounds) is more likely to be used to express the mass. The pound of mass is officiallydefined precisely in terms of the kilogram as shown in Table A.3. Since one pound of mass weighs one pound of force where the standard sea-level acceleration of gravity(g 0=9.80665 m/s2) exists, we can use Newton’s second law to relate the pound of force to the Newton: 1 lb (force) =0.4536 kg ×9.807 m /s2 =4.448 N The slug is the quantity of matter accelerated at one foot per second2by a force of one pound. We can again use Newton’s second law to relate the slug to the kilogram.Noting the relationship between feet and meters in Table A.3, we find 1s l u g =1l b 1f t/s2=4.448 N 0.3048 m /s2=14.59kg·m/s2 m/s2 =14.59 kg 12 Chapter 1 Dynamics of point masses Example 1.5On a NASA mission the space shuttle Atlantis orbiter was reported to weigh 239 255 lb just prior to lift-off. On orbit 18 at an altitude of about 350 km, the orbiter’s weight was reported to be 236 900 lb. (a) What was the mass, in kilograms, of Atlantis on the launch pad and in orbit? (b) If no mass were lost between launch and orbit 18, whatwould have been the weight of Atlantis in pounds? (a) The given data illustrates the common use of weight in pounds as a measure of mass. The ‘weights’ given are actually the mass in pounds of mass. Therefore, prior to launch mlaunch pad =239 255 lb (mass) ×0.4536 kg 1 lb (mass)=108 500 kg In orbit, morbit 18 =236 900 lb (mass) ×0.4536 kg 1 lb (mass)=107 500 kg The decrease in mass is the propellant expended by the orbital maneuvering and reaction control rockets on the orbiter. (b) Since the space shuttle launch pad at Kennedy Space Center is essentially at sea level, the launch-pad weight of Atlantis in lb (force) is numerically equal to its mass in lb (mass). With no change in mass, the force of gravity at 350 km would be, according to Equation 1.8, W=239 255 lb (force) ×/parenleftBigg 1 1+350 6378/parenrightBigg2 =215 000 lb (force) The integral of a force Fover a time interval is called the impulse Iof the force, I=/integraldisplayt2 t1Fdt (1.11) From Equation 1.10 it is apparent that if the mass is constant, then Inet=/integraldisplayt2 t1mdv dtdt=mv2−mv1 (1.12) That is, the net impulse on a body yields a change m/Delta1vin its linear momentum, so that /Delta1v=Inet m(1.13) IfFnetis constant, then Inet=Fnet/Delta1t, in which case Equation 1.13 becomes /Delta1v=Fnet m/Delta1t(ifFnetis constant) (1.14) 1.4 Newton’s law of motion 13 Let us conclude this section by introducing the concept of angular momentum. The moment of the net force about Oin Figure 1.6 is MOnet=r×Fnet Substituting Equation 1.10 yields MOnet=r×ma=r×mdv dt(1.15) But, keeping in mind that the mass is constant, r×mdv dt=d dt(r×mv)−/parenleftbiggdr dt×mv/parenrightbigg =d dt(r×mv)−(v×mv) Since v×mv=m(v×v)=0, it follows that Equation 1.15 can be written MOnet=dHO dt(1.16) where HOis the angular momentum about O, HO=r×mv (1.17) Thus, just as the net force on a particle changes its linear momentum mv, the moment of that force about a fixed point changes the moment of its linear momentum aboutthat point. Integrating Equation 1.16 with respect to time yields /integraldisplay t2 t1MOnetdt=HO2−HO1 (1.18) The integral on the left is the net angular impulse. This angular impulse–momentum equation is the rotational analog of the linear impulse–momentum relation givenabove in Equation 1.12. Example 1.6A particle of mass mis attached to point Oby an inextensible string of length l. Initially the string is slack when mis moving to the left with a speed voin the position shown. Calculate the speed of mjust after the string becomes taut. Also, compute the xy dυ0 υlm Oc Figure 1.7 Particle attached to Oby an inextensible string. 14 Chapter 1 Dynamics of point masses (Example 1.6 continued)average force in the string over the small time interval /Delta1trequired to change the direction of the particle’s motion. Initially, the position and velocity of the particle are r1=cˆi+dˆjv 1=−v0ˆi The angular momentum is H1=r1×mv1=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆiˆjˆk cd 0 −mv 000/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=mv 0ˆk (a) Just after the string becomes taut r2=−/radicalbig l2−d2ˆi+dˆjv 2=vxˆi+vyˆj (b) and the angular momentum is H2=r2×mv2=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆi ˆjˆk −√ l2−d2d 0 vx vy0/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/parenleftBig −mv xd−mvy/radicalbig l2−d2/parenrightBig ˆk (c) Initially the force exerted on mby the slack string is zero. When the string becomes taut, the force exerted on mpasses through O. Therefore, the moment of the net force onmabout Oremains zero. According to Equation 1.18, H2=H1 Substituting (a) and (c) yields vxd+/radicalbig l2−d2vy=−v0d (d) The string is inextensible, so the component of the velocity of malong the string must be zero: v2·r2=0 Substituting v2and r2from (b) and solving for vywe get vy=vx/radicalbigg l2 d2−1( e ) Solving (d) and (e) for vxandvyleads to vx=−d2 l2v0vy=−/radicalbigg 1−d2 l2d lv0 (f) Thus, the speed, v=/radicalBig v2x+v2y, after the string becomes taut is v=d lv0 1.5 Time derivatives of moving vectors 15 From Equation 1.12, the impulse on mduring the time it takes the string to become taut is I=m(v 2−v1)=m/bracketleftBigg/parenleftBigg −d2 l2v0ˆi−/radicalbigg 1−d2 l2d lv0ˆj/parenrightBigg −(−v 0ˆi)/bracketrightBigg =/parenleftbigg 1−d2 l2/parenrightbigg mv0ˆi−/radicalbigg 1−d2 l2d lmv0ˆj The magnitude of this impulse, which is directed along the string, is I=/radicalbigg 1−d2 l2mv0 Hence, the average force in the string during the small time interval /Delta1trequired to change the direction of the velocity vector turns out to be Favg=I /Delta1t=/radicalbigg 1−d2 l2mv0 /Delta1t 1.5 Time derivatives of moving vectors Figure 1.8(a) shows a vector Ainscribed in a rigid body Bthat is in motion relative to an inertial frame of reference (a rigid, cartesian coordinate system which is fixedrelative to the fixed stars). The magnitude of Ais fixed. The body Bis shown at two times, separated by the differential time interval dt. At time t+dtthe orientation of ω fdθ AdAA /H11001 dA (a) (b) A(t /H11001 dt) XYZ tt /H11001 dtA(t)Rigid body B Inertial frameInstantaneous axis of rotation Figure 1.8 Displacement of a rigid body. 16 Chapter 1 Dynamics of point masses vector Adiffers slightly from that at time t, but its magnitude is the same. According to one of the many theorems of the prolific eighteenth century Swiss mathematicianLeonhard Euler (1707–1783), there is a unique axis of rotation about which Band, therefore, Arotates during the differential time interval. If we shift the two vectors A(t) and A(t+dt) to the same point on the axis of rotation, so that they are tail-to-tail as shown in Figure 1.8(b), we can assess the difference dAbetween them caused by the infinitesimal rotation. Remember that shiftin g a vector to a parallel line does not change the vector. The rotation of the body Bis measured in the plane perpendicular to the instantaneous axis of rotation. The amount of rotation is the angle dθthrough which a line element normal to the rotation axis turns in the time interval dt.I n Figure 1.8(b) that line element is the component of Anormal to the axis of rotation. We can express the difference dAbetween A(t) and A(t+dt)a s dA=magnitude of dA/bracehtipdownleft/bracehtipupright/bracehtipupleft /bracehtipdownright [(/bardblA/bardbl·sinφ)dθ]ˆn (1.19) where ˆnis the unit normal to the plane defined by Aand the axis of rotation, and it points in the direction of the rotation. The angle φis the inclination of Ato the rotation axis. By definition, dθ=/bardblω/bardbldt (1.20) where ωis the angular velocity vector, which points along the instantaneous axis of rotation and its direction is given by the right-hand rule. That is, wrapping the righthand around the axis of rotation, with the fingers pointing in the direction of dθ, results in the thumb’s defining the direction of ω. This is evident in Figure 1.8(b). It should be pointed out that the time derivative of ωis the angular acceleration, usually given the symbol α. Thus, α=dω dt(1.21) Substituting Equation 1.20 into Equation 1.19, we get dA=/bardbl A/bardbl·sinφ/bardblω/bardbldt·ˆn=(/bardblω/bardbl·/bardbl A/bardbl·sinφ)ˆndt (1.22) By definition of the cross product, ω×Ais the product of the magnitude of ω, the magnitude of A, the sine of the angle between ωand Aand the unit vector normal to the plane of ωand A, in the rotation direction. That is, ω×A=/bardblω/bardbl·/bardbl A/bardbl·sinφ·ˆn (1.23) Substituting Equation 1.23 into Equation 1.22 yields dA=ω×Adt Dividing through by dt, we finally obtain dA dt=ω×A (1.24) Equation 1.24 is a formula we can use to compute the time derivative of any vector of constant magnitude. 1.5 Time derivatives of moving vectors 17 Example 1.7Calculate the second time derivative of a vector Aof constant magnitude, expressing the result in terms of ωand its derivatives and A. Differentiating Equation 1.24 with respect to time, we get d2A dt2=d dtdA dt=d dt(ω×A)=dω dt×A+ω×dA dt Using Equations 1.21 and 1.24, this can be written d2A dt2=α×A+ω×(ω×A) (1.25) Example 1.8Calculate the third derivative of a vector Aof constant magnitude, expressing the result in terms of ωand its derivatives and A. d3A dt3=d dtd2A dt2=d dt[α×A+ω×(ω×A)] =d dt(α×A)+d dt[ω×(ω×A)] =/parenleftbiggdα dt×A+α×dA dt/parenrightbigg +/bracketleftbiggdω dt×(ω×A)+ω×d dt(ω×A)/bracketrightbigg =/bracketleftbiggdα dt×A+α×(ω×A)/bracketrightbigg +/bracketleftbigg α×(ω×A)+ω×/parenleftbiggdω dt×A+ω×dA dt/parenrightbigg/bracketrightbigg =/bracketleftbiggdα dt×A+α×(ω×A)/bracketrightbigg +{α×(ω×A)+ω×[α×A+ω×(ω×A)]} =dα dt×A+α×(ω×A)+α×(ω×A)+ω×(α×A)+ω×[ω×(ω×A)] =dα dt×A+2α×(ω×A)+ω×(α×A)+ω×[ω×(ω×A)] d3A dt3=dα dt×A+2α×(ω×A)+ω×[α×A+ω×(ω×A)] LetXYZ be a rigid inertial frame of reference and xyz a rigid moving frame of reference, as shown in Figure 1.9. The moving frame can be moving (translating androtating) freely of its own accord, or it can be imagined to be attached to a physicalobject, such as a car, an airplane or a spacecraft. Kinematic quantities measuredrelative to the fixed inertial frame will be called absolute (e.g., absolute acceleration),and those measured relative to the moving system will be called relative (e.g., relativeacceleration). The unit vectors along the inertial XYZ system are ˆI,ˆJandˆK, whereas those of the moving xyzsystem are ˆi,ˆjandˆk. The motion of the moving frame is arbitrary, and its absolute angular velocity is /Omega1. If, however, the moving frame is rigidly attached to an object, so that it not only translates but rotates with it, then the 18 Chapter 1 Dynamics of point masses XYZ xyz Inertial frameMoving frameOQ Qy QxQz J IK ij k ˆ ˆ ˆˆˆ ˆ Figure 1.9 Fixed (inertial) and moving rigid frames of reference. frame is called a body frame and the axes are referred to as body axes. A body frame clearly has the same angular velocity as the body to which it is bound. LetQbe any time-dependent vector. Resolved into components along the inertial frame of reference, it is expressed analytically as Q=QXˆI+QYˆJ+QZˆK where QX,QYand QZare functions of time. Since ˆI,ˆJandˆKare fixed, the time derivative of Qis simply given by dQ dt=dQX dtˆI+dQY dtˆJ+dQZ dtˆK dQX/dt,dQY/dtanddQZ/dtare the components of the absolute time derivative of Q. Qmay also be resolved into components along the moving xyzframe, so that, at any instant, Q=Qxˆi+Qyˆj+Qzˆk (1.26) Using this expression to calculate the time derivative of Qyields dQ dt=dQx dtˆi+dQy dtˆj+dQz dtˆk+Qxdˆi dt+Qydˆj dt+Qzdˆk dt(1.27) The unit vectors ˆi,ˆjandˆkare not fixed in space, but are continuously changing direction; therefore, their time derivatives are not zero. They obviously have a constant 1.5 Time derivatives of moving vectors 19 magnitude (unity) and, being attached to the xyzframe, they all have the angular velocity /Omega1. It follows from Equation 1.24 that dˆi dt=/Omega1׈idˆj dt=/Omega1׈jdˆk dt=/Omega1׈k Substituting these on the right-hand side of Equation 1.27 yields dQ dt=dQx dtˆi+dQy dtˆj+dQz dtˆk+Qx(/Omega1׈i)+Qy(/Omega1׈j)+Qz(/Omega1׈k) =dQx dtˆi+dQy dtˆj+dQz dtˆk+(/Omega1×Qxˆi)+(/Omega1×Qyˆj)+(/Omega1×Qzˆk) =dQx dtˆi+dQy dtˆj+dQz dtˆk+/Omega1×(Qxˆi+Qyˆj+Qzˆk) In view of Equation 1.26, this can be written dQ dt=dQ dt/parenrightbigg rel+/Omega1×Q (1.28) where dQ dt/parenrightbigg rel=dQx dtˆi+dQy dtˆj+dQz dtˆk (1.29) dQ/dt)relis the time derivative of Qrelative to the moving frame. Equation 1.28 shows how the absolute time derivative is obtained from the relative time derivative. Clearly,dQ/dt=dQ/dt) relonly when the moving frame is in pure translation ( /Omega1=0). Equation 1.28 can be used recursively to compute higher order time derivatives. Thus, differentiating Equation 1.28 with respect to t,w eg e t d2Q dt2=d dtdQ dt/parenrightbigg rel+d/Omega1 dt×Q+/Omega1×dQ dt Using Equation 1.28 in the last term yields d2Q dt2=d dtdQ dt/parenrightbigg rel+d/Omega1 dt×Q+/Omega1×/bracketleftbiggdQ dt/parenrightbigg rel+/Omega1×Q/bracketrightbigg (1.30) Equation 1.28 also implies that d dtdQ dt/parenrightbigg rel=d2Q dt2/parenrightbigg rel+/Omega1×dQ dt/parenrightbigg rel(1.31) where d2Q dt2/parenrightbigg rel=d2Qx dt2ˆi+d2Qy dt2ˆj+d2Qz dt2ˆk Substituting Equation 1.31 into Equation 1.30 yields d2Q dt2=/bracketleftbiggd2Q dt2/parenrightbigg rel+/Omega1×dQ dt/parenrightbigg rel/bracketrightbigg +d/Omega1 dt×Q+/Omega1×/bracketleftbiggdQ dt/parenrightbigg rel+/Omega1×Q/bracketrightbigg (1.32) 20 Chapter 1 Dynamics of point masses Collecting terms, this becomes d2Q dt2=d2Q dt2/parenrightbigg rel+˙/Omega1×Q+/Omega1×(/Omega1×Q)+2/Omega1×dQ dt/parenrightbigg rel where ˙/Omega1≡d/Omega1/dtis the absolute angular acceleration of the xyzframe. Formulas for higher order time derivatives are found in a similar fashion. 1.6 Relative motion LetPbe a particle in arbitrary motion. The absolute position vector of Pisrand the position of Prelative to the moving frame is rrel.I frOis the absolute position of the origin of the moving frame, then it is clear from Figure 1.10 that r=rO+rrel (1.33) Since rrelis measured in the moving frame, rrel=xˆi+yˆj+zˆk (1.34) where x,yand zare the coordinates of Prelative to the moving reference. The absolute velocity vofPisdr/dt, so that from Equation 1.33 we have v=vO+drrel dt(1.35) where vO=drO/dtis the (absolute) velocity of the origin of the xyzframe. From Equation 1.28, we can write drrel dt=vrel+/Omega1×rrel (1.36) XYZ xyz r Inertial frame (non-rotating, non-accelerating)Moving frameOP r0rrel J IKj k i ˆˆˆˆˆ Figure 1.10 Absolute and relative position vectors. 1.6 Relative motion 21 where vrelis the velocity of Prelative to the xyzframe: vrel=drrel dt/parenrightbigg rel=dx dtˆi+dy dtˆj+dz dtˆk (1.37) Substituting Equation 1.36 into Equation 1.35 yields v=vO+/Omega1×rrel+vrel (1.38) The absolute acceleration aofPisdv/dt, so that from Equation 1.35 we have a=aO+d2rrel dt2(1.39) where aO=dvO/dtis the absolute acceleration of the origin of the xyzframe. We evaluate the second term on the right using Equation 1.32: d2rrel dt2=d2rrel dt2/parenrightbigg rel+˙/Omega1×rrel+/Omega1×(/Omega1×rrel)+2/Omega1×drrel dt/parenrightbigg rel(1.40) Since vrel=drrel/dt)reland arel=d2rrel/dt2)rel, this can be written d2rrel dt2=arel+˙/Omega1×rrel+/Omega1×(/Omega1×rrel)+2/Omega1×vrel (1.41) Upon substituting this result into Equation 1.39, we find a=aO+˙/Omega1×rrel+/Omega1×(/Omega1×rrel)+2/Omega1×vrel+arel (1.42) The cross product 2 /Omega1×vrelis called the Coriolis acceleration after Gustave Gaspard de Coriolis (1792–1843), the French mathematician who introduced this term (Coriolis,1835). For obvious reasons, Equation 1.42 is sometimes referred to as the five-termacceleration formula. Example 1.9At a given instant, the absolute position, velocity and acceleration of the origin Oof a moving frame are rO=100ˆI+200ˆJ+300ˆK(m) vO=− 50ˆI+30ˆJ−10ˆK(m/s) (given) (a) aO=− 15ˆI+40ˆJ+25ˆK(m/s2) The angular velocity and acceleration of the moving frame are /Omega1=1.0ˆI−0.4ˆJ+0.6ˆK(rad/s)(given) (b)˙/Omega1=− 1.0ˆI+0.3ˆJ−0.4ˆK(rad/s2) The unit vectors of the moving frame are ˆi=0.5571 ˆI+0.7428 ˆJ+0.3714 ˆK ˆj=− 0.06331 ˆI+0.4839 ˆJ−0.8728 ˆK(given) (c) ˆk=− 0.8280 ˆI+0.4627 ˆJ+0.3166 ˆK 22 Chapter 1 Dynamics of point masses (Example 1.9 continued)The absolute position, velocity and acceleration of Pare r=300ˆI−100ˆJ+150ˆK(m) v=70ˆI+25ˆJ−20ˆK(m/s) (given) (d) a=7.5ˆI−8.5ˆJ+6.0ˆK(m/s2) Find the velocity vreland acceleration arelofPrelative to the moving frame. First use Equations (c) to solve for ˆI,ˆJandˆKin terms of ˆi,ˆjandˆk(three equations in three unknowns): ˆI=0.5571ˆi−0.06331ˆj−0.8280ˆk ˆJ=0.7428ˆi+0.4839ˆj+0.4627ˆk (e) ˆK=0.3714ˆi−0.8728ˆj+0.3166ˆk The relative position vector is rrel=r−rO=(300ˆI−100ˆJ+150ˆK)−(100ˆI+200ˆJ+300ˆK) =200ˆI−300ˆJ−150ˆK(m) (f) From Equation 1.38, the relative velocity vector is vrel=v−vO−/Omega1×rrel =(70ˆI+25ˆJ−20ˆK)−(−50ˆI+30ˆJ−10ˆK)−/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆI ˆJ ˆK 1.0−0.40 .6 200−300−150/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =(70ˆI+25ˆJ−20ˆK)−(−50ˆI+30ˆJ−10ˆK)−(240ˆI+270ˆJ−220ˆK) or vrel=− 120ˆI−275ˆJ+210ˆK(m/s) (g) T o obtain the components of the relative velocity along the axes of the moving frame, substitute Equations (e) into Equation (g). vrel=− 120(0.5571 i−0.06331 j−0.8280 k) −275(0.7428 i+0.4839 j+0.4627 k)+210(0.3714 i−0.8728 j+0.3166 k) so that vrel=− 193.1ˆi−308.8ˆj+38.60ˆk(m/s) (h) Alternatively, vrel=366.2ˆuv(m/s),w h e r e ˆuv=− 0.5272ˆi−0.8432ˆj+0.1005ˆk (i) 1.6 Relative motion 23 T o find the relative acceleration, we use the five-term acceleration formula, Equation 1.42: arel=a−aO−˙/Omega1×rrel−/Omega1×(/Omega1×rrel)−2(/Omega1×vrel) =a−aO−/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆI ˆJ ˆK −1.00 .3−0.4 200−300 −150/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle−/Omega1 ×/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆI ˆJ ˆK 1.0−0.40 .6 200−300 −150/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle−2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆI ˆJ ˆK 1.0−0.40 .6 −120 −275 210/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =a−a O−(−165 ˆI−230ˆJ+240ˆK)−/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆI ˆJ ˆK 1.0−0.40 .6 240 270 −220/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle −(162ˆI−564ˆJ−646ˆK) =(7.5ˆI−8.5ˆJ+6ˆK)−(−15ˆI+40ˆJ+25ˆK) −(−165 ˆI−230ˆJ+240ˆK)−(−74ˆI+364ˆJ+366ˆK) −(162ˆI−564ˆJ−646ˆK) arel=99.5ˆI+381.5ˆJ+21.0ˆK(m/s2) (j) The components of the relative acceleration along the axes of the moving frame are found by substituting Equations (e) into Equation (j): arel=99.5(0.5571 ˆi−0.06331 ˆj−0.8280 ˆk) +381.5(0.7428 ˆi+0.4839 ˆj+0.4627 ˆk)+21.0(0.3714 ˆi−0.8728 ˆj+0.3166 ˆk) arel=346.6ˆi+160.0ˆj+100.8ˆk(m/s2) (k) or arel=394.8ˆua(m/s2),w h e r e ˆua=0.8778 ˆi+0.4052 ˆj+0.2553 ˆk (l) Figure 1.11 shows the non-rotating inertial frame of reference XYZ with its origin at the center Cof the earth, which we shall assume to be a sphere. That assumption will be relaxed in Chapter 5. Embedded in the earth and rotating with it is theorthogonal x /primey/primez/primeframe, also centered at C, with the z/primeaxis parallel to Z, the earth’s axis of rotation. The x/primeaxis intersects the equator at the prime meridian (zero degrees longitude), which passes through Greenwich in London, England. The angle betweenXandx /primeisθg, and the rate of increase of θgis just the angular velocity /Omega1of the earth. Pis a particle (e.g., an airplane, spacecraft, etc.), which is moving in an arbitrary fashion above the surface of the earth. rrelis the position vector of Prelative to Cin the rotating x/primey/primez/primesystem. At a given instant, Pis directly over point O, which lies on 24 Chapter 1 Dynamics of point masses XY Λ (East longitude)Ω x (East)y (North) z (Zenith) EarthC x′ y′Z, z′ P REO θgGreenwich meridian Equatorrrelj k i φ (North latitude)ˆ ˆˆ Figure 1.11 Earth-centered inertial frame ( XYZ ); earth-centered non-inertial x/primey/primez/primeframe embedded in and rotating with the earth; and a non-inertial, topocentric-horizon frame xyzattached to a point Oon the earth’s surface. the earth’s surface at longitude /Lambda1and latitude φ. Point Ocoincides instantaneously with the origin of what is known as a topocentric-horizon coordinate system xyz. For our purposes xand yare measured positive eastward and northward along the local latitude and meridian, respectively, through O. The tangent plane to the earth’s surface at Ois the local horizon. The zaxis is the local vertical (straight up) and it is directed radially outward from the center of the earth. The unit vectors of the xyzframe are ˆiˆjˆk, as indicated in Figure 1.11. Keep in mind that Oremains directly below P, so that as Pmoves, so do the xyzaxes. Thus, the ˆiˆjˆktriad, which are the unit vectors of a spherical coordinate system, vary in direction as Pchanges location, thereby accounting for the curvature of the earth. Let us find the absolute velocity and acceleration of P. It is convenient first to obtain the velocity and acceleration of Prelative to the non-rotating earth, and then use Equations 1.38 and 1.42 to calculate their inertial values. The relative position vector can be written rrel=(RE+z)ˆk (1.43) 1.6 Relative motion 25 where REis the radius of the earth and zis the height of Pabove the earth (i.e., its altitude). The time derivative of rrelis the velocity vrelrelative to the non-rotating earth, vrel=drrel dt=˙zˆk+(RE+z)dˆk dt(1.44) T o calculate dˆk/dt, we must use Equation 1.24. The angular velocity ωof the xyz frame relative to the non-rotating earth is found in terms of the rates of change oflatitude φand longitude /Lambda1, ω=−˙φˆi+˙/Lambda1cosφˆj+˙/Lambda1sinφˆk (1.45) Thus, dˆk dt=ω׈k=˙/Lambda1cosφˆi+˙φˆj (1.46) Let us also record the following for future use: dˆj dt=ω׈j=−˙/Lambda1sinφˆj−˙φˆk (1.47) dˆi dt=ω׈i=˙/Lambda1sinφˆj−˙/Lambda1cosφˆk (1.48) Substituting Equation 1.46 into Equation 1.44 yields vrel=˙xˆi+˙yˆj+˙zˆk (1.49a) where ˙x=(RE+z)˙/Lambda1cosφ ˙y=(RE+z)˙φ (1.49b) It is convenient to use these results to express the rates of change of latitude and longitude in terms of the components of relative velocity over the earth’s surface, ˙φ=˙y RE+z˙/Lambda1=˙x (RE+z)c o sφ(1.50) The time derivatives of these two expressions are ¨φ=(RE+z)¨y−˙y˙z (RE+z)2¨/Lambda1=(RE+z)¨xcosφ−(˙zcosφ−˙ysinφ)˙x (RE+z)2cos2φ(1.51) The acceleration of Prelative to the non-rotating earth is found by taking the time derivative of vrel. From Equation 1.49 we thereby obtain arel=¨xˆi+¨yˆj+¨zˆk+˙xdˆi dt+˙ydˆj dt+˙zdˆk dt =[˙z˙/Lambda1cosφ+(RE+z)¨/Lambda1cosφ−(RE+z)˙φ˙/Lambda1sinφ]ˆi +[˙z˙φ+(RE+z)¨φ]ˆj+¨zˆk+(RE+z)˙/Lambda1cosφ(ω׈i) +(RE+z)˙φ(ω׈j)+˙z(ω׈k) 26 Chapter 1 Dynamics of point masses Substituting Equations 1.46 through 1.48 together with 1.50 and 1.51 into this expression yields, upon simplification, arel=/bracketleftbigg ¨x+˙x(˙z−˙ytanφ) RE+z/bracketrightbigg ˆi+/parenleftbigg ¨y+˙y˙z+˙x2tanφ RE+z/parenrightbigg ˆj+/parenleftbigg ¨z−˙x2+˙y2 RE+z/parenrightbigg ˆk(1.52) Observe that the curvature of the earth’s surface is neglected by letting RE+zbecome infinitely large, in which case arel)neglecting earth/primes curvature =¨xˆi+¨yˆj+¨zˆk That is, for a ‘flat earth’ , the components of the relative acceleration vector are just the derivatives of the components of the relative velocity vector. For the absolute velocity we have, according to Equation 1.38, v=vC+/Omega1×rrel+vrel (1.53) From Figure 1.11 it can be seen that ˆK=cosφˆj+sinφˆk, which means the angular velocity of the earth is /Omega1=/Omega1ˆK=/Omega1cosφˆj+/Omega1sinφˆk (1.54) Substituting this, together with Equations 1.43 and 1.49a and the fact that vC=0, into Equation 1.53 yields v=[˙x+/Omega1(RE+z)c o sφ]ˆi+˙yˆj+˙zˆk (1.55) From Equation 1.42 the absolute acceleration of Pis a=aC+˙/Omega1×rrel+/Omega1×(/Omega1×rrel)+2/Omega1×vrel+arel Since aC=˙/Omega1=0, we find, upon substituting Equations 1.43, 1.49a, 1.52 and 1.54, that a=/bracketleftbigg ¨x+˙x(˙z−˙ytanφ) RE+z+2/Omega1(˙zcosφ−˙ysinφ)/bracketrightbigg ˆi +/braceleftbigg ¨y+˙y˙z+˙x2tanφ RE+z+/Omega1sinφ[/Omega1(RE+z)c o sφ+2˙x]/bracerightbigg ˆj +/braceleftbigg ¨z−˙x2+˙y2 RE+z−/Omega1cosφ[/Omega1(RE+z)c o sφ+2˙x]/bracerightbigg ˆk (1.56) Some special cases of Equations 1.55 and 1.56 follow. 1.6 Relative motion 27 Straight and level, unaccelerated flight: ˙z=¨z=¨x=¨y=0 v=[˙x+/Omega1(R E+z)c o sφ]ˆi+˙yˆj (1.57a) a=−/bracketleftbigg˙x˙ytanφ RE+z+2/Omega1˙ysinφ/bracketrightbigg ˆi +/braceleftbigg˙x2tanφ RE+z+/Omega1sinφ[/Omega1(R E+z)c o sφ+2˙x]/bracerightbigg ˆj −/braceleftbigg˙x2+˙y2 RE+z+/Omega1cosφ[/Omega1(R E+z)c o sφ+2˙x]/bracerightbigg ˆk (1.57b) Flight due north (y) at constant speed and altitude: ˙z=¨z=˙x=¨x=¨y=0 v=/Omega1(R E+z)c o sφˆi+˙yˆj (1.58a) a=− 2/Omega1˙ysinφˆi+/Omega12(RE+z) sinφcosφˆj −/bracketleftbigg˙y2 RE+z+/Omega12(RE+z)c o s2φ/bracketrightbigg ˆk (1.58b) Flight due east (x) at constant speed and altitude: ˙z=¨z=¨x=˙y=¨y=0 v=[˙x+/Omega1(R E+z)c o sφ]ˆi (1.59a) a=/braceleftbigg˙x2tanφ RE+z+/Omega1sinφ[/Omega1(R E+z)c o sφ+2˙x]/bracerightbigg ˆj −/braceleftbigg˙x2 RE+z+/Omega1cosφ[/Omega1(R E+z)c o sφ+2˙x]/bracerightbigg ˆk (1.59b) Flight straight up (z): ˙x=¨x=˙y=¨y=0 v=/Omega1(R E+z)c o sφˆi+˙zˆk (1.60a) a=2/Omega1(˙zcosφ)ˆi+/Omega12(RE+z) sinφcosφˆj +/bracketleftbig ¨z−/Omega12(RE+z)c o s2φ/bracketrightbigˆk (1.60b) Stationary: ˙x=¨x=˙y=¨y=˙z=¨z=0 v=/Omega1(R E+z)c o sφˆi (1.61a) a=/Omega12(RE+z) sinφcosφˆj−/Omega12(RE+z)c o s2φˆk (1.61b) Example 1.10An airplane of mass 70 000 kg is traveling due north at latitude 30◦north, at an altitude of 10 km (32 800 ft) with a speed of 300 m/s (671 mph). Calculate (a) the components of the absolute velocity and acceleration along the axes of the topocentric-horizonreference frame, and (b) the net force on the airplane. 28 Chapter 1 Dynamics of point masses (Example 1.10 continued)(a) First, using the sidereal rotation period of the earth in Table A.1, we note that theearth’s angular velocity is /Omega1=2πrad sidereal day=2πrad 23.93 hr=2πrad 86 160 s=7.292×10−5rad/s( a ) From Equation 1.58a, the absolute velocity is v=/Omega1(RE+z)c o sφˆi+˙yˆj=/bracketleftbig (7.292×10−5)·(6378 +10)·103cos 30◦/bracketrightbigˆi+300ˆj or v=403.4ˆi+300ˆj(m/s) The 403.4 m/s (901 mph) component of velocity to the east ( xdirection) is due entirely to the earth’s rotation. From Equation 1.58b 2, the absolute acceleration is a=− 2/Omega1˙ysinφˆi+/Omega12(RE+z) sinφcosφˆj−/bracketleftbigg˙y2 RE+z+/Omega12(RE+z)c o s2φ/bracketrightbigg ˆk =− 2(7.292×10−5)·300·sin 30◦ˆi +(7.292×10−5)2·(6378 +10)·103·sin 30◦·cos 30◦ˆj −/bracketleftbigg3002 (6378 +10)·103+(7.292×10−5)2·(6378 +10)·103·cos230◦/bracketrightbigg ˆk or a=− 0.02187ˆi+0.01471ˆj−0.03956 ˆk(m/s2) (a) The westward acceleration of 0.02187 m/s2is the Coriolis acceleration. (b) Since the acceleration in part (a) is the absolute acceleration, we can use it in Newton’s law to calculate the net force on the airplane, Fnet=ma=70 000( −0.02187ˆi+0.01471ˆj−0.03956 ˆk) =− 1531ˆi+1029ˆj−2769ˆk(N) Figure 1.12 shows the components of this relatively small force. The forward and downward forces are in the directions of the airplane’s centripetal acceleration, caused by the earth’s rotation and, in the case of the downward force, by the earth’s curvature as well. The westward force is in the direction of the Coriolis acceleration, which is due to the combined effects of the earth’s rotation and the motion of the airplane. These net external forces must exist if the airplane is tofly the prescribed path. In the vertical direction, the net force is that of the upward lift Lof the wings plus the downward weight Wof the aircraft, so that Fnet)z=L−W=− 2769 ⇒ L=W−2769 (N) Problems 29 2769 N (622 lb)1531 N (344 lb) 1029 N (231 lb)x EastzUp y North Figure 1.12 Components of the net force on the airplane . Thus, the effect of the earth’s rotation and curvature is to apparently produce an outward centrifugal force , reducing the weight of the airplane a bit, in this case by about 0.4 percent. The fictitious centrifugal force also increases the apparent drag in the flight direction by 1029 N. That is, in the flight direction Fnet)y=T−D=− 2769 N where Tis the thrust and Dis the drag. Hence T=D+1029 (N) The 1531 N force to the left, produced by crabbing the airplane very slightly in that direction, is required to balance the fictitious Coriolis force which wouldotherwise cause the airplane to deviate to the right of its flight path. Problems 1.1 Given the three vectors A=Axˆi+Ayˆj+AzˆkB=Bxˆi+Byˆj+BzˆkC =Cxˆi+Cyˆj+Czˆk show, analytically, that (a) A·A=A2 (b) A·(B×C)=(A×B)·C(interchangeability of the ‘dot’ and ‘cross’) (c) A×(B×C)=B(A·C)−C(A·B) (the bac – cab rule) 30 Chapter 1 Dynamics of point masses (Simply compute the expressions on each side of the =signs and demonstrate conclusively that they are the same. Do notsubstitute numbers to‘prove’your point. Use the fact that the cartesian coordinate unit vectors ˆi,ˆjandˆkform a right-handed orthogonal triad, so that ˆi·ˆj=ˆi·ˆk=ˆj·ˆk=0ˆi·ˆi=ˆj·ˆj=ˆk·ˆk=1 ˆi׈j=ˆk ˆj׈k=ˆiˆk׈i=ˆj (ˆi׈k=−ˆjˆj׈i=−ˆkˆk׈j=−ˆi) Also, ˆi׈i=ˆj׈j=ˆk׈k=0 1.2 Use just the vector identities in parts (a) and (b) of Exercise 1.1 to show that (A×B)·(C×D)=(A·C)(B·D)−(A·D)(B·C) 1.3 The absolute position, velocity and acceleration of Oare rO=300ˆI+200ˆJ+100ˆK(m) vO=− 10ˆI+30ˆJ−50ˆK(m/s) aO=25ˆI+40ˆJ−15ˆK(m/s2) The angular velocity and acceleration of the moving frame are /Omega1=0.6ˆI−0.4ˆJ+1.0ˆK(rad/s) ˙/Omega1=− 0.4ˆI+0.3ˆJ−1.0ˆK(rad/s2) The unit vectors of the moving frame are ˆi=0.57735ˆI+0.57735ˆJ+0.57735 ˆK ˆj=− 0.74296ˆI+0.66475ˆJ+0.078206 ˆK ˆk=− 0.33864ˆI−0.47410ˆJ+0.81274 ˆK The absolute position of Pis r=150ˆI−200ˆJ+300ˆK(m) The velocity and acceleration of Prelative to the moving frame are vrel=− 20ˆi+25ˆj+70ˆk(m/s) arel=7.5ˆi−8.5ˆj+6.0ˆk(m/s2) Calculate the absolute velocity vPand acceleration aPofP. {Ans.: vP=478.7ˆuv(m/s),ˆuv=0.5352ˆI−0.5601ˆJ−0.6324ˆK; aP=616.3ˆua(m/s2),ˆua=0.1655ˆI+0.9759ˆJ+0.1424ˆK} 1.4 F is a force vector of fixed magnitude embedded on a rigid body in plane motion (in the xyplane). At a given instant, ω=3ˆkrad/s, ˙ω=− 2ˆkrad/s2,¨ω=0and F=10ˆiN. At that instant, calculate...F. {Ans.:...F=180ˆi−270ˆjN/s3} Problems 31 XYZ xyz r Inertial frameMoving frameOP r0rrel Figure P .1.3 υ θrx yh XY A Figure P .1.5 1.5 An airplane in level flight at an altitude hand a uniform speed vpasses directly over a radar tracking station A. Calculate the angular velocity ˙θand angular acceleration of the radar antenna ¨θas well as the rate ˙rat which the airplane is moving away from the antenna. Use the equations of this chapter (rather than polar coordinates, which you can use to check your work). Attach the inertial frame of reference to the ground and assumea non-rotating earth. Attach the moving frame to the antenna, with the xaxis pointing always from the antenna towards the airplane.{Ans.: (a) ˙θ=vcos 2θ/h; (b) ¨θ=− 2v2cos3θsinθ/h2; (c) vrel=vsinθ} 32 Chapter 1 Dynamics of point masses 1.6 At 30◦north latitude, a 1000 kg (2205 lb) car travels due north at a constant speed of 100 km/hr (62 mph) on a level road at sea level. Taking into account the earth’s rotation,calculate the lateral (sideways) force of the road on the car, and the normal force of theroad on the car.{Ans.: F lateral=2.026 N, to the left (west); N=9784 N} 1.7 At 29◦north latitude, what is the deviation dfrom the vertical of a plumb bob at the end of a 30 m string, due to the earth’s rotation?{Ans.: 44.1 mm to the south} θ g yz NorthdL /H11005 30 m Figure P .1.7 2Chapter The two-body problem Chapter outline 2.1 Introduction 33 2.2 Equations of motion in an inertial frame 34 2.3 Equations of relative motion 37 2.4 Angular momentum and the orbit formulas 42 2.5 The energy law 50 2.6 Circular orbits (e =0)5 1 2.7 Elliptical Orbits (0 <e<1)5 5 2.8 Parabolic trajectories (e =1)6 5 2.9 Hyperbolic trajectories (e >1)6 9 2.10 Perifocal frame 76 2.11 The Lagrange coefficients 78 2.12 Restricted three-body problem 89 2.12.1 Lagrange points 92 2.12.2 Jacobi constant 96 Problems 101 2.1 Introduction This chapter presents the vector-based approach to the classical problem of deter- mining the motion of two bodies due solely to their own mutual gravitational attraction. We show that the path of one of the masses relative to the other is aconic section (circle, ellipse, parabola or hyperbola) whose shape is determined by the eccentricity. Several fundamental properties of the different types of orbits are 33 34 Chapter 2 The two-body problem developed with the aid of the laws of conservation of angular momentum and energy. These properties include the period of el liptical orbits, the escape velocity associated with parabolic paths and the characteristic energy of hyperbolic trajectories. Follow-ing the presentation of the four types of orbits, the perifocal frame is introduced. Thisframe of reference is used to describe orbits in three dimensions, which is the subjectof Chapter 4. In this chapter the perifocal frame provides the backdrop for developing the Lagrange fandgcoefficients. By means of the Lagrange fandgcoefficients, the posi- tion and velocity on a trajectory can be found in terms of the position and velocity atan initial time. These functions are needed in the orbit determination algorithms ofLambert and Gauss presented in Chapter 5. The chapter concludes with a discussion of the restricted three-body problem in order to provide a basis for understanding of the concepts of Lagrange points as wellas the Jacobi constant. This material is optional. In studying this chapter it would be well from time to time to review the road map provided in Appendix B. 2.2 Equations of motion in an inertial frame Figure 2.1 shows two point masses acted upon only by the mutual force of gravitybetween them. The positions of their centers of mass are shown relative to an inertialframe of reference XYZ . The origin Oof the frame may move with constant velocity (relative to the fixed stars), but the axes do not rotate. Each of the two bodies is actedupon by the gravitational attraction of the other. F 12is the force exerted on m1bym2, and F21is the force exerted on m2bym1. The position vector RGof the center of mass Gof the system in Figure 2.1(a) is, defined by the formula RG=m1R1+m2R2 m1+m2(2.1) XYZ Or Inertial frame of reference (fixed with respect to the fixed stars)GR1 R2RG m2m1 r r XYZ OR1 R2m2m1 F12 F21 (a) (b)ur /H11005ˆ Figure 2.1 (a) Two masses located in an inertial frame. (b) Free-body diagrams. 2.2 Equations of motion in an inertial frame 35 Therefore, the absolute velocity and the absolute acceleration of Gare vG=˙RG=m1˙R1+m2˙R2 m1+m2(2.2) aG=¨RG=m1¨R1+m2¨R2 m1+m2(2.3) The adjective ‘absolute’ means that the quantities are measured relative to an inertial frame of reference. Letrbe the position vector of m2relative to m1. Then r=R2−R1 (2.4) Furthermore, let ˆurbe the unit vector pointing from m1towards m2, so that ˆur=r r(2.5) where r=/bardblr/bardbl, the magnitude of r. The body m1i sa c t e du p o no n l yb yt h ef o r c eo f gravitational attraction towards m2. The force of gravitational attraction, Fg, which acts along the line joining the centers of mass of m1andm2, is given by Equation 1.3. The force exerted on m2bym1is F21=Gm 1m2 r2(−ˆur)=−Gm 1m2 r2ˆur (2.6) where −ˆuraccounts for the fact that the force vector F21is directed from m2towards m1. (Do not confuse the symbol G, used in this context to represent the universal gravitational constant, with its use elsewhere in the book to denote the center ofmass.) Newton’s second law of motion as applied to body m 2isF21=m2¨R2,w h e r e ¨R2is the absolute acceleration of m2.T h u s −Gm 1m2 r2ˆur=m2¨R2 (2.7) By Newton’s third law (the action–reaction principle), F12=− F21, so that for m1we have Gm 1m2 r2ˆur=m1¨R1 (2.8) Equations 2.7 and 2.8 are the equations of motion of the two bodies in inertial space. By adding each side of these equations together, we find m1¨R1+m2¨R2=0. According to Equation 2.3, that means the acceleration of the center of mass Gof the system of two bodies m1and m2is zero. Gmoves with a constant velocity vGin a straight line, so that its position vector relative to XYZ given by RG=RG0+vGt (2.9) where RG0is the position of Gat time t=0. The center of mass of a two-body system may therefore serve as the origin of an inertial frame. 36 Chapter 2 The two-body problem Example 2.1Use the equations of motion to show why orbiting astronauts experience weightlessness. We sense weight by feeling the contact forces that develop wherever our body is supported. Consider an astronaut of mass mAstrapped into the space shuttle of mass mS, in orbit about the earth. The distance between the center of the earth and the spacecraft is r, and the mass of the earth is ME. Since the only external force on the space shuttle is that of gravity, FS)g, the equation of motion of the shuttle is FS)g=mSaS (a) According to Equation 2.6, FS)g=−GM EmS r2ˆur (b) where ˆuris the unit vector pointing outward from the earth to the orbiting space shuttle. Thus, (a) and (b) imply aS=−GM E r2ˆur (c) The equation of motion of the astronaut is FA)g+CA=mAaA (d) where FA)gis the force of gravity on (i.e., the weight of) the astronaut, CAis the net contact force on the astronaut from restraints (e.g., seat, seat belt), and aAis the astronaut’s acceleration. According to Equation 2.6, FA)g=−GM EmA r2ˆur (e) Since the astronaut is moving with the shuttle we have, noting (c), aA=aS=−GM E r2ˆur (f) Substituting (e) and (f) into (d) yields −GM EmA r2ˆur+CA=mA/parenleftbigg −GM E r2ˆur/parenrightbigg from which it is clear that CA=0. The net contact force on the astronaut is zero. With no reaction to the force of gravity exerted on the body, there is no sensation of weight. The potential energy Vof the gravitational force in Equation 2.6 is given by V=−Gm 1m2 r(2.10) A force can be obtained from its potential energy function by means of the gradient operator, F=− ∇ V (2.11) 2.3 Equations of relative motion 37 where, in cartesian coordinates, ∇=∂ ∂xˆi+∂ ∂yˆj+∂ ∂zˆk (2.12) In Appendix E it is shown that the gravitational potential, and hence the gravitational force, outside of a sphere with a spherically symmetric mass distribution Mis the same as that of a point mass Mlocated at the center of the sphere. Therefore, the two- body problem applies not just to point masses but also to spherical bodies (as long,of course, as they do not come into contact!). 2.3 Equations of relative motion Let us now multiply Equation 2.7 by m1and Equation 2.8 by m2to obtain −Gm2 1m2 r2ˆur=m1m2¨R2 Gm 1m2 2 r2ˆur=m1m2¨R1 Subtracting the second of these two equations from the first yields m1m2/parenleftbig¨R2−¨R1/parenrightbig =−Gm 1m2 r2/parenleftbig m1+m2/parenrightbig ˆur Canceling the common factor m1m2and using Equation 2.4 yields ¨r=−G(m1+m2) r2ˆur (2.13) Let the gravitational µparameter be defined as µ=G(m1+m2) (2.14) The units of µare km3s−2. Using Equation 2.14 together with Equation 2.5, we can write Equation 2.13 as ¨r=−µ r3r (2.15) This is the second order differential equation that governs the motion of m2relative to m1. It has two vector constants of integration, each having three scalar components. Therefore, Equation 2.15 has six constants of integration. Note that interchangingthe roles of m 1and m2in all of the above amounts to simply multiplying Equation 2.15 through by −1, which, of course, changes nothing. Thus, the motion of m2as seen from m1is precisely the same as the motion of m1as seen from m2. The relative position vector rin Equation 2.15 was defined in the inertial frame (Equation 2.4). It is convenient, however, to measure the components of rin a frame of reference attached to and moving with m1. In a co-moving reference frame, such as the xyzsystem illustrated in Figure 2.2, rhas the expression r=xˆi+yˆj+zˆk 38 Chapter 2 The two-body problem XYZ OR1m2m1 xyz r R2ˆ i ˆ j ˆ k Figure 2.2 Moving reference frame xyzattached to the center of mass of m1. The relative velocity ˙rreland acceleration ¨rrelin the co-moving frame are found by simply taking the derivatives of the coefficients of the unit vectors, which themselvesa r efi x e di nt h e xyzsystem. Thus ˙r rel=˙xˆi+˙yˆj+˙zˆk¨rrel=¨xˆi+¨yˆj+¨zˆk From Equation 1.40 we know that the relationship between absolute acceleration ¨r and relative acceleration ¨rrelis ¨r=¨rrel+˙/Omega1×r+/Omega1×(/Omega1×r)+2/Omega1×˙rrel where /Omega1and ˙/Omega1are the angular velocity and angular acceleration of the moving frame of reference. Thus ¨r=¨rrelonly if /Omega1=˙/Omega1=0. That is to say, the relative acceleration may be used on the left of Equation 2.15 as long as the co-moving frame in which itis measured is not rotating. As an example of two-body motion, consider two identical, isolated bodies m 1 and m2positioned in an inertial frame of reference, as shown in Figure 2.3. At time t=0,m1is at rest at the origin of the frame, whereas m2, to the right of m1, has a velocity vodirected upward to the right, making a 45◦angle with the Xaxis. The subsequent motion of the two bodies, which is due solely to their mutual gravitationalattraction, is determined relative to the inertial frame by means of Equations 2.7 and2.8. Figure 2.3 is a computer-generated sol ution of those equations. The motion is rather complex. Nevertheless, at any time t,m 1andm2lie in the XYplane, equidistant and in opposite directions from their center of mass G, whose straight-line path is also shown in Figure 2.3. The very same motion appears rather less complex whenviewed from m 1, as the computer simulation reveals in Figure 2.4(a). Figure 2.4(a) 2.3 Equations of relative motion 39 45°vo G m 1 (initially at rest)m2G m1m2 Path of GPath of m 1 Path of m 2 XY Inertial frame Figure 2.3 The motion of two identical bodies acted on only by their mutual gravitational attraction, as viewed from the inertial frame of reference. represents the solution to Equation 2.15, and we see that, relative to m1,m2follows what appears to be an elliptical path. (So does the center of mass.) Figure 2.4(b) reveals that both m1and m2follow elliptical paths around the center of mass. Since the center of mass has zero acceleration, we can use it as an inertial reference frame. Let r1and r2be the position vectors of m1and m2, respectively, relative to the center of mass Gin Figure 2.1. The equation of motion of m2relative to the center of mass is −Gm1m2 r2ˆur=m2¨r2 (2.16) where, as before, ris the position vector of m2relative to m1.I nt e r m so f r1and r2, r=r2−r1 Since the position vector of the center of mass relative to itself is zero, it follows from Equation 2.1 that m1r1+m2r2=0 Therefore, r1=−m2 m1r2 so that r=m1+m2 m1r2 Substituting this back into Equation 2.16 and using the fact that ˆur=r2/r2,w eg e t −Gm3 1m2 (m1+m2)2r3 2r2=m2¨r2 40 Chapter 2 The two-body problem Gm2 m1XY Non-rotating frame attached to m1 (a) m1XY Gm2 Non-rotating frame attached to G (b) Figure 2.4 The motion in Figure 2.3, (a) as viewed relative to m1(orm2); (b) as viewed from the center of mass. which, upon simplification, becomes −/parenleftbiggm1 m1+m2/parenrightbiggµ r3 2r2=¨r2 (2.17) where µis given by Equation 2.14. If we let µ/prime=/parenleftbiggm1 m1+m2/parenrightbigg3 µ 2.3 Equations of relative motion 41 vo m1m2m3m1m1m1 m2m2m3 m3m2 G XY Inertial frame Figure 2.5 The motion of three identical masses as seen from the inertial frame in which m1and m3are initially at rest, while m2has an initial velocity v0directed upwards and to the right, as shown. then Equation 2.17 reduces to ¨r2=−µ/prime r3 2r2 which is identical in form to Equation 2.15. In a similar fashion, the equation of motion of m1relative to the center of mass is found to be ¨r1=−µ/prime/prime r3 1r1 in which µ/prime/prime=/parenleftbiggm2 m1+m2/parenrightbigg3 µ Since the equations of motion of either particle relative to the center of mass have the same form as the equations of motion relative to either one of the bodies, m1orm2, it follows that the relative motion as viewed from these different perspectives must besimilar, as illustrated in Figure 2.4. One may wonder what the motion looks like if there are more than two bodies moving under the influence only of their mutual gravitational attraction. The n-body problem with n>2 has no closed form solution, which is complex and chaotic in nature. We can use a computer simulation (see Appendix C.1) to get an idea of themotion for some special cases. Figure 2.5 shows the motion of three equal masses, 42 Chapter 2 The two-body problem m1 m1m1 m2m2 m3m3m3 XY Non-rotating frame attached to G Figure 2.6 The same motion as Figure 2.5, as viewed from the inertial frame attached to the center of mass G. equally spaced initially along the Xaxis of an inertial frame. The center mass has an initial velocity, while the other two are at rest. As time progresses, we see no periodicbehavior as was evident in the two-body motion in Figure 2.3. The chaos is moreobvious if the motion is viewed from the center of mass of the three-body system, asshown in Figure 2.6. The computer simulation from which these figures were takenshows that the masses eventually collide. 2.4 Angular momentum and the orbit formulas The angular momentum of body m2relative to m1is the moment of m2’s relative linear momentum m2˙r(cf. Equation 1.17), H2/1=r×m2˙r where ˙r=vis the velocity of m2relative to m1. Let us divide this equation through bym2and let h=H2/1/m2, so that h=r×˙r (2.18) his the relative angular momentum of m2per unit mass, that is, the specific relative angular momentum. The units of hare km2s−1. Taking the time derivative of hyields dh dt=˙r×˙r+rרr But˙r×˙r=0. Furthermore, ¨r=− (µ/r3)r, according to Equation 2.15, so that rרr=r×/parenleftBig −µ r3r/parenrightBig =−µ r3(r×r)=0 2.4 Angular momentum and the orbit formulas 43 m1 m2rr ·r·r ˆh/H11005h hˆh/H11005h h Figure 2.7 The path of m2around m1lies in a plane whose normal is defined by h. m1 m2rˆu⊥ ˆur υrυ⊥ Pathr· Figure 2.8 Components of the velocity of m2, viewed above the plane of the orbit. Therefore, dh dt=0( o r r×˙r=constant) (2.19) At any given time, the position vector rand the velocity vector ˙rlie in the same plane, as illustrated in Figure 2.7. Their cross product r×˙ris perpendicular to that plane. Since r×˙r=h, the unit vector normal to the plane is ˆh=h h(2.20) But, according to Equation 2.19, this unit vector is constant. Thus, the path of m2 around m1lies in a single plane. Since the orbit of m2around m1forms a plane, it is convenient to orient oneself above that plane and look down upon the path, as shown in Figure 2.8. Let us resolvethe relative velocity vector ˙rinto components v r=vrˆurand v⊥=v⊥ˆu⊥along the outward radial from m1and perpendicular to it, respectively, where ˆurandˆu⊥are the radial and perpendicular (azimuthal) unit vectors. Then we can write Equation 2.18 44 Chapter 2 The two-body problem dAr(t /H11001 dt) r(t)m1m2f fvdt Path r sin fυdt Figure 2.9 Differential area dAswept out by the relative position vector rduring time interval dt. as h=rˆur×(vrˆur+v⊥ˆu⊥)=rv⊥ˆh That is, h=rv⊥ (2.21) Clearly, the angular momentum depends only on the azimuth component of the relative velocity. During the differential time interval dtthe position vector rsweeps out an area dA, as shown in Figure 2.9. From the figure it is clear that the triangular area dAis given by dA=1 2×base×altitude =1 2×vdt×rsinφ=1 2r(vsinφ)dt=1 2rv⊥dt Therefore, using Equation 2.21 we have dA dt=h 2(2.22) dA/dtis called the areal velocity, and according to Equation 2.22 it is constant. Named after the German astronomer Johannes Kepler (1571–1630), this result is known asKepler’s second law: equal areas are swept out in equal times. Before proceeding with an effort to integrate Equation 2.15, recall the vector identity known as the bac−cabrule: A×(B×C)=B(A·C)−C(A·B) (2.23) Recall as well that r·r=r 2(2.24) 2.4 Angular momentum and the orbit formulas 45 so that d dt(r·r)=2rdr dt But d dt(r·r)=r·dr dt+dr dt·r=2r·dr dt Thus, we obtain the important identity r·˙r=r˙r (2.25a) Since˙r=vand r=/bardblr/bardbl, this can be written alternatively as r·v=/bardbl r/bardbld/bardblr/bardbl dt(2.25b) Now let us take the cross product of both sides of Equation 2.15 [ ¨r=− (µ/r3)r] with the specific angular momentum h: ¨r×h=−µ r3r×h (2.26) Sinced dt(˙r×h)=¨r×h+˙r×˙h, the left-hand side can be written ¨r×h=d dt(˙r×h)−˙r×˙h But according to Equation 2.19, the angular momentum is constant ( ˙h=0), so this reduces to ¨r×h=d dt(˙r×h) (2.27) The right-hand side of Equation 2.26 can be transformed by the following sequence of substitutions: 1 r3r×h=1 r3[r×(r×˙r)] (Equation 2.18 [h =r×˙r]) =1 r3[r(r·˙r)−˙r(r·r)] (Equation 2.23 [bac −cabrule]) =1 r3[r(r˙r)−˙rr2] (Equations 2.24 and 2.25) =r˙r−˙rr r2 But d dt/parenleftBigr r/parenrightBig =r˙r−r˙r r2=−r˙r−r˙r r2 Therefore 1 r3r×h=−d dt/parenleftBigr r/parenrightBig (2.28) 46 Chapter 2 The two-body problem Substituting Equations 2.27 and 2.28 into Equation 2.26, we get d dt(˙r×h)=d dt/parenleftBig µr r/parenrightBig or d dt/parenleftBig ˙r×h−µr r/parenrightBig =0 That is, ˙r×h−µr r=C (2.29) where the vector Cis an arbitrary constant of integration having the dimensions of µ. Equation 2.29 is the first integral of the equation of motion, ¨r=− (µ/r3)r. Taking the dot product of both sides of Equation 2.29 with the vector hyields (˙r×h)·h−µr·h r=C·h Since˙r×his perpendicular to both ˙rand h, it follows that ( ˙r×h)·h=0. Likewise, since h=r×˙ris perpendicular to both rand˙r, it is true that r·h=0. Therefore, we have C·h=0, i.e., Cis perpendicular to h, which is normal to the orbital plane. That of course means Cmust lie in the orbital plane. Let us rearrange Equation 2.29 and write it as r r+e=˙r×h µ(2.30) where e=C/µ. The dimensionless vector eis called the eccentricity vector. The line defined by the vector eis commonly called the apse line. In order to obtain a scalar equation, let us take the dot product of both sides of Equation 2.30 with r: r·r r+r·e=r·(˙r×h) µ(2.31) In order to simplify the right-hand side, we can employ the useful vector identity, known as the interchange of the dot and the cross, A·(B×C)=(A×B)·C (2.32) to obtain r·(˙r×h)=(r×˙r)·h=h·h=h2(2.33) Substituting this expression into the right-hand side of Equation 2.31, and substituting r·r=r2on the left yields r+r·e=h2 µ(2.34) Observe that by following the steps leading from Equation 2.30 to 2.34 we have lost track of the variable time. This occurred at Equation 2.33, because his constant. Finally, from the definition of the dot product we have r·e=recosθ 2.4 Angular momentum and the orbit formulas 47 r eθ m1m2 Figure 2.10 The true anomaly θis the angle between the eccentricity vector eand the position vector r. in which eis the eccentricity (the magnitude of the eccentricity vector e) and θis the true anomaly. θis the angle between the fixed vector eand the variable position vector r, as illustrated in Figure 2.10. (Other symbols used to represent true anomaly includeν,f,vandφ.) In terms of the eccentricity and the true anomaly, we may therefore write Equation 2.34 as r+recosθ=h 2 µ or r=h2 µ1 1+ecosθ(2.35) This is the orbit equation, and it defines the path of the body m2around m1, relative tom1. Remember that µ,h, and eare constants. Observe as well that there is no significance to negative values of eccentricity; i.e., e≥0. Since the orbit equation describes conic sections, including ellipses, it is a mathematical statement of Kepler’sfirst law, namely, that the planets follow elliptical paths around the sun. Two-bodyorbits are often referred to as Keplerian orbits. In Section 2.3 it was pointed out that integration of the equation of relative motion, Equation 2.15, leads to six constants of integration. In this section it wouldseem that we have arrived at those constants, namely the three components of the angular momentum hand the three components of the eccentricity vector e.H o w e v e r , we showed that his perpendicular to e. This places a condition, namely h·e=0, on the components of hand e, so that we really have just five independent constants of integration. The sixth constant of the motion will arise when we work time back intothe picture in the next chapter. The angular velocity of the position vector ris˙θ, the rate of change of the true anomaly. The component of velocity normal to the position vector is found in termsof the angular velocity by the formula v ⊥=r˙θ (2.36) Substituting this into Equation 2.21 (h =rv⊥) yields the specific angular momentum in terms of the angular velocity, h=r2˙θ (2.37) 48 Chapter 2 The two-body problem It is convenient to have formulas for computing the radial and azimuth components of velocity shown in Figure 2.11. From h=rv⊥we of course obtain v⊥=h r Substituting rfrom Equation 2.35 readily yields v⊥=µ h(1+ecosθ) (2.38) Sincevr=˙r, we take the derivative of Equation 2.35 to get ˙r=dr dt=h2 µ/bracketleftbigg −e(−˙θsinθ) (1+ecosθ)2/bracketrightbigg =h2 µesinθ (1+ecosθ)2h r2 where we made use of the fact that ˙θ=h/r2, from Equation 2.37. Substituting Equation 2.35 once again and simplifying finally yields vr=µ hesinθ (2.39) Apse lineeγ Periapsis rpθr m1vr m2 r · v⊥ Figure 2.11 Position and velocity of m2in polar coordinates centered at m1, with the eccentricity vector being the reference for true anomaly (polar angle) θ.γis the flight path angle. 2.4 Angular momentum and the orbit formulas 49 We see from Equation 2.35 that m2comes closest to m1(ris smallest) when θ=0 (unless e=0, in which case the distance between m1and m2is constant). The point of closest approach lies on the apse line and is called periapsis. The distance rpto periapsis, as shown in Figure 2.11, is obtained by setting the true anomaly equal tozero, r p=h2 µ1 1+e(2.40) Clearly, vr=0 at periapsis. The flight path angle γis also illustrated in Figure 2.11. It is the angle that the velocity vector v=˙rmakes with the normal to the position vector. The normal to the position vector points in the direction of v⊥, and it is called the local horizon. From Figure 2.11 it is clear that tanγ=vr v⊥(2.41) Substituting Equations 2.38 and 2.39 leads at once to the expression tanγ=esinθ 1+ecosθ(2.42) Since cos( −θ)=cosθ, the trajectory described by the orbit equation is symmetric about the apse line, as illustrated in Figure 2.12, which also shows a chord, the straightline connecting any two points on the orbit. The latus rectum is the chord throughthe center of attraction perpendicular to the apse line. By symmetry, the center ofattraction divides the latus rectum into two equal parts, each of length p, known historically as the semi-latus rectum. In modern parlance, pis called the parameter of the orbit. From Equation 2.35 it is apparent that p=h 2 µ(2.43) Chord 90°p Latus rectumm1Periapsis Apse lineA A'P1P2 Figure 2.12 Illustration of latus rectum, semi-latus rectum p, and the chord between any two points on an orbit. 50 Chapter 2 The two-body problem Since the path of m2around m1lies in a plane, for the time being we will for simplicity continue to view the trajectory from above the plane. Unless there is reason to dootherwise, we will assume that the eccentricity vector points to the right and that m 2 moves counterclockwise around m1, which means that the true anomaly is measured positive counterclockwise, consistent with the usual polar coordinate sign convention. 2.5 The energy law By taking the cross product of Equation 2.15, ¨r=− (µ/r3)r( N e w t o n ’ ss e c o n dl a wo f motion), with the specific relative angular momentum per unit mass h,w ew e r el e d to the vector Equation 2.29, and from that we obtained the orbit formula, Equation 2.35. Now let us see what results from taking the dotproduct of Equation 2.15 with the relative linear momentum per unit mass. The relative linear momentum per unitmass is just the relative velocity, m 2˙r m2=˙r Thus, carrying out the dot product in Equation 2.15 yields ¨r·˙r=−µr·˙r r3(2.44) For the left-hand side we observe that ¨r·˙r=1 2d dt(˙r·˙r)=1 2d dt(v·v)=1 2d dt(v2)=d dt/parenleftbiggv2 2/parenrightbigg (2.45) For the right-hand side of Equation 2.44 we have, recalling that r·r=r2and d(1/r)/dt=(−1/r2)(dr/dt), µr·˙r r3=µr˙r r3=µ˙r r2=−d dt/parenleftBigµ r/parenrightBig (2.46) Substituting Equations 2.45 and 2.46 into Equation 2.44 yields d dt/parenleftbiggv2 2−µ r/parenrightbigg =0 or v2 2−µ r=ε(constant) (2.47) where εis a constant. v2/2 is the relative kinetic energy per unit mass. ( −µ/r)i st h e potential energy per unit mass of the body m2in the gravitational field of m1. The total mechanical energy per unit mass εis the sum of the kinetic and potential energies per unit mass. Equation 2.47 is a statement of conservation of energy, namely, that thespecific mechanical energy is the same at all points of the trajectory. Equation 2.47 is 2.6 Circular orbits ( e=0) 51 also known as the vis-viva(‘living force’) equation. Since εis constant, let us evaluate it at periapsis ( θ=0), ε=εp=v2 p 2−µ rp(2.48) where rpandvpare the position and speed at periapsis. Since vr=0 at periapsis, we havevp=v⊥=h/rp. Thus, ε=1 2h2 r2 p−µ rp(2.49) Substituting Equation 2.40 into 2.49 yield s a formula for the orbital specific energy in terms of the orbital constants hand e, ε=−1 2µ2 h2(1−e2) (2.50) Clearly, the orbital energy is not an independent orbital parameter. Note that the mechanical energy Eof a satellite of mass m1is obtained from the specific energy εby the formula E=m1ε (2.51) 2.6 Circular orbits (e =0) Setting e=0 in the orbital equation r=(h2/µ)/(1 +ecosθ) yields r=h2 µ(2.52) That is, r=constant, which means the orbit of m2around m1is a circle. Since ˙r=0, it follows that v=v⊥so that the angular momentum formula h=rv⊥becomes simply h=rvfor a circular orbit. Substituting this expression for hinto Equation 2.52 and solving for vyields the velocity of a circular orbit, vcircular =/radicalbiggµ r(2.53) The time Trequired for one orbit is known as the period. Because the speed is constant, the period of a circular orbit is easy to compute: T=circumference speed=2πr/radicalbiggµ r so that Tcircular =2π√µr3 2 (2.54) 52 Chapter 2 The two-body problem The specific energy of a circular orbit is found by setting e=0 in Equation 2.50, ε=−1 2µ2 h2 Employing Equation 2.52 yields εcircular =−µ 2r(2.55) Obviously, the energy of a circular orbit is negative. As the radius goes up, the energy becomes less negative, i.e., it increases. In other words, the higher the orbit, the greaterits energy. T o launch a satellite from the surface of the earth into a circular orbit requires increasing its specific mechanical energy ε. This energy comes from the rocket motors of the launch vehicle. Since the mechanical energy of a satellite of mass mis E=mε,a propulsion system that can place a large mass in a low earth orbit can place a smallermass in a higher earth orbit. The space shuttle orbiters are the largest man-made satellites so far placed in orbit with a single launch vehicle. For example, on NASA mission STS-82 in February1997, the orbiter Discovery rendezvoused with the Hubble space telescope to repairand refurbish it. The altitude of the nearly circular orbit was 580 km (360 miles).Discovery’s orbital mass early in the mission was 106 000 kg (117 tons). That was only6 percent of the total mass of the shuttle prior to launch (comprising the orbiter’sdry mass, plus that of its payload and fuel, plus the two solid rocket boosters, plusthe external fuel tank filled with liquid hydrogen and oxygen). This mass of about2 million kilograms (2200 tons) was lifted off the launch pad by a total thrust inthe vicinity of 35 000 kN (7.8 million pounds). Eighty-five percent of the thrust wasfurnished by the solid rocket boosters (SRBs), w hich were depleted and jettisoned about two minutes into the flight. The remaining thrust came from the three liquidrockets (space shuttle main engines, or SSMEs) on the orbiter. These were fueled bythe external tank which was jettisoned just after the SSMEs were shut down at MECO(main engine cut off), about eight and a half minutes after lift-off. Manned orbital spacecraft and a host of unmanned remote sensing, imaging and navigation satellites occupy nominally circular, low-earth orbits. A low-earthorbit (LEO) is one whose altitude lies between about 150 km (100 miles) and about 1000 km (600 miles). An LEO is well above the nominal outer limits of the drag-producing atmosphere (about 80 km or 50 miles), and well below the hazardous Van Allen radiation belts, the innermost of which begins at about 2400 km (1500 miles). Nearly all of our applications of the orbital equations will be to the analysis of man-made spacecraft, all of which have a mass that is insignificant compared tothe sun and planets. For example, since the earth is nearly 20 orders of magnitudemore massive than the largest conceivable artificial satellite, the center of mass of thetwo-body system lies at the center of the earth and µin Equation 3.14 becomes m /H11005 G (mearth /H11001 msatellite ) /H11005 Gm earth The value of the earth’s gravitational parameter to be used throughout this book is found in Table A.2, µearth=398 600 km3/s2(2.56) 2.6 Circular orbits ( e=0) 53 Example 2.2Plot the speed vand period Tof a satellite in circular LEO as a function of altitude z. Equations 2.53 and 2.54 give the speed and period, respectively, of the satellite: v=/radicalbiggµ r=/radicalbiggµ RE+z=/radicalbigg 398 600 6378+zT=2π√µr3 2=2π√ 398 600(6378 +z)3 2 These relations are graphed in Figure 2.13. 200 400 600 8007.47.67.8 200 400 600 80090100υ, km/s 1000 z, km7.2 80T, min110 8.0 z, km1000 (a) (b) Figure 2.13 Circular orbital speed (a) and period (b) as a function of altitude. If a satellite remains always above the same point on the earth’s equator, then it is in a circular, geostationary equatorial orbit or GEO . For GEO, the radial from the center of the earth to the satellite must have the same angular velocity as the earth itself, namely,2πradians per sidereal day. The sidereal day is the time it takes the earth to complete one rotation relative to inertial space (the fixed stars). The ordinary 24-hour day, orsynodic day, is the time it takes the sun to apparently rotate once around the earth,from high noon one day to high noon the next. The synodic and sidereal days would beidentical if the earth stood still in space. However, while the earth makes one absoluterotation around its axis, it advances 2 π/365.26 radians along its solar orbit. Therefore, its inertial angular velocity ω Eis [(2π +2π/365.26)radians]/(24 hours); i.e., ωE=72.9217 ×10−6rad/s (2.57) Communications satellites and global weather satellites are placed in geostationary orbit because of the large portion of the earth’s surface visible from that altitudeand the fact that ground stations do not have to track the satellite, which appearsmotionless in the sky. Example 2.3Calculate the altitude zGEO and speed vGEO of a geostationary earth satellite. The speed of the satellite in its circular GEO of radius rGEO is vGEO=/radicalbiggµ rGEO(a) 54 Chapter 2 The two-body problem On the other hand, the speed vGEO along its circular path is related to the absolute angular velocity ωEof the earth by the kinematics formula vGEO=ωErGEO Equating these two expressions and solving for rGEO yields rGEO=3/radicalBigg µ ω2 E Substituting Equation 2.56, we get rGEO=3/radicalBigg 398 600 (72.9217×10−6)2=42 164 km (2.58) Therefore, the distance of the satellite above the earth’s surface is zGEO=rGEO−RE=42 164 −6378=35 786 km (22 241 mi) Substituting Equation 2.58 into (a) yields the speed, vGEO=/radicalbigg 398 600 42 164=3.075 km /s (2.59)(Example 2.3 continued) Example 2.4Calculate the maximum latitude and the percentage of the earth’s surface visible from GEO. T o find the maximum viewable latitude φ, use Figure 2.14, from which it is apparent that φ=cos−1RE r(a) RE r N SEquatorφ Figure 2.14 Satellite in GEO. 2.7 Elliptical orbits (0 <e<1) 55 where RE=6378 km and, according to Equation 2.57, r=42 164 km. Therefore φ=cos−16378 42 164=81.30◦Maximum visible north or south latitude. (b) The surface area Svisible from GEO is the shaded region illustrated in Figure 2.15. It can be shown that the area Sis given by S=2πR2 E(1−cosφ) Therefore, the percentage of the hemisphere visible from GEO is S 2πR2 E×100=(1−cos 81.30◦)×100=84.9% which of course means that 42.4 percent of the total surface of the earth can be seen from GEO. N EquatorRES φ Figure 2.15 Surface area Svisible from GEO. Figure 2.16 is a photograph taken from geosynchronous equatorial orbit by one of the National Oceanic and Atmospheric Administat ion’s Geostationary Operational Environmental Satellites (GOES). 2.7 Elliptical orbits (0 <e<1) If 0<e<1, then the denominator of Equation 2.35 varies with the true anomaly θ, but it remains positive, never becoming zero. Therefore, the relative position vector remains bounded, having its smallest magnitude at periapsis rp,g i v e nb y Equation 2.40. The maximum value of ris reached when the denominator of r=(h2/µ)/(1 +ecosθ) obtains its minimum value, which occurs at θ=180◦. That 56 Chapter 2 The two-body problem Figure 2.16 The view from GEO. NASA-Goddard Space Flight Center, data from NOAA GOES. a a CF F'B P Ab aeb Apse line rprarB Figure 2.17 Elliptical orbit. m1is at the focus F.F/primeis the unoccupied empty focus. point is called the apoapsis, and its radial coordinate, denoted ra,i s ra=h2 µ1 1−e(2.60) The curve defined by Equation 2.35 in this case is an ellipse. Let 2 abe the distance measured along the apse line from periapsis Pto apoapsis A, as illustrated in Figure 2.17. Then 2a=rp+ra 2.7 Elliptical orbits (0 <e<1) 57 Substituting Equations 2.40 and 2.61 into this expression we get a=h2 µ1 1−e2(2.61) ais the semimajor axis of the ellipse. Solving Equation 2.61 for h2/µand putting the result into Equation 2.35 yields an alternative form of the orbit equation, r=a1−e2 1+ecosθ(2.62) In Figure 2.17, let Fdenote the location of the body m1, which is the origin of the r,θpolar coordinate system. The center Cof the ellipse is the point lying midway between the apoapsis and periapsis. The distance CFfrom CtoFis CF=a−FP=a−rp But from Equation 2.62, rp=a(1−e) (2.63) Therefore, CF=ae, as indicated in Figure 2.17. LetBbe the point on the orbit which lies directly above C, on the perpendicular bisector of AP. The distance bfrom CtoBis the semiminor axis. If the true anomaly of point Bisβ, then according to Equation 2.62, the radial coordinate of Bis rB=a1−e2 1+ecosβ(2.64) The projection of rBonto the apse line is ae; i.e., ae=rBcos(180 −β)=− rBcosβ=−/parenleftbigg a1−e2 1+ecosβ/parenrightbigg cosβ Solving this expression for e, we obtain e=− cosβ (2.65) Substituting this result into Equation 2.64 reveals the interesting fact that rB=a According to the Pythagorean theorem, b2=r2 B−(ae)2=a2−a2e2 which means the semiminor axis is found in terms of the semimajor axis and the eccentricity of the ellipse as b=a/radicalbig 1−e2 (2.66) Let an xycartesian coordinate system be centered at C, as shown in Figure 2.18. In terms of randθ, we see from the figure that the xcoordinate of a point on 58 Chapter 2 The two-body problem xy aer(x, y) Pb C aθ Figure 2.18 Cartesian coordinate description of the orbit. the orbit is x=ae+rcosθ=ae+/parenleftbigg a1−e2 1+ecosθ/parenrightbigg cosθ=ae+cosθ 1+ecosθ From this we have x a=e+cosθ 1+ecosθ(2.67) For the ycoordinate we have, making use of Equation 2.66, y=rsinθ=/parenleftbigg a1−e2 1+ecosθ/parenrightbigg sinθ=b√ 1−e2 1+ecosθsinθ Therefore, y b=√ 1−e2 1+ecosθsinθ (2.68) Using Equations 2.67 and 2.68, we find x2 a2+y2 b2=1 (1+ecosθ)2/bracketleftbig (e+cosθ)2+(1−e2) sin2θ/bracketrightbig =1 (1+ecosθ)2/bracketleftbig e2+2ecosθ+cos2θ+sin2θ−e2sin2θ/bracketrightbig =1 (1+ecosθ)2/bracketleftbig e2+2ecosθ+1−e2sin2θ/bracketrightbig =1 (1+ecosθ)2/bracketleftbig e2(1−sin2θ)+2ecosθ+1/bracketrightbig =1 (1+ecosθ)2/bracketleftbig e2cos2θ+2ecosθ+1/bracketrightbig =1 (1+ecosθ)2(1+ecosθ)2 That is, x2 a2+y2 b2=1 (2.69) 2.7 Elliptical orbits (0 <e<1) 59 This is the familiar cartesian coordinate formula for an ellipse centered at the origin, with xintercepts at ±aand yintercepts at ±b.I fa=b, Equation 2.69 describes a circle, which is really an ellipse whose eccentricity is zero. The specific energy of an elliptical orbit i s negative, and it is found by substituting the specific angular momentum and eccentricity into Equation 2.50, ε=−1 2µ2 h2(1−e2) However, according to Equation 2.61, h2=µa(1−e2), so that ε=−µ 2a(2.70) This shows that the specific energy is independent of the eccentricity and depends only on the semimajor axis of the ellipse. For an e lliptical orbit, the conservation of energy (Equation 2.47) may therefore be written v2 2−µ r=−µ 2a(2.71) The area of an ellipse is found in terms of its semimajor and semiminor axes by the formula A=πab(which reduces to the formula for the area of a circle if a=b). T o find the period Tof the elliptical orbit, we employ Kepler’s second law, dA/dt =h/2, to obtain /Delta1A=h 2/Delta1t For one complete revolution, /Delta1A=πaband/Delta1t=T. Thus, πab=(h/2)T ,o r T=2πab h Substituting Equations 2.61 and 2.66, we get T=2π ha2/radicalbig 1−e2=2π h/parenleftbiggh2 µ1 1−e2/parenrightbigg2/radicalbig 1−e2 so that the formula for the period of an elliptical orbit, in terms of the orbital parameters hand e, becomes T=2π µ2/parenleftbiggh√ 1−e2/parenrightbigg3 (2.72) We can once again appeal to Equation 2.61 to substitute h=/radicalbig µa(1−e2) into this equation, thereby obtaining an alternative expression for the period, T=2π√µa3 2 (2.73) This expression, which is identical to that of a circular orbit of radius a(Equa- tion 2.54), reveals that, like the energy, the period of an elliptical orbit is independent 60 Chapter 2 The two-body problem 1 2 3 4 5 12345 Figure 2.19 Since all five ellipses have the same major axis, their periods and energies are identical. of the eccentricity (see Figure 2.19). Equation 2.73 embodies Kepler’s third law: the period of a planet is proportional to the three-halves power of its semimajor axis. Finally, observe that dividing Equation 2.40 by Equation 2.60 yields rp ra=1−e 1+e Solving this for eresults in a useful formula for calculating the eccentricity of an elliptical orbit, namely, e=ra−rp ra+rp(2.74) From Figure 2.17 it is apparent that ra−rp=F/primeF, the distance between the foci. As previously noted, ra+rp=2a. Thus, Equation 2.74 has the geometrical interpretation, eccentricity =distance between the foci length of the major axis What is the average distance of m2from m1in the course of one complete orbit? T o answer this question, we divide the range of the true anomaly (2 π) into nequal segments /Delta1θ, so that n=2π /Delta1θ We then use r=(h2/µ)/(1+ecosθ)t oe v a l u a t e r(θ)a tt h e nequally spaced values of true anomaly, starting at periapsis: θ1=0,θ2=/Delta1θ,θ3=2/Delta1θ,...,θn=(n−1)/Delta1θ 2.7 Elliptical orbits (0 <e<1) 61 The average of this set of nvalues of ris given by ¯rθ=1 nn/summationdisplay i=1r(θi)=/Delta1θ 2πn/summationdisplay i=1r(θi)=1 2πn/summationdisplay i=1r(θi)/Delta1θ (2.75) Now let nbecome very large, so that /Delta1θbecomes very small. In the limit as n→∞ , Equation 2.75 becomes ¯rθ=1 2π/integraldisplay2π 0r(θ)dθ (2.76) Substituting Equation 2.62 into the integrand yields ¯rθ=1 2πa(1−e2)/integraldisplay2π 0dθ 1+ecosθ The integral in this expression can be found in integral tables (e.g., Beyer, 1991), from which we obtain ¯rθ=1 2πa(1−e2)/parenleftbigg2π√ 1−e2/parenrightbigg =a/radicalbig 1−e2 (2.77) Comparing this result with Equation 2.66, we see that the true-anomaly-averaged orbital radius equals the length of the semiminor axis bof the ellipse. Thus, the semimajor axis, which is the average of the maximum and minimum distances fromthe focus, is not the mean distance. Since, from Equation 2.62, r p=a(1−e) and ra=a(1+e), Equation 2.77 also implies that ¯rθ=√rpra (2.78) The mean distance is the one-half power of the product of the maximum and mini- mum distances from the focus and not one-half their sum. Example 2.5An earth satellite is in an orbit with perigee altitude zp=400 km and an eccentricity e=0.6. Find (a) the perigee velocity, vp; (b) the apogee radius, ra; (c) the semimajor axis, a; (d) the true-anomaly-averaged radius ¯rθ; (e) the apogee velocity; (f) the period of the orbit; (g) the true anomaly when r=¯rθ; (h) the satellite speed when r=¯rθ; (i) the flight path angle γwhen r=¯rθ; (j) the maximum flight path angle γmaxand the true anomaly at which it occurs. The strategy is always to go after the primary orbital parameters, eccentricity andangular momentum, first. In this problem we are given the eccentricity, so we will first seek h. Recall from Equation 2.56 that µ=398 600 km3/s2and also that RE=6378 km. (a) The perigee radius is rp=RE+zp=6378+400=6778 km Evaluating the orbit formula, Equation 2.35, at θ=0 (perigee), we get rp=h2 µ1 1+e 62 Chapter 2 The two-body problem (Example 2.5 continued)We use this to evaluate the angular momentum 6778=h2 398 6001 1+0.6 h=65 750 km2/s Now we can find the perigee velocity using the angular momentum formula, Equation 2.21: vp=v⊥)perigee =h rp=65 750 6778=9.700 km /s (b) The apogee radius is found by evaluating the orbit equation at θ=180◦(apogee): ra=h2 µ1 1−e=65 7502 398 6001 1−0.6=27 110 km (c) The semimajor axis is the average of the perigee and apogee radii: a=rp+ra 2=6778+27 110 2=16 940 km (d) The azimuth-averaged radius is given by Equation 2.78: ¯rθ=√rpra=√ 6778·27 110 =13 560 km (e) The apogee velocity, like that at perigee, is obtained from the angular momentum formula, va=v⊥)apogee=h ra=65 750 27 110=2.425 km /s (f) T o find the orbit period, use Equation 2.73 T=2π µ2/parenleftbiggh√ 1−e2/parenrightbigg3 =2π 398 6002/parenleftbigg65 750√ 1−0.62/parenrightbigg3 =21 950 s =6.098 hr (g) T o find the true anomaly when r=¯rθ, we again use the orbit formula ¯rθ=h2 µ1 1+ecosθ 13 560 =65 7502 398 6001 1+0.6c o sθ cosθ=− 0.3333 This means θ=109.5◦, where the satellite passes through ¯rθon its way from perigee and θ=250.5◦, where the satellite passes through ¯rθon its way towards perigee 2.7 Elliptical orbits (0 <e<1) 63 (h) T o find the speed of the satellite when r=¯rθ, we first calculate the radial and transverse components of velocity: v⊥=h ¯rθ=65 750 13 560=4.850 km /s For the radial velocity component, use Equation 2.38, vr=µ hesinθ=398 600 65 750·0.6·sin(109.5◦)=3.430 km /s or vr=µ hesinθ=398 600 65 750·0.6·sin(250.5◦)=− 3.430 km /s The magnitude of the velocity can now be found as v=/radicalBig v2r+v2 ⊥=/radicalbig 3.4302+4.8502=5.940 km /s We could have obtained the speed vmore directly by using conservation of energy (Equation 2.71), since the semimajor axis is available from part (c) above. However, we would still need to compute vrandv⊥in order to solve the next part of this problem. (i) Use Equation 2.39 to calculate the flight path angle at r=¯rθ, γ=tan−1vr v⊥=tan−13.430 4.850=35.26◦atθ=109.5◦ γis positive, meaning the velocity vector is above the local horizon, indicating the spacecraft is flying away from the attracting force. At θ=250.5◦, where the space- craft is flying towards perigee, γ=− 35.26◦. Since the satellite is approaching the attracting body, the velocity vector lies below the local horizon, as indicated by the minus sign. (j) Equation 2.42 gives the flight path angle in terms of the true anomaly, γ=tan−1 esinθ 1+ecosθ(a) T o find where γis a maximum, we must take the derivative of this expression with respect to θand set the result equal to zero. Using the rules of calculus, dγ dθ=1 1+/parenleftbiggesinθ 1+ecosθ/parenrightbigg2d dθ/parenleftbiggesinθ 1+ecosθ/parenrightbigg =e(e+cosθ) (1+ecosθ)2+e2sin2θ Fore<1, the denominator is positive for all values of θ. Therefore, dγ/dθ=0 only if the numerator vanishes, that is, if cos θ=− e. Recall from Equation 2.65 that this true anomaly locates the end-point of the minor axis of the ellipse. The maximum positive flight path angle therefore occurs at the true anomaly, θ=cos−1(−0.6) =126.9◦ 64 Chapter 2 The two-body problem Substituting this into (a), we find the value of the flight path angle to be γmax=tan−10.6 sin 126 .9◦ 1+0.6 cos 126 .9◦=36.87◦ After attaining this greatest magnitude, the flight path angle starts to decrease steadily towards its value at apogee (zero).(Example 2.5 continued) Example 2.6At two points on a geocentric orbit the altitude and true anomaly are z1= 1545 km, θ1=126◦and z2=852 km, θ2=58◦, respectively. Find (a) the eccentricity; (b) the altitude of perigee; (c) the semimajor axis; and (d) the period. (a) The radii of the two points are r1=RE+z1=6378+1545=7923 km r2=RE+z2=6378+852=7230 km Applying the orbit formula, Equation 2.35, to both of these points yields two equations for the two primary orbital parameters, angular momentum hand eccentricity e: r1=h2 µ1 1+ecosθ1 7923=h2 398 6001 1+ecos 126◦ h2=3.158×109−1.856×109e (a) r2=h2 µ1 1+ecosθ2 7230=h2 398 6001 1+ecos 58◦ h2=2.882×109+1.527×109e (b) Equating (a) and (b), the two expressions for h2, yields a single equation for the eccentricity e, 3.158×109−1.856×109e=2.882×109+1.527×109e⇒ 3.384×109e =276.2×106 Therefore, e=0.08164 (an ellipse) (c) (b) By substituting the eccentricity back into (a) [or (b)] we find the angular momentum, h2=3.158×109−1.856×109·0.08164 ⇒ h=54 830 km2/s( d ) 2.8 Parabolic trajectories ( e=1) 65 Now we can use the orbit equation to obtain the perigee radius rp=h2 µ1 1+ecos(0)=54 8302 398 6001 1+0.08164=6974 km and perigee altitude zp=rp−RE=6974−6378=595.5k m (c) The semimajor axis can be found after we calculate the apogee radius by means of the orbit equation, just as we did for perigee radius: ra=h2 µ1 1+ecos(180◦)=54 8302 398 6001 1−0.08164=8213 km Hence a=rp+ra 2=8213+6974 2=7593 km (d) Since the semimajor axis is available, it is convenient to use Equation 2.74 to find the period: T=2π√µa3 2=2π√ 398 60075933 2=6585 s =1.829 hr 2.8 Parabolic trajectories (e =1) If the eccentricity equals 1, then the orbit equation (Equation 2.35) becomes r=h2 µ1 1+cosθ(2.79) As the true anomaly θapproaches 180◦, the denominator approaches zero, so that rtends towards infinity. According to Equation 2.50, the energy of a trajectory for which e=1 is zero, so that for a parabolic trajectory the conservation of energy, Equation 2.47, is v2 2−µ r=0 In other words, the speed anywhere on a parabolic path is v=/radicalbigg 2µ r(2.80) If the body m2is launched on a parabolic trajectory, it will coast to infinity, arriving there with zero velocity relative to m1. It will not return. Parabolic paths are therefore 66 Chapter 2 The two-body problem called escape trajectories. At a given distance rfrom m1, the escape velocity is given by Equation 2.80, vesc=/radicalbigg 2µ r(2.81) Letvobe the speed of a satellite in a circular orbit of radius r. Then from Equations 2.53 and 2.81 we have vesc=√ 2vo (2.82) That is, to escape from a circular orbit requires a velocity boost of 41.4 percent. However, remember our assumption is that m1and m2are the only objects in the universe. A spacecraft launched from earth with velocity vesc(relative to the earth) will not coast to infinity (i.e., leave the solar system) because it will eventually succumb tothe gravitational influence of the sun and, in fact, end up in the same orbit as earth.This will be discussed in more detail in Chapter 8. For the parabola, Equation 2.42 for the flight path angle takes the form tanγ=sinθ 1+cosθ Using the trigonometric identities sinθ=2 sinθ 2cosθ 2 cosθ=cos2θ 2−sin2θ 2=2c o s2θ 2−1 we can write tanγ=2 sinθ 2cosθ 2 2c o s2θ 2=sinθ 2 cosθ 2=tanθ 2 It follows that γ=θ 2(2.83) That is, on parabolic trajectories the flight path angle is one-half the true anomaly. Recall that the parameter pof an orbit is given by Equation 2.43. Let us substitute that expression into Equation 2.79 and then plot r=2a/(1+cosθ) in a cartesian coordinate system centered at the focus, as illustrated in Figure 2.21. From the figureit is clear that x=rcosθ=pcosθ 1+cosθ(2.84a) y=rsinθ=psinθ 1+cosθ(2.84b) 2.8 Parabolic trajectories ( e=1) 67 v θ FPApse lineγ/H11005θ 2υr υ Figure 2.20 Parabolic trajectory around the focus F. xy p 0θ r (x, y) /H11002pp/2 Figure 2.21 Parabola with focus at the origin of the cartesian coordinate system. 68 Chapter 2 The two-body problem Therefore x p/2+/parenleftbiggy p/parenrightbigg2 =2cosθ 1+cosθ+sin2θ (1+cosθ)2 Working to simplify the right-hand side, we get x p/2+/parenleftbiggy p/parenrightbigg2 =2c o sθ(1+cosθ)+sin2θ (1+cosθ)2=2c o sθ+2c o s2θ+(1−cos2θ) (1+cosθ)2 =1+2c o sθ+cos2θ (1+cosθ)2=(1+cosθ)2 (1+cosθ)2=1 It follows that x=p 2−y2 2p(2.85) This is the equation of a parabola in a cartesian coordinate system whose origin serves as the focus. Example 2.7The perigee of a satellite in a parabolic geocentric trajectory is 7000 km. Find the distance dbetween points P1and P2on the orbit which are 8000 km and 16 000 km, respectively, from the center of the earth. First, let us calculate the angular momentum of the satellite by evaluating the orbitequation at perigee, r p=h2 µ1 1+cos(0)=h2 2µ P1P2 7000 kmEarth16 000 km8000 kmd ∆θ Figure 2.22 Parabolic geocentric trajectory. 2.9 Hyperbolic trajectories ( e>1) 69 from which h=/radicalbig 2µr p=√ 2·398 600 ·7000=74 700 km2/s( a) T o find the length of the chord P1P2, we must use the law of cosines from trigonometry, d2=80002+16 0002−2·8000·16 000 cos /Delta1θ (b) The true anomalies of points P1and P2are found using the orbit equation: 8000=74 7002 398 6001 1+cosθ1⇒ cosθ1=0.75⇒θ1=41.41◦ 16 000 =74 7002 398 6001 1+cosθ2⇒ cosθ2=− 0.125 ⇒θ2=97.18◦ Therefore, /Delta1θ=97.18◦−41.41◦=55.78◦, so that (b) yields d=13 270 km (c) 2.9 Hyperbolic trajectories ( e>1) Ife>1, the orbit formula, r=h2 µ1 1+ecosθ(2.86) describes the geometry of the hyperbola shown in Figure 2.23. The system consists of two symmetric curves. One of them is occupied by the orbiting body, the other oneis its empty, mathematical image. Clearly, the denominator of Equation 2.86 goes tozero when cos θ=− 1/e. We denote this value of true anomaly θ ∞=cos−1(−1/e ) (2.87) since the radial distance approaches infinity as the true anomaly approaches θ∞.θ∞ is known as the true anomaly of the asymptote. Observe that θ∞lies between 90◦and 180◦. From trigonometry it follows that sinθ∞=√ e2−1 e(2.88) For−θ∞<θ<θ ∞, the physical trajectory is the occupied hyperbola Ishown on the left in Figure 2.23. For θ∞<θ<(360◦−θ∞), hyperbola II– the vacant orbit around the empty focus F/prime– is traced out. (The vacant orbit is physically impossible, because it would require a repulsive gravitational force.) Periapsis Plies on the apse line on the physical hyperbola I, whereas apoapsis Alies on the apse line on the vacant orbit. The point halfway between periapsis and apoapsis is the center Cof the hyperbola. The asymptotes of the hyperbola are the straight lines towards which the curves tend 70 Chapter 2 The two-body problem F F' P A rp a aEmpty focusApse line I IIVacant orbit Cδ β βbM raAsymptote Asymptote θ∞∆ Figure 2.23 Hyperbolic trajectory. as they approach infinity. The asymptotes intersect at C, making an acute angle β with the apse line, where β=180◦−θ∞. Therefore, cos β=− cosθ∞, which means β=cos−1(1/e) (2.89) The angle δbetween the asymptotes is called the turn angle. This is the angle through which the velocity vector of the orbiting body is rotated as it rounds the attractingbody at Fand heads back towards infinity. From the figure we see that δ=180 ◦−2β, so that sinδ 2=sin/parenleftbigg180◦−2β 2/parenrightbigg =sin(90◦−β)=cosβEq.2.89 /bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright =1 e or δ=2 sin−1(1/e) (2.90) The distance rpfrom the focus Fto the periapsis is given by Equation 2.40, ra=h2 µ1 1+e(2.91) Just as for an ellipse, the radial coordinate raof apoapsis is found by setting θ=180◦ in Equation 2.35, ra=h2 µ1 1−e(2.92) Observe that rais negative, since e>1 for the hyperbola. That means the apoapse lies to the right of the focus F. From Figure 2.23 we see that the distance 2 afrom periapse 2.9 Hyperbolic trajectories ( e>1) 71 Pto apoapse Ais 2a=|ra|−r p=− ra−rp Substituting Equations 2.91 and 2.92 yields 2a=−h2 µ/parenleftbigg1 1−e+1 1+e/parenrightbigg From this it follows that a, the semimajor axis of the hyperbola, is given by an expression which is nearly identical to that for an ellipse (Equation 2.62), a=h2 µ1 e2−1(2.93) Therefore, Equation 2.86 may be written for the hyperbola r=ae2−1 1+ecosθ(2.94) This formula is analogous to Equation 2.63 for the elliptical orbit. Furthermore, from Equation 2.94 it follows that rp=a(e−1) (2.95a) ra=− a(e+1) (2.95b) The distance bfrom periapsis to an asymptote, measured perpendicular to the apse line, is the semiminor axis of the hyperbola. From Figure 2.23, we see that the lengthbof the semiminor axis PM is b=atanβ=asinβ cosβ=asin (180 −θ∞) cos (180 −θ∞)=asinθ∞ −cosθ∞=a√ e2−1 e −/parenleftbigg −1 e/parenrightbigg so that for the hyperbola, b=a/radicalbig e2−1 (2.96) This relation is analogous to Equation 2.67 for the semiminor axis of an ellipse. The distance /Delta1between the asymptote and a parallel line through the focus is called the aiming radius, which is illustrated in Figure 2.23. From that figure we seethat /Delta1=(r p+a) sinβ =aesinβ (Equation 2.95a) =ae√ e2−1 e(Equation 2.89) =aesinθ∞ (Equation 2.88) =ae/radicalbig 1−cos2θ∞ (trig identity) =ae/radicalbigg 1−1 e2(Equation 2.87) 72 Chapter 2 The two-body problem F F' rp a aθ r x y Oxy Figure 2.24 Plot of Equation 2.93 in a cartesian coordinate system with origin Omidway between the two foci. or /Delta1=a/radicalbig e2−1 (2.97) Comparing this result with Equation 2.96, it is clear that the aiming radius equals the length of the semiminor axis of the hyperbola. As with the ellipse and the parabola, we can express the polar form of the equation of the hyperbola in a cartesian coordinate system whose origin is in this case midwaybetween the two foci, as illustrated in Figure 2.24. From the figure it is apparent that x=− a−r p+rcosθ (2.98a) y=rsinθ (2.98b) Using Equations 2.94 and 2.95a in 2.98a, we obtain x=− a−a(e−1)+ae2−1 1+ecosθcosθ=− ae+cosθ 1+ecosθ Substituting Equations 2.94 and 2.96 into 2.98b yields y=b√ e2−1e2−1 1+ecosθsinθ=b√ e2−1 sinθ 1+ecosθ It follows that x2 a2−y2 b2=/parenleftbigge+cosθ 1+ecosθ/parenrightbigg2 −/parenleftBigg√ e2−1 sinθ 1+ecosθ/parenrightBigg2 =e2+2ecosθ+cos2θ−(e2−1)(1−cos2θ) (1+ecosθ)2 =1+2ecosθ+e2cos2θ (1+ecosθ)2=(1+ecosθ)2 (1+ecosθ)2 2.9 Hyperbolic trajectories ( e>1) 73 That is, x2 a2−y2 b2=1 (2.99) This is the familiar equation of a hyperbola which is symmetric about the xand y axes, with intercepts on the xaxis. The specific energy of the hyperbolic trajectory is given by Equation 2.50. Substituting Equation 2.93 into that expression yields ε=µ 2a(2.100) The specific energy of a hyperbolic orbit i s clearly positive and independent of the eccentricity. The conservation of energy for a hyperbolic trajectory is v2 2−µ r=µ 2a(2.101) Letv∞denote the speed at which a body on a hyperbolic path arrives at infinity. According to Equation 2.101 v∞=/radicalbiggµ a(2.102) v∞is called the hyperbolic excess speed. In terms of v∞we may write Equa- tion 2.101 as v2 2−µ r=v2 ∞ 2 Substituting the expression for escape speed, vesc=√2µ/r (Equation 2.81), we obtain for a hyperbolic trajectory v2=v2 esc+v2 ∞ (2.103) This equation clearly shows that the hyperbolic excess speed v∞represents the excess kinetic energy over that which is required to simply escape from the centerof attraction. The square of v ∞is denoted C3, and is known as the characteristic energy, C3=v2 ∞ (2.104) C3is a measure of the energy required for an interplanetary mission and C3is also a measure of the maximum energy a launch vehicle can impart to a spacecraft of agiven mass. Obviously, to match a launch vehicle with a mission, C 3)launchvehicle > C3)mission . Note that the hyperbolic excess speed can also be obtained from Equations 2.39 and 2.88, v∞=µ hesinθ∞=µ h/radicalbig e2−1 (2.105) Finally, for purposes of comparison, Figure 2.25 shows a range of trajectories, from a circle through hyperbolas, all having a common focus and periapsis. The parabolais the demarcation between the closed, nega tive energy orbits (ellipses) and open, positive energy orbits (hyperbolas). 74 Chapter 2 The two-body problem 0.5 0.7 0.8 0.85 0.9e /H11005 1.0 1.3 1.52.5 0.3 0 FP1.1 Figure 2.25 Orbits of various eccentricities, having a common focus Fand periapsis P. Example 2.8At a given point of a spacecraft’s geocentric trajectory, the radius is 14 600 km, the speed is 8.6 km/s, and the flight path angle is 50◦. Show that the path is a hyper- bola and calculate the following: (a) C3, (b) angular momentum, (c) true anomaly, (d) eccentricity, (e) radius of perigee, (f) turn angle, (g) semimajor axis, and (h) aiming radius. T o determine the type of the trajectory, calculate the escape speed at the given radius: vesc=/radicalbigg 2µ r=/radicalbigg 2·398 600 14 600=7.389 km /s Since the escape speed is less than the spacecraft’s speed of 8.6 km/s, the path is a hyperbola. (a) The hyperbolic excess velocity v∞is found from Equation 2.103, v2 ∞=v2−v2 esc=8.62−7.3892=19.36 km2/s2 From Equation 2.104 it follows that C3=19.36 km2/s2 (b) Knowing the speed and the flight path angle, we can obtain both vrandv⊥: vr=vsinγ=8.6 sin 50◦=6.588 km /s( a ) 2.9 Hyperbolic trajectories ( e>1) 75 v⊥=vcosγ=8.6·cos 50◦=5.528 km /s (b) Then Equation 2.21 provides us with the angular momentum, h=rv⊥=14 600 ·5.528=80 710 km2/s (c) (c) Evaluating the orbit equation at the given location on the trajectory, we get 14 600 =80 7102 398 6001 1+ecosθ from which ecosθ=0.1193 (d) The radial component of velocity is given by Equation 2.39, vr=µesinθ/h,s o that with (a) and (c), we obtain 6.588=398 600 80 170esinθ or esinθ=1.334 (e) Computing the ratio of (e) to (d) yields tanθ=1.334 0.1193=11.18 ⇒θ=84.89◦ (d) We substitute the true anomaly back into either (d) or (e) to find the eccentricity, e=1.339 (e) The radius of perigee can now be found from the orbit equation, rp=h2 µ1 1+ecos(0)=80 7102 398 6001 1+1.339=6986 km (f) The formula for turn angle is Equation 2.90, from which δ=2 sin−1/parenleftbigg1 e/parenrightbigg =2 sin−1/parenleftbigg1 1.339/parenrightbigg =96.60◦ (g) The semimajor axis of the hyperbola is found in Equation 2.93, a=h2 µ1 e2−1=80 7102 398 6001 1.3392−1=20 590 km (h) According to Equations 2.96 and 2.97, the aiming radius is /Delta1=a/radicalbig e2−1=20 590/radicalbig 1.3392−1=18 340 km 76 Chapter 2 The two-body problem 2.10 Perifocal frame The perifocal frame is the ‘natural frame’ for an orbit. It is centered at the focus of the orbit. Its xyplane is the plane of the orbit, and its xaxis is directed from the focus through periapse, as illustrated in Figure 2.26. The unit vector along the xaxis (the apse line) is denoted ˆp.T h e yaxis, with unit vector ˆq, lies at 90◦true anomaly to the xaxis. The zaxis is normal to the plane of the orbit in the direction of the angular momentum vector h.T h e¯zunit vector is ˆw, ˆw=h h(2.106) In the perifocal frame, the position vector ris written (see Figure 2.27) r=xˆp+yˆq (2.107) where x=rcosθ y=rsinθ (2.108) and r, the magnitude of r, is given by the orbit equation, r=(h2/µ)[1/(1+ecosθ)]. Thus, we may write Equation 2.107 as r=h2 µ1 1+ecosθ(c o sθˆp+sinθˆq) (2.109) The velocity is found by taking the time derivative of r, v=˙r=˙xˆp+˙yˆq (2.110) ˆ p ˆ q ˆ w FocusPeriapseSemilatus rectum z y x Figure 2.26 Perifocal frame ˆpˆqˆw. 2.10 Perifocal frame 77 From Equations 2.108 we obtain ˙x=˙rcosθ−r˙θsinθ ˙y=˙rsinθ+r˙θcosθ (2.111) ˙ris the radial component of velocity, vr. Therefore, according to Equation 2.39, ˙r=µ hesinθ (2.112) From Equations 2.36 and 2.38 we have r˙θ=v⊥=µ h(1+ecosθ) (2.113) Substituting Equations 2.112 and 2.113 into 2.111 and simplifying the results yields ˙x=−µ hsinθ ˙y=µ h(e+cosθ) (2.114) Hence, Equation 2.110 becomes v=µ h[−sinθˆp+(e+cosθ)ˆq] (2.115) Formulating the kinematics of orbital motion in the perifocal frame, as we have done here, is a prelude to the study of orbits in three dimensions (Chapter 4). We also needEquations 2.107 and 2.110 in the next section. ˆq θrv ˆpˆw Periapsey x Figure 2.27 Position and velocity relative to the perifocal frame. 78 Chapter 2 The two-body problem 2.11 The Lagrange coefficients In this section we will establish what may s eem intuitively obvious: if the position and velocity of an orbiting body are known at a given instant, then the position andvelocity at any later time are found in terms of the initial values. Let us start withEquations 2.107 and 2.110, r= xˆp+yˆq (2.116) v=˙r=˙xˆp+˙yˆq (2.117) Attach a subscript ‘zero’ to quantities evaluated at time t=t0. Then the expressions forrand vevaluated at t=t0are r0=x0ˆp+y0ˆq (2.118) v0=˙x0ˆp+˙y0ˆq (2.119) The angular momentum his constant, so let us calculate it using the initial conditions. Substituting Equations 2.118 and 2.119 into Equation 2.18 yields h=r0×v0=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆpˆqˆw x0y00 ˙x0˙y00/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=ˆw( x0˙y0−y0˙x0) (2.120) Recall that ˆwis the unit vector in the direction of h(Equation 2.106). Therefore, the coefficient of ˆwon the right of Equation 2.120 must be the magnitude of the angular momentum. That is, h=x0˙y0−y0˙x0 (2.121) Now let us solve the two vector equations (2.118) and (2.119) for the unit vectors ˆp andˆqin terms of r0and v0. From (2.118) we get ˆq=1 y0r0−x0 y0ˆp (2.122) Substituting this into Equation (2.119), combining terms and using Equation 2.121 yields v0=˙x0ˆp+˙y0/parenleftbigg1 y0r0−x0 y0ˆp/parenrightbigg =y0˙x0−x0˙y0 y0ˆp+˙y0 y0r0=−h y0ˆp+˙y0 y0r0 Solve this for ˆpto obtain ˆp=˙y0 hr0−y0 hv0 (2.123) Putting this result back into Equation 2.122 gives ˆq=1 y0r0−x0 y0/parenleftBigg˙y0 hr0−y0 hv0/parenrightBigg =h−x0˙y0 y0r0+x0 hv0 2.11 The Lagrange coefficients 79 Upon replacing hby the right-hand side of Equation 2.121 we get ˆq=−˙x0 hr0+x0 hv0 (2.124) Equations 2.123 and 2.124 give ˆpandˆqin terms of the initial position and veloc- ity. Substituting those two expressions back into Equations 2.116 and 2.117 yields,respectively r= x/parenleftBigg˙y0 hr0−y0 hv0/parenrightBigg +y/parenleftBigg −˙x0 hr0+x0 hv0/parenrightBigg =x˙y0−y˙x0 hr0+−xy0+yx0 hv0 v=˙x/parenleftBigg˙y0 hr0−y0 hv0/parenrightBigg +˙y/parenleftBigg −˙x0 hr0+x0 hv0/parenrightBigg =˙x˙y0−˙y˙x0 hr0+−˙xy0+˙yx0 hv0 Therefore, r=fr0+gv0 (2.125) v=˙fr0+˙gv0 (2.126) where fand gare given by f=x˙y0−y˙x0 h(2.127a) g=−xy0+yx0 h(2.127b) together with their time derivatives ˙f=˙x˙y0−˙y˙x0 h(2.128a) ˙g=−˙xy0+˙yx0 h(2.128b) The fand gfunctions are referred to as the Lagrange coefficients after Joseph- Louis Lagrange (1736–1813), a French mathematical physicist whose numerouscontributions include calculations of planetary motion. From Equations 2.125 and 2.126 we see that the position and velocity vectors r and vare indeed linear combinations of the initial position and velocity vectors. The Lagrange coefficients and their time derivatives in these expressions are themselvesfunctions of time and the initial conditions. Before proceeding, let us show that the conservation of angular momentum h imposes a condition on fand gand their time derivatives ˙fand˙g. Calculate husing Equations 2.125 and 2.126, h=r×v=(fr 0+gv0)×(˙fr0+˙gv0) Expanding the right-hand side yields h=(fr0×˙fr0)+(fr0×˙gv0)+(gv0×˙fr0)+(gv0×˙gv0) 80 Chapter 2 The two-body problem Factoring out the scalars f,g,˙fand˙g,w eg e t h=f˙f(r0×r0)+f˙g(r0×v0)+˙fg(v0×r0)+g˙g(v0×v0) But r0×r0=v0×v0=0, so h=f˙g(r0×v0)+˙fg(v0×r0) Since v0×r0=− (r0×v0) this reduces to h=(f˙g−˙fg)(r0×v0) or h=(f˙g−˙fg)h0 where h0=r0×v0, which is the angular momentum at t=t0. But the angular momentum is constant (recall Equation 2.19), which means h=h0, so that h=(f˙g−˙fg)h Since hcannot be zero (unless the body is traveling in a straight line towards the center of attraction), it follows that f˙g−˙fg=1 (conservation of angular momentum) (2.129) Thus, if any three of the functions f,g,˙fand˙gare known, the fourth may be found from Equation 2.129. Let us use Equations 2.127 and 2.128 to evaluate the Lagrange coefficients and their time derivatives in terms of the true anomaly. First of all, note that evaluatingEquations 2.108 at time t=t 0yields x0=r0cosθ0(2.130) y0=r0sinθ0 Likewise, from Equations 2.114 we get ˙x0=−µ hsinθ0 (2.131) ˙y0=µ h(e+cosθ0) T o evaluate the function f, we substitute Equations 2.108 and 2.131 into Equation 2.127a, f=x˙y0−y˙x0 h =1 h/braceleftBig [rcosθ]/bracketleftBigµ h(e+cosθ0)/bracketrightBig −[rsinθ]/bracketleftBig −µ hsinθ0/bracketrightBig/bracerightBig =µr h2[ecosθ+(c o sθcosθ0+sinθsinθ0)] (2.132) 2.11 The Lagrange coefficients 81 If we invoke the trig identity cos(θ−θ0)=cosθcosθ0+sinθsinθ0 (2.133) and let /Delta1θrepresent the difference between the current and initial true anomalies, /Delta1θ=θ−θ0 (2.134) then Equation 2.132 reduces to f=µr h2(ecosθ+cos/Delta1θ) (2.135) Finally, from Equation 2.35, we have ecosθ=h2 µr−1 (2.136) Substituting this into Equation 2.135 leads to f=1−µr h2(1−cos/Delta1θ) (2.137) We obtain rfrom the orbit formula, Equation 2.35, in which the true anomaly θ appears, whereas the difference in the true anomalies occurs on the right-hand side ofEquation 2.137. However, we can express the orbit equation in terms of the differencein true anomalies as follows. From Equation 2.134 we have θ=θ 0+/Delta1θ, which means we can write the orbit equation as r=h2 µ1 1+ecos(θ0+/Delta1θ)(2.138) By replacing θ0by−/Delta1θ in Equation 2.133, Equation 2.138 becomes r=h2 µ1 1+ecosθ0cos/Delta1θ−esinθ0sin/Delta1θ(2.139) To re m ove θ0from this expression, observe first of all that Equation 2.136 implies that, at t=t0, ecosθ0=h2 µr0−1 (2.140) Furthermore, from Equation 2.39 for the radial velocity we obtain esinθ0=hvr0 µ(2.141) Substituting Equations 2.140 and 2.141 into 2.139 yields r=h2 µ1 1+/parenleftbiggh2 µr0−1/parenrightbigg cos/Delta1θ−hvr0 µsin/Delta1θ(2.142) 82 Chapter 2 The two-body problem Using this form of the orbit equation, we can find rin terms of the initial conditions and the change in the true anomaly. Thus fin Equation 2.137 depends only on /Delta1θ. The Lagrange coefficient gis found by substituting Equations 2.108 and 2.130 into Equation 2.127b, g=−xy0+yx0 h =1 h[(−rcosθ)(r0sinθ0)+(rsinθ)(rcosθ0)] =rr0 h( sinθcosθ0−cosθsinθ0) (2.143) Making use of the trig identity sin(θ−θ0)=sinθcosθ0−cosθsinθ0 together with Equation 2.134, we find g=rr0 hsin(/Delta1θ) (2.144) T o obtain ˙g, substitute Equations 2.114 and 2.130 into Equation 2.128b, ˙g=−˙xy0+˙yx0 h =1 h/braceleftBig −/bracketleftBig −µ hsinθ/bracketrightBig [r0sinθ0]+/bracketleftBigµ h(e+cosθ)/bracketrightBig (r0cosθ0)/bracerightBig =µr0 h2[ecosθ0+(c o sθcosθ0+sinθsinθ0)] With the aid of Equations 2.133 and 2.140, this reduces to ˙g=1−µr0 h2(1−cos/Delta1θ) (2.145) ˙fcan be found using Equation 2.129. Thus ˙f=1 g(f˙g−1) (2.146) Substituting Equations 2.137, 2.143 and 2.145 results in ˙f=1 rr0 hsin/Delta1θ/braceleftBig/bracketleftBig 1−µr h2(1−cos/Delta1θ)/bracketrightBig/bracketleftBig 1−µr0 h2(1−cos/Delta1θ)/bracketrightBig −1/bracerightBig =1 rr0 hsin/Delta1θh2µrr0 h4/bracketleftbigg (1−cos/Delta1θ)2µ h2−(1−cos/Delta1θ)/parenleftbigg1 r0+1 r/parenrightbigg/bracketrightbigg or ˙f=µ h1−cos/Delta1θ sin/Delta1θ/bracketleftbiggµ h2(1−cos/Delta1θ)−1 r0−1 r/bracketrightbigg (2.147) 2.11 The Lagrange coefficients 83 T o summarize, the Lagrange coefficients in terms of the change in true anomaly are f=1−µr h2(1−cos/Delta1θ) (2.148a) g=rr0 hsin/Delta1θ (2.148b) ˙f=µ h1−cos/Delta1θ sin/Delta1θ/bracketleftbiggµ h2(1−cos/Delta1θ)−1 r0−1 r/bracketrightbigg (2.148c) ˙g=1−µr0 h2(1−cos/Delta1θ) (2.148d) where ris given by Equation 2.142. Observe that using the Lagrange coefficients to determine the position and veloc- ity from the initial conditions does not require knowing the type of orbit we aredealing with (ellipse, parabola, hyperbola), since the eccentricity does not appear inEquations 2.142 and 2.148. However, the initial position and velocity give us thatinformation. From r 0and v0we obtain the angular momentum h=|r0×v0|.T h e initial radius r0is just the magnitude of the vector r0. The initial radial velocity vr0is the projection of v0onto the direction of r0, vr0=v0·r0 r0 From Equations 2.35 and 2.39 we have r0=h2 µ1 1+ecosθ0vr0=µ hesinθ0 (2.149) These two equations can be solved for the eccentricity eand the true anomaly of the initial point θ0. Example 2.9An earth satellite moves in the xyplane of an inertial frame with origin at the earth’s center. Relative to that frame, the position and velocity of the satellite at time t0are r0=8182.4 ˆi−6865.9 ˆj(km) (a) v0=0.47572 ˆi+8.8116 ˆj(km/s) Compute the position and velocity vectors after the satellite has traveled through a true anomaly of 120◦. First, use r0and v0to calculate the angular momentum of the satellite: h=r0×v0=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆi ˆj ˆk 8182.4 −6865.90 0.47572 8.8116 0/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=75 366 ˆk(km 2/s) so that h=75 366 km2/s (b) 84 Chapter 2 The two-body problem (Example 2.9 continued) xy 120° r0v0rv ˆiˆ j C Figure 2.28 The initial and final position and velocity vectors. The magnitude of the position vector r0is r0=√r0·r0=10 861 km (c) The initial radial velocity vr0is found by projecting the velocity v0onto the unit vector in the radial direction r0, vr0=v0·r0 r0=(0.47572ˆi+8.8116ˆj)·(8182.4ˆi−6865.9ˆj) 10 681=− 5.2996 km /s( d ) The final distance ris obtained from Equation 2.142, r=h2 µ1 1+/parenleftbiggh2 µr0−1/parenrightbigg cos/Delta1θ−hvr0 µsin/Delta1θ =75 3662 398 6001 1+/parenleftbigg75 3662 398 600 ·10 681−1/parenrightbigg cos 120◦−75 366 ·(−5.2995) 398 600sin 120◦ so that r=8378.8k m ( e ) 2.11 The Lagrange coefficients 85 Now we can evaluate the Lagrange coefficients in Equations 2.148: f=1−µr h2(1−cos/Delta1θ) =1−398 600 ·8378.9 75 3662(1−cos 120◦)=0.11802 (dimensionless) (f) g=rr0 hsin(/Delta1θ ) =8378.9 ·10 681 75 366sin(120◦)=1028.4 s (g) ˙f=µ h1−cos/Delta1θ sin/Delta1θ/bracketleftbiggµ h2(1−cos/Delta1θ)−1 r0−1 r/bracketrightbigg =398 600 75 3661−cos 120◦ sin 120◦/bracketleftbigg398 600 75 3662(1−cos 120◦)−1 10 681−1 8378.9/bracketrightbigg (h) =− 9.8665 ×10−4(dimensionless) ˙g=1−µr0 h2(1−cos/Delta1θ) =1−398 600 ·10 681 75 3662(1−cos 120◦)=− 0.12432 (dimensionless) (i) At this point we have all that is required to find the final position and velocity vectors. From Equation 2.125 we have r=fr0+gv0 Substituting Equations (a), (f) and (g), we get r=0.11802(8182.4 ˆi−6865.9 ˆj)+1028.4(0.47572 ˆi+8.8116 ˆj) =1454.9 ˆi+8251.6 ˆj(km) Likewise, according to Equation 2.126, v=˙fr0+˙gv0 Substituting Equations (a), (h) and (i) yields v=(−9.8665 ×10−4)(8182.4 ˆi−6865.9 ˆj)+(−0.12435)(0.47572 ˆi+8.8116 ˆj) or v=− 8.1323 ˆi+5.6785 ˆj(km/s) In order to use the Lagrange coefficients to find the position and velocity as a function of time, we need to come up with a relation between /Delta1θand time. We will deal with that complex problem in the next chapter. Meanwhile, for times twhich are close to 86 Chapter 2 The two-body problem the initial time t0, we can obtain polynomial expressions for fand gin which the variable /Delta1θis replaced by the time interval /Delta1t=t−t0. T o do so, we expand the position vector r(t), considered to be a function of time, in a Taylor series about t=t0. By definition, the Taylor series is given by r(t)=∞/summationdisplay n=01 n!r(n)(t0)(t−t0)n(2.150) where r(n)(t0)i st h e nth time derivative of r(t), evaluated at t0, r(n)(t0)=/parenleftbiggdnr dtn/parenrightbigg t=t0(2.151) Let us truncate this infinite series at five terms. Then, to that degree of approximation, r(t)=r(t0)+/parenleftbiggdr dt/parenrightbigg t=t0/Delta1t+1 2/parenleftbiggd2r dt2/parenrightbigg t=t0/Delta1t2+1 6/parenleftbiggd3r dt3/parenrightbigg t=t0/Delta1t3 +1 24/parenleftbiggd4r dt4/parenrightbigg t=t0/Delta1t4(2.152) where /Delta1t=t−t0. T o evaluate the four derivatives, we note first that ( dr/dt)t=t0is just the velocity v0att=t0, /parenleftbiggdr dt/parenrightbigg t=t0=v0 (2.153) (d2r/dt2)t=t0is evaluated using Equation 2.15, ¨r=−µ r3r (2.154) Thus, /parenleftbiggd2r dt2/parenrightbigg t=t0=−µ r3 0r0 (2.155) (d3r/dt3)t=t0is evaluated by differentiating Equation 2.154, d3r dt3=−µd dt/parenleftBigr r3/parenrightBig =−µ/parenleftbiggr3v−3rr2˙r r6/parenrightbigg =−µv r3+3µ˙rr r4(2.156) From Equation 2.25a we have ˙r=r·v r(2.157) Hence, Equation 2.156, evaluated at t=t0,i s /parenleftbiggd3r dt3/parenrightbigg t=t0=−µv0 r3 0+3µr0·v0 r5 0r0 (2.158) 2.11 The Lagrange coefficients 87 Finally, ( d4r/dt4)t=t0is found by first differentiating Equation 2.156, d4r dt4=d dt/parenleftbigg −µ˙r r3+3µ˙rr r4/parenrightbigg =−µ/parenleftbiggr3¨r−3r2˙r˙r r6/parenrightbigg +3µ/bracketleftbiggr4(¨rr+˙r˙r)−4r3˙r2r r8/bracketrightbigg (2.159) ¨ris found in terms of rand vby differentiating Equation 2.157 and making use of Equation 2.154. This leads to the expression ¨r=d dt/parenleftbiggr·˙r r/parenrightbigg =v2 r−µ r2−(r·v)2 r3(2.160) Substituting Equations 2.154, 2.157 and 2.160 into Equation 2.159, combining terms and evaluating the result at t=t0yields /parenleftbiggd4r dt4/parenrightbigg t=t0=/bracketleftbigg −2µ2 r6 0+3µv2 0 r5 0−15µ(r0·v0)2 r7 0/bracketrightbigg r0+6µ(r0·v0) r5 0v0 (2.161) After substituting Equations 2.153, 2.155, 2.158 and 2.161 into Equation 2.152 and rearranging terms, we obtain r(t)=/braceleftbigg 1−µ 2r3 0/Delta1t2+µ 2r0·v0 r5 0/Delta1t3+µ 24/bracketleftbigg −2µ r6 0+3v2 0 r5 0−15(r0·v0)2 r7 0/bracketrightbigg /Delta1t4/bracerightbigg r0 +/bracketleftbigg /Delta1t−1 6µ r3 0/Delta1t3+µ 4(r0·v0) r5 0/Delta1t4/bracketrightbigg v0 (2.162) Comparing this expression with Equation 2.125, we see that, to the fourth order in/Delta1t, f=1−µ 2r3 0/Delta1t2+µ 2r0·v0 r5 0/Delta1t3+µ 24/bracketleftbigg −2µ r6 0+3v2 0 r5 0−15(r0·v0)2 r7 0/bracketrightbigg /Delta1t4 (2.163) g=/Delta1t−1 6µ r3 0/Delta1t3+µ 4(r0·v0) r5 0/Delta1t4 For small values of elapsed time /Delta1tthese fand gseries may be used to calculate the position of an orbiting body from the initial conditions. Example 2.10The orbit of an earth satellite has an eccentricity e=0.2 and a perigee radius of 7000 km. Starting at perigee, plot the radial distance as a function of time using the f and gseries and compare the curve with the exact solution. Since the satellite starts at perigee, t0=0 and we have, using the perifocal frame, r0=7000ˆp(km) (a) The orbit equation evaluated at perige e is Equation 2.40, which in the present case becomes 7000=h2 398 6001 1+0.2 88 Chapter 2 The two-body problem (Example 2.10 continued)Solving for the angular momentum, we get h=57 864 km2/s. Then, using the angular momentum formula, Equation 2.21, we find that the speed at perigee is v0=8.2663 km/s, so that v0=8.2663ˆq(km/s) (b) Clearly, r0·v0=0. Hence, with µ=398 600 km3/s2, the Lagrange series in Equation 2.163 become f=1−5.8105(10−7)t2+9.0032(10−14)t4 g=t−1.9368(10−7)t3 where the units of tare seconds. Substituting fand ginto Equation 2.125 yields r=[1−5.8105(10−7)t2+9.0032(10−14)t4](7000 ˆp)+[t−1.9368(10−7)t3](8.2663ˆq) From this we obtain r=/bardbl r/bardbl =/radicalbig 49(106)+11.389t2−1.103(10−6)t4−2.5633(10−12)t6+3.9718(10−19)t8 (c) For the exact solution of rversus time we must appeal to the methods presented in the next chapter. The exact solution and the series solution [Equation (c)] are plotted in Figure 2.29. As can be seen, the series solution begins to seriously diverge from the exact solution after about ten minutes. 180 360 540 720720074007600 7000 900 10 minr (km) t (sec)f and g seriesExact Figure 2.29 Exact and series solutions for the radial position of the satellite. If we include terms of fifth and higher order in the fandgseries, Equations 2.163, then the approximate solution in the above example will agree with the exact solution 2.12 Restricted three-body problem 89 for a longer time interval than that indicated in Figure 2.29. However, there is a time interval beyond which the series solution will diverge from the exact one no matterhow many terms we include. This time interval is called the radius of convergence. According to Bond and Allman (1996), for the elliptical orbit of Example 2.10, the radius of convergence is 1700 seconds (not quite half an hour), which is one-fifth of theperiod of that orbit. This further illustrates the fact that the series form of the Lagrangecoefficients is applicable only over small time intervals. For arbitrary time intervalsthe closed form of these functions, presented in Chapter 3, must be employed. 2.12 Restricted three-body problem Consider two bodies m1and m2moving under the action of just their mutual grav- itation, and let their orbit around each other be a circle of radius r12. Consider a non-inertial, co-moving frame of reference xyzwhose origin lies at the center of mass Gof the two-body system, with the xaxis directed towards m2, as shown in Figure 2.30. The yaxis lies in the orbital plane, to which the zaxis is perpendicular. In this frame of reference, m1and m2appear to be at rest. The constant, inertial angular velocity /Omega1is given by /Omega1=/Omega1ˆk (2.164) where /Omega1=2π T and Tis the period of the orbit (Equation 2.54), T=2πr123 2 √µ m1 xr r2Gz (x, y, z) ym Plane of motion of m 1 and m 2(x1, 0, 0) (x2, 0, 0)Co-moving xyz frame m2r12(0, 0, 0)r1 Figure 2.30 Primary bodies m1and m2in circular orbit around each other, plus a secondary mass m. 90 Chapter 2 The two-body problem Thus /Omega1=/radicalbiggµ r3 12(2.165) Recall that if Mis the total mass of the system, M=m1+m2 (2.166) then µ=GM (2.167) m1andm2lie in the orbital plane, so their yandzcoordinates are zero. T o determine their locations on the xaxis, we use the definition of the center of mass (Equation 2.1) to write m1x1+m2x2=0 Since m2is at a distance r12from m1in the positive xdirection, it is also true that x2=x1+r12 From these two equations we obtain x1=−π2r12 (2.168a) x2=π1r12 (2.168b) where the dimensionless mass ratios π1andπ2are given by π1=m1 m1+m2 π2=m2 m1+m2(2.169) We now introduce a third body of mass m, which is vanishingly small compared to the primary masses m1and m2– like the mass of a spacecraft compared to that of a planet or moon of the solar system. This is called the restricted three-body problem,because the mass mis assumed to be so small that it has no effect on the motion of the primary bodies. We are interested in the motion of mdue to the gravitational fields of m 1and m2. Unlike the two-body problem, there is no general, closed form solution for this motion. However, we can set up the equations of motion and drawsome general conclusions from them. In the co-moving coordinate system, the position vector of the secondary mass m relative to m 1is given by r1=(x−x1)ˆi+yˆj+zˆk=(x+π2r12)ˆi+yˆj+zˆk (2.170) Relative to m2the position of mis r2=(x−π1r12)ˆi+yˆj+zˆk (2.171) 2.12 Restricted three-body problem 91 Finally, the position vector of the secondar y body relative to the center of mass is r=xˆi+yˆj+zˆk (2.172) The inertial velocity of mis found by taking the time derivative of Equation 2.172. However, relative to inertial space, the xyzcoordinate system is rotating with the angular velocity /Omega1, so that the time derivatives of the unit vectors ˆiandˆjare not zero. T o account for the rotating frame, we use Equation 1.38 to obtain ˙r=vG+/Omega1×r+vrel (2.173) vGis the inertial velocity of the center of mass (the origin of the xyzframe), and vrel is the velocity of mas measured in the moving xyzframe, namely, vrel=˙xˆi+˙yˆj+˙zˆk (2.174) The absolute acceleration of mis found using the ‘five-term’ relative acceleration formula, Equation 1.42, ¨r=aG+˙/Omega1×r+/Omega1×(/Omega1×r)+2/Omega1×vrel+arel (2.175) Recall from Section 2.2 that the velocity vGof the center of mass is constant, so that aG=0. Furthermore, ˙/Omega1=0 since the angular velocity of the circular orbit is constant. Therefore, Equation 2.175 reduces to ¨r=/Omega1×(/Omega1×r)+2/Omega1×vrel+arel (2.176) where arel=¨xˆi+¨yˆj+¨zˆk (2.177) Substituting Equations 2.164, 2.172, 2.174 and 2.177 into Equation 2.176 yields ¨r=(/Omega1k)×/bracketleftBig (/Omega1ˆk)×(xˆi+yˆj+zˆk)/bracketrightBig +2(/Omega1ˆk)×(˙xˆi+˙yˆj+˙zˆk)+¨xˆi+¨yˆj+¨zˆk =/bracketleftBig −/Omega12(xˆi+yˆj)/bracketrightBig +(2/Omega1˙xˆj−2/Omega1˙yˆi)+¨xˆi+¨yˆj+¨zˆk Collecting terms, we find ¨r=(¨x−2/Omega1˙y−/Omega12x)ˆi+(¨y+2/Omega1˙x−/Omega12y)ˆj+¨zˆk (2.178) Now that we have an expression for the inertial acceleration in terms of quantities measured in the rotating frame, let us observe that Newton’s second law for thesecondary body is m¨r=F 1+F2 (2.179) F1and F2are the gravitational forces exerted on mbym1and m2, respectively. Recalling Equation 2.6, we have F1=−Gm 1m r2 1ur1=−µ1m r3 1r1 F2=−Gm 2m r2 2ur2=−µ2m r3 2r2(2.180) 92 Chapter 2 The two-body problem where µ1=Gm 1µ2=Gm 2 (2.181) Substituting Equations 2.180 into 2.179 and canceling out myields ¨r=−µ1 r3 1r1−µ2 r3 2r2 (2.182) Finally, we substitute Equation 2.178 on the left and Equations 2.170 and 2.171 on the right to obtain (¨x−2/Omega1˙y−/Omega12x)ˆi+(¨y+2/Omega1˙x−/Omega12y)ˆj+¨zˆk=−µ1 r3 1/bracketleftBig (x+π2r12)ˆi+yˆj+zˆk/bracketrightBig −µ2 r3 2/bracketleftBig (x−π1r12)ˆi+yˆj+zˆk/bracketrightBig Equating the coefficients of ˆi,ˆjandˆkon each side of this equation yields the three scalar equations of motion for the restricted three-body problem: ¨x−2/Omega1˙y−/Omega12x=−µ1 r3 1(x+π2r12)−µ2 r3 2(x−π1r12) (2.183a) ¨y+2/Omega1˙x−/Omega12y=−µ1 r3 1y−µ2 r3 2y (2.183b) ¨z=−µ1 r3 1z−µ2 r3 2z (2.183c) 2.12.1 Lagrange points Although Equations 2.183 have no closed form analytical solution, we can use them to determine the location of the equilibrium points. These are the locations in spacewhere the secondary mass mwould have zero velocity and zero acceleration, i.e., where mwould appear permanently at rest relative to m 1and m2(and therefore appear to an inertial observer to move in circular orbits around m1and m2). Once placed at an equilibrium point (also called librati on point or Lagrange point), a body will presumably stay there. The equilibrium points are therefore defined by the conditions ˙x=˙y=˙z=0 and ¨x=¨y=¨z=0 Substituting these conditions into Equations 2.183 yields −/Omega12x=−µ1 r3 1(x+π2r12)−µ2 r3 2(x−π1r12) (2.184a) −/Omega12y=−µ1 r3 1y−µ2 r3 2y (2.184b) 0=−µ1 r3 1z−µ2 r3 2z (2.184c) 2.12 Restricted three-body problem 93 From Equation 2.184c we have /parenleftbiggµ1 r3 1+µ2 r3 2/parenrightbigg z=0 (2.185) Sinceµ1/r3 1>0 and µ2/r3 2>0, it must therefore be true that z=0. That is, the equilibrium points lie in the orbital plane. From Equations 2.169 it is clear that π1=1−π2 (2.186) Using this, along with Equation 2.165, and assuming y/negationslash=0, we can write Equations 2.184a and 2.184b as (1−π2)(x+π2r12)1 r3 1+π2(x+π2r12−r12)1 r3 2=x r3 12(2.187) (1−π2)1 r3 1+π21 r3 2=1 r3 12 where we made use of the fact that π1=µ1/µ π 2=µ2/µ (2.188) Treating Equations 2.187 as two linear equations in 1 /r3 1and 1/r3 2, we solve them simultaneously to find that 1 r3 1=1 r3 2=1 r3 12 or r1=r2=r12 (2.189) Using this result, together with z=0 and Equation 2.186, we obtain from Equations 2.170 and 2.171, respectively, r2 12=(x+π2r12)2+y2(2.190) r2 12=(x+π2r12−r12)2+y2(2.191) Equating the right-hand sides of these two equations leads at once to the conclusion that x=r12 2−π2r12 (2.192) Substituting this result into Equation 2.190 or 2.191 and solving for yyields y=±√ 3 2r12 We have thus found two of the equilibrium points, the Lagrange points L4and L5. As Equation 2.189 shows, these points are the same distance r12from the primary 94 Chapter 2 The two-body problem bodies m1andm2that the primary bodies are from each other, and in the co-moving coordinate system their coordinates are L4,L5:x=r12 2−π2r12,y=±√ 3 2r12,z=0 (2.193) Therefore, the two primary bodies and these two Lagrange points lie at the vertices of equilateral triangles, as illustrated in Figure 2.32. The remaining equilibrium points are found by setting y=0a sw e l la s z=0, which satisfy both Equations 2.184b and 2.184c. For these values, Equations 2.170and 2.171 become r 1=(x+π2r12)ˆi r2=(x−π1r12)ˆi=(x+π2r12−r12)ˆi Therefore r1=|x+π2r12| r2=|x+π2r12−r12| Substituting these together with Equations 2.165, 2.186 and 2.188 into Equation 2.184a yields 1−π2 |x+π2r12|3(x+π2r12)+π2 |x+π2r12−r12|3(x+π2r12−r12)−1 r3 12x=0 (2.194) Further simplification is obtained by non-dimensionalizing x, ξ=x r12 In terms of ξ, Equation 2.194 becomes f(ξ)=0, where f(ξ)=1−π2 |ξ+π2|3(ξ+π2)+π2 |ξ+π2−1|3(ξ+π2−1)−ξ (2.195) The roots of f(ξ)=0 yields the other equilibrium points besides L4and L5.T o find them first requires specifying a value for the mass ratio π2, and then using a numerical technique to obtain the roots for that particular value. For example, let thetwo primary bodies m 1and m2be the earth and the moon, respectively. Then m1=5.974×1024kg m2=7.348×1022kg r12=3.844×105km(2.196) (from Table A.1) using this data, we find π2=m2 m1+m2=0.01215 Substituting this value of π2into Equation 2.195 and plotting the function yields the curves shown in Figure 2.31. By carefully determining where various branches of the 2.12 Restricted three-body problem 95 /H110021 /H110020.5 0.5 1 /H11002101Earth–moon center of mass L1L2 L3/H110021.0050.8369 1.156f(ξ) ξ Figure 2.31 Graph of Equation 2.195 for earth–moon data ( π2=0.01215), showing the three real roots. L2 326 400 kmL1 L3449 100 km381 600 km Earth MoonL4 L5Apse line60° 60°Moon's orbit relative to earth 384 400 km 384 400 km384 400 km384 400 km Figure 2.32 Location of the five Lagrange points of the earth–moon system. These points orbit the earth with the same period as the moon. curve cross the ξaxis, we find the real roots, which are the three additional Lagrange points for the earth–moon system, all lying on the apse line: L1:x=0.8369r 12=3.217×105km L2:x=1.156r 12=4.444×105km (2.197) L3:x=− 1.005r 12=− 3.863×105km The locations of the five Lagrange points for the earth–moon system are shown in Figure 2.32. For convenience, all of their positions are shown relative to the center of the earth, instead of the center of mass. As can be seen from Equation 2.168a, the 96 Chapter 2 The two-body problem center of mass of the earth–moon system is only 4670 km from the center of the earth. That is, it lies within the earth at 73 percent of its radius. Since the Lagrange pointsare fixed relative to the earth and moon, they follow circular orbits around the earthwith the same period as the moon. If an equilibrium point is stable, then a small mass occupying that point will tend to return to that point if nudged out of position. The perturbation results in a smalloscillation (orbit) about the equilibrium po int. Thus, objects can be placed in small orbits (called halo orbits) around stable equilibrium points without requiring much in the way of station keeping. On the other hand, if a body located at an unstableequilibrium point is only slightly perturbed, it will oscillate in a divergent fashion,drifting eventually completely away from that point. It turns out that the Lagrangepoints L 1,L2and L3on the apse line are unstable, whereas L4and L5–6 0◦ahead of and behind the moon in its orbit – are stable. However, L4and L5are destabilized by the influence of the sun’s gravity, so that in actuality station keeping would berequired to maintain position in the neighborhood of those points. Solar observation spacecraft have been placed in halo orbits around the L 1point of the sun–earth system. L1lies about 1.5 million kilometers from the earth (1/100 the distance to the sun) and well outside the earth’s magnetosphere. Three such missionswere the International Sun–Earth Explorer 3 (ISSE-3) launched in August 1978; theSolar and Heliocentric Observatory (SOHO) launched in December 1995; and the Advanced Composition Explorer (ACE) launched in August 1997. 2.12.2 Jacobi constant Multiply Equation 2.183a by ˙x, Equation 2.183b by ˙yand Equation 2.183c by ˙zto obtain ¨x˙x−2/Omega1˙x˙y−/Omega12x˙x=−µ1 r3 1(x˙x+π2r12˙x)−µ2 r3 2(x˙x−π1r12˙x) ¨y˙y+2/Omega1˙x˙y−/Omega12y˙y=−µ1 r3 1y˙y−µ2 r3 2y˙y ¨z˙z=−µ1 r3 1z˙z−µ2 r3 2z˙z Sum the left and right sides of these equations to get ¨x˙x+¨y˙y+¨z˙z−/Omega12(x˙x+y˙y)=−/parenleftbiggµ1 r3 1+µ2 r3 2/parenrightbigg/parenleftbig x˙x+y˙y+z˙z/parenrightbig +r12/parenleftbiggπ1µ2 r3 2−π2µ1 r3 1/parenrightbigg ˙x or, rearranging terms, ¨x˙x+¨y˙y+¨z˙z−/Omega12(x˙x+y˙y)=−µ1 r3 1(x˙x+y˙y+z˙z+π2r12˙x) −µ2 r3 2(x˙x+y˙y+z˙z−π1r12˙x) (2.198) Note that ¨x˙x+¨y˙y+¨z˙z=1 2d dt(˙x2+˙y2+˙z2)=1 2dv2 dt(2.199) 2.12 Restricted three-body problem 97 where vis the speed of the secondary mass relative to the rotating frame. Similarly, x˙x+y˙y=1 2d dt(x2+y2) (2.200) From Equation 2.170 we obtain r2 1=(x+π2r12)2+y2+z2 Therefore 2r1dr1 dt=2(x+π2r12)˙x+2y˙y+2z˙z or dr1 dt=1 r1(π2r12˙x+x˙x+y˙y+z˙z) It follows that d dt1 r1=−1 r2 1dr1 dt=−1 r3 1(x˙x+y˙y+z˙z+π2r12˙x) (2.201) In a similar fashion, starting with Equation 2.171, we find d dt1 r2=−1 r3 2(x˙x+y˙y+z˙z−π1r12˙x) (2.202) Substituting Equations 2.199, 2.200, 2.201 and 2.202 into Equation 2.198 yields 1 2dv2 dt−1 2/Omega12d dt(x2+y2)=µ1d dt1 r1+µ2d dt1 r2 Alternatively, upon rearranging terms d dt/bracketleftbigg1 2v2−1 2/Omega12(x2+y2)−µ1 r1−µ2 r2/bracketrightbigg =0 which means the bracketed expression is a constant 1 2v2−1 2/Omega12(x2+y2)−µ1 r1−µ2 r2=C (2.203) v2/2 is the kinetic energy per unit mass relative to the rotating frame. −µ 1/r1 and−µ 2/r2are the gravitational potential energies of the two primary masses. −/Omega12(x2+y2)/2 may be interpreted as the potential energy of the centrifugal force per unit mass /Omega12(xˆi+yˆj) induced by the rotation of the reference frame. The con- stant Cis known as the Jacobi constant, after the German mathematician Carl Jacobi (1804–1851), who discovered it in 1836. Jacobi’s constant may be interpreted as thetotal energy of the secondary particle relative to the rotating frame. Cis a constant of the motion of the secondary mass just like the energy and angular momentum areconstants of the relative motion in the two-body problem. Solving Equation 2.203 for v 2yields v2=/Omega12(x2+y2)+2µ1 r1+2µ2 r2+2C (2.204) 98 Chapter 2 The two-body problem If we restrict the motion of the secondary mass to lie in the plane of motion of the primary masses, then r1=/radicalBig (x+π2r12)2+y2r2=/radicalBig (x−π1r12)2+y2 (2.205) For a given value of the Jacobi constant, v2is a function only of position in the rotating frame. Since v2cannot be negative, it must be true that /Omega12(x2+y2)+2µ1 r1+2µ2 r2+2C≥0 (2.206) Trajectories of the secondary body in regions where this inequality is violated are not allowed. The boundaries between forbidden and allowed regions of motion are found by setting v2=0, i.e., /Omega12(x2+y2)+2µ1 r1+2µ2 r2+2C=0 (2.207) For a given value of the Jacobi constant the curves of zero velocity are determined by this equation. These boundaries cannot be crossed by a secondary mass (spacecraft)moving within an allowed region. Since the first three terms on the left of Equation 2.207 are all positive, it follows that the zero velocity curves correspond to negative values of the Jacobi constant.Large negative values of Cmean that the secondary body is far from the system center of mass ( x 2+y2is large) or that the body is close to one of the primary bodies ( r1is small or r2is small). Let us consider again the earth–moon system. From Equations 2.165, 2.166, 2.167, 2.181 and 2.196, together with Table A.2, we have µ1=µearth=398 600 km3/s2 µ2=µmoon=4903.02 km3/s2 (2.208) /Omega1=/radicalBigg µ1+µ2 r3 12=/radicalbigg 398 600 +4903 384 4003 =2.66538 ×10−6rad/s Substituting these values into Equation 2.207, we can plot the zero velocity curves for different values of Jacobi’s constant. The curves bound regions in which the motionof a spacecraft is not allowed. ForC=− 1.8k m 2/s2, the allowable regions are circles surrounding the earth and the moon, as shown in Figure 2.33(a). A spacecraft launched from the earth with thisvalue of Ccannot reach the moon, to say nothing of escaping the earth–moon system. Substituting the coordinates of the Lagrange points L 1,L2and L3into Equation 2.207, we obtain the successively larger values of the Jacobi constants C1,C2and C3 which are required to arrive at those points with zero velocity. These are shown along with the allowable regions in Figure 2.33. From part (c) of that figure we see that C2 represents the minimum energy for a spacecraft to escape the earth–moon system via a narrow corridor around the moon. Increasing Cwidens that corridor and at C3 escape becomes possible in the opposite direc tion from the moon. The last vestiges of 2.12 Restricted three-body problem 99 L1L2 L3L4 L5xEarth Moony (a) C 0 /H11005 /H110021.8 L1 L2 L3 EarthMoonL4 L5xy (b) C 1 /H11005 /H110021.6735 L1 L2L3L4 L5xEarthMoony (c) C 2 /H11005 /H110021.6649L1L2 L3L4 L5xEarth Moony (d) C 3 /H11005 /H110021.5810 L1L2 L3xEarth MoonL4 L5y (e) C 4 /H11005 /H110021.5683L1L2 L3xEarth MoonL4 L5y (f ) C 5 /H11005 /H110021.5600 Figure 2.33 Forbidden regions (shaded) within the earth–moon system for increasing values of Jacobi’s constant (km2/s2). 100 Chapter 2 The two-body problem the forbidden regions surround L4andL5. Further increase in Jacobi’s constant makes the entire earth–moon system and beyond accessible to an earth-launched spacecraft. For a given value of the Jacobi constant, the relative speed at any point within an allowable region can be found using Equation 2.204. Example 2.11A spacecraft has a burnout velocity vboat a point on the earth–moon line with an altitude of 200 km. Find the value of vbofor each of the scenarios depicted in Figure 2.33. From Equations 2.168 and 2.196 we have π1=m1 m1+m2=5.974×1024 6.047×1024=0.9878 π2=1−π1=0.1215 x1=−π1r12=− 0.9878·384 400 =− 4670.6k m Therefore, the coordinates of the burnout point are x=6578−4670.6=1907.3k m y=0 4671 km 6578 kmCOγ Moon ( m2)υboy x S Earth ( m1) Figure 2.34 Spacecraft Sburnout position and velocity relative to the rotating earth–moon frame. Substituting these values along with the Jacobi constant into Equations 2.204 and 2.205 yields the burnout velocity vbo. For the six Jacobi constants in Figure 2.33 we obtain C0:vbo=10.845 km /s C1:vbo=10.857 km /s C2:vbo=10.858 km /s C3:vbo=10.866 km /s C4:vbo=10.867 km /s C5:vbo=10.868 km /s Problems 101 These velocities are not substantially differ ent from the escape velocity (Equation 2.81) at 200 km altitude, vesc=/radicalbigg 2µ r=/radicalbigg 2·398 600 6578=11.01 km /s It is remarkable that a change in vboon the order of only 10 m/s or less can have a significant influence on the regions of earth–moon space accessible to the spacecraft. Problems For man-made earth satellites use µ=398 600 km2/s2.RE=6378 km (Tables A.1 and A.2). 2.1 Ifr, in meters, is given by r=3t4ˆI+2t3ˆJ+9t2ˆK,w h e r e tis time in seconds, calculate ˙r(where r=/bardblr/bardbl) and /bardbl˙r/bardblatt=2s . {Ans.: ˙r=101.3 m/s, /bardbl˙r/bardbl=105.3 m/s} 2.2 Show that, in general, if ˆur=r/r, then ˆur·dˆur/dt=0. 2.3 Two particles of identical mass mare acted on only by the gravitational force of one upon the other. If the distance dbetween the particles is constant, what is the angular velocity of the line joining them?{Ans.: ω=/radicalbig 2Gm/d3} 2.4 Three particles of identical mass mare acted on only by their mutual gravitational attraction. They are located at the vertices of an equilateral triangle with sides of lengthd. Consider the motion of any one of the particles about the system center of mass and use Newton’s second law to determine the angular velocity ωrequired for dto remain constant.{Ans.: ω=/radicalbig 3Gm/d3} 2.5 A satellite is in a circular, 350 km orbit (i.e., it is 350 km above the earth’s surface). Calculate (a) the speed in km/s; (b) the period. {Ans.: (a) 7.697 km/s; (b) 91 min 32 s} 2.6 A spacecraft is in a circular orbit of the moon at an altitude of 80 km. Calculate its speed and its period.{Ans.: 1.642 km/s; 1 hr 56 min} 2.7 It is desired to place a satellite in earth polar orbit such that successive ground tracks at the equator are spaced 3000 km apart. Determine the required altitude of the circularorbit.{Ans.: 1440 km} 2.8 Find the minimum additional speed required to escape from GEO. {Ans.: 1.274 km/s} 2.9 What velocity, relative to the earth, is required to escape the solar system on a parabolic path from the earth’s orbit? {12.34 km/s} 2.10 Calculate the area Aswept out during the time t=T/3 since periapsis, where Tis the period of the elliptical orbit.{Ans.: 1.047ab } 102 Chapter 2 The two-body problem b aF PA Figure P .2.10 2.11 Show that v=µ h√ 1+2ecosθ+e2for any orbit. 2.12 Determine the true anomaly θof the point(s) on an elliptical orbit at which the speed equals the speed of a circular orbit with the same radius, i.e., vellipse=vcircle. {Ans.: θ=cos−1(−e), where eis the eccentricity of the ellipse} θ F' Fυcircleυellipse r Figure P .2.12 2.13 Calculate the flight path angle at the locations found in Exercise 2.12./braceleftBig Ans.:γ=tan−1/parenleftBig e/√ 1−e2/parenrightBig/bracerightBig 2.14 An unmanned satellite orbits the earth with a perigee radius of 7000 km and an apogee radius of 70 000 km. Calculate (a) the eccentricity of the orbit; (b) the semimajor axis of the orbit (km); (c) the period of the orbit (hours); (d) the specific energy of the orbit (km2/s2); (e) the true anomaly at which the altitude is 1000 km (degrees); Problems 103 (f)vrandv⊥at the points found in part (e) (km/s); (g) the speed at perigee and apogee (km/s). {Partial ans.: (c) 20.88 hr; (e) 27.61◦; (g) 10.18 km/s, 1.018 km/s} 2.15 A spacecraft is in a 250 km by 300 km low earth orbit. How long (in minutes) does it t a k et ofl yf r o mp e r i g e et oa p o g e e ?{Ans.: 45.00 min} 2.16 The altitude of a satellite in an elliptical orbit around the earth is 1600 km at apogee and 600 km at perigee. Determine (a) the eccentricity of the orbit; (b) the orbital speeds at perigee and apogee; (c) the period of the orbit. {Ans.: (a) 0.06686; (b) v P=7.81 km/s; (c) vA=6.83 km /s; (d) T=107.2 min} 2.17 A satellite is placed into an earth orbit at perigee at an altitude of 1270 km with a speed of 9 km/s. Calculate the flight path angle γand the altitude of the satellite at a true anomaly of 100◦. {Ans.: γ=31.1◦;z=6774 km} 2.18 A satellite is launched into earth orbit at an altitude of 640 km with a speed of 9.2 km/s and a flight path angle of 10◦. Calculate the true anomaly of the launch point and the period of the orbit.{Ans.: θ=29.8 ◦;T=4.46 hr} 2.19 A satellite has perigee and apogee altitudes of 250 km and 42 000 km. Calculate the orbit period, eccentricity, and the maximum speed.{Ans.: 12 hr 36 min, 0.759, 10.3 km/s} 2.20 A satellite is launched parallel to the earth’s surface with a speed of 8 km/s at an altitude of 640 km. Calculate the apogee altitude and the period.{Ans.: 2679 km, 1 hr 59 min 30 s} 2.21 A satellite in orbit around the earth has a perigee velocity of 8 km/s. Its period is 2 hours. Calculate its altitude at perigee.{Ans.: 648 km} 2.22 A satellite in polar orbit around the earth comes within 150 km of the North Pole at its point of closest approach. If the satellite passes over the pole once every 90 minutes,calculate the eccentricity of its orbit.{Ans.: 0.0187} 2.23 A hyperbolic earth departure trajectory has a perigee altitude of 300 km and a perigee speed of 15 km/s. (a) Calculate the hyperbolic excess speed (km/s); (b) Find the radius (km) when the true anomaly is 100 ◦; {Ans.: 48 497 km} (c) Find vrandv⊥(km/s) when the true anomaly is 100◦. 2.24 A meteoroid is first observed approaching the earth when it is 402 000 km from the center of the earth with a true anomaly of 150◦. If the speed of the meteoroid at that time is 2.23 km/s, calculate (a) the eccentricity of the trajectory; (b) the altitude at closest approach; (c) the speed at closest approach. {Ans.: (a) 1.086; (b) 5088 km; (c) 8.516 km/s} 2.25 Calculate the radius rat which the speed on a hyperbolic trajectory is 1.1 times the hyperbolic excess speed. Express your result in terms of the periapse radius rpand the eccentricity e. {Ans.: r=9.524r p/(e−1)} 104 Chapter 2 The two-body problem 150 Earth402 000 km2.23 km/s Figure P .2.24 2.26 A hyperbolic trajectory has an eccentricity e=3.0 and an angular momentum h=105 000 km2/s. Without using the energy equation, calculate the hyperbolic excess speed.{Ans.: 10.7 km/s} 2.27 The following position data for an earth orbiter is given: Altitude =1700 km at a true anomaly of 130 ◦. Altitude =500 km at a true anomaly of 50◦. Calculate (a) the eccentricity; (b) the perigee altitude (km); (c) the semimajor axis (km). {Ans.: (c) 7547 km} 2.28 An earth satellite has a speed of 7 km/s and a flight path angle of 15◦when its radius is 9000 km. Calculate (a) the true anomaly (degrees); (b) the eccentricity of the orbit. {Ans.: (a) 83.35◦; (b) 0.2785} 2.29 If, for an earth satellite, the specific angular momentum is 60 000 km2/s and the specific energy is −20 km2/s2, calculate the apogee and perigee altitudes. {Ans.: 6637 km and 537.2 km} 2.30 A rocket launched from the surface of the earth has a speed of 8.85 km/s when powered flight ends at an altitude of 550 km. The flight path angle at this time is 6◦. Determine (a) the eccentricity of the trajectory; (b) the period of the orbit. {Ans.: (a) e=0.3742; (b) T=187.4 min} 2.31 A space vehicle has a velocity of 10 km/s in the direction shown when it is 10 000 km from the center of the earth. Calculate its true anomaly.{Ans.: 51 ◦} Problems 105 120° Earth10 km/s Apse line10 000 kmradius Figure P .2.31 2.32 A space vehicle has a velocity of 10 km/s and a flight path angle of 20◦when it is 15 000 km from the center of the earth. Calculate its true anomaly.{Ans.: 27.5 ◦} 2.33 For a spacecraft trajectory around the earth, r=10 000 km when θ=30◦, and r=30 000 km when θ=105◦. Calculate the eccentricity. {Ans.: 1.22} 2.34 A spacecraft in a 500 km altitude circular orbit is given a delta-v equal to one-half its orbital speed. Use the energy equation to calculate the hyperbolic excess velocity.{Ans.: 3.806 km/s} 2.35 A satellite is in a circular orbit at an altitude of 320 km above the earth’s surface. If an onboard rocket provides a delta-v of 500 m/s in the direction of the satellite’s motion,calculate the altitude of the new orbit’s apogee.{Ans.: 2390 km} 2.36 A spacecraft is in a circular orbit of radius rand speed varound an unspecified planet. A rocket on the spacecraft is fired, instantaneously increasing the speed in the direction of motion by the amount /Delta1v=α,w h e r e α>0. Calculate the eccentricity of the new orbit.{Ans.: e=α(α+2)} 2.37 A satellite is in a circular earth orbit of altitude 400 km. Determine the new perigee and apogee altitudes if the satellite on-board engine (a) increases the speed of the satellite in the flight direction by 240 m/s; (b) gives the satellite a radial (outward) component of velocity of 240 m/s. {Ans.: (a) z A=1230 km, zP=400 km; (b) zA=621 km, zP=196 km} 2.38 For the sun–earth system, find the distance of the L1,L2and L3Lagrange points from the center of mass of the sun–earth system.{Ans.: x 1=151.101 ×106km, x2=148.108 ×106km, x3=− 149.600 ×106km (oppo- site side of the sun)} This page int entionally left blank 3Chapter Orbital position as a function of time Chapter outline 3.1 Introduction 107 3.2 Time since periapsis 108 3.3 Circular orbits 108 3.4 Elliptical orbits 109 3.5 Parabolic trajectories 124 3.6 Hyperbolic trajectories 125 3.7 Universal variables 134 Problems 145 3.1 Introduction In Chapter 2 we found the relationship between position and true anomaly for the two-body problem. The only place time appeared explicitly was in the expression for the period of an ellipse. Obtaining position as a function of time is a simplematter for circular orbits. For elliptical, parabolic and hyperbolic paths we are led tothe various forms of Kepler’s equation relating position to time. These transcendentalequations must be solved iteratively using a procedure like Newton’s method, whichis presented and illustrated in the chapter. The different forms of Kepler’s equation are combined into a single universal Kepler’s equation by introducing univer sal variables. Implementation of this appeal- ing notion is accompanied by the introduction of an unfamiliar class of functionsknown as Stumpff functions. The universal variable formulation is required for theLambert and Gauss orbit determination algorithms in Chapter 5. 107 108 Chapter 3 Orbital position as a function of time The road map of Appendix B may aid in grasping how the material presented here depends on that of Chapter 2. 3.2 Time since periapsis The orbit formula, r=(h2/µ)/(1+ecosθ), gives the position of body m2in its orbit around m1as a function of the true anomaly. For many practical reasons we need to be able to determine the position of m2as a function of time. For elliptical orbits, we have a formula for the period T(Equation 2.72), but we cannot yet calculate the time required to fly between any two true anomalies. The purpose of this section isto come up with the formulas that allow us to do that calculation. The one equation we have which relates tr ue anomaly directly to time is Equation 2.37, h=r 2˙θ, which can be written dθ dt=h r2 Substituting r=(h2/µ)/(1+ecosθ), we find, after separating variables, µ2 h3dt=dθ (1+ecosθ)2 Integrating both sides of this equation yields µ2 h3(t−tp)=/integraldisplayθ 0dϑ (1+ecosϑ)2(3.1) in which the constant of integration tpis the time at periapse passage, where by definition θ=0.tpis the sixth constant of the motion that was missing in Chapter 2. The origin of time is arbitrary. It is convenient to measure time from periapse passage,so we will usually set t p=0. In that case we have µ2 h3t=/integraldisplayθ 0dϑ (1+ecosϑ)2(3.2) The integral on the right may be found in any standard mathematical handbook. See, for example, Beyer (1991), integrals 341, 366 and 372. The specific form of theintegral depends on whether the value of the eccentricity ecorresponds to a circle, ellipse, parabola or hyperbola. 3.3 Circular orbits For a circle, e=0, so the integral in Equation 3.2 is simply/integraltextθ 0dϑ.T h u sw eh a v e t=h3 µ2θ 3.4 Elliptical orbits 109 Pt /H11005 0 Apse lineur C, FD Figure 3.1 Time since periapsis is directly proportional to true anomaly in a circular orbit. Recall that for a circle (Equation 2.52), r=h2/µ. Therefore h3=r3 2µ3 2, so that t=r3 2 √µθ Finally, substituting the formula (Equation 2.54) for the period Tof a circular orbit, T=2πr3 2/√µ, yields t=θ 2πT or θ=2π Tt The reason that tis directly proportional to θin a circular orbit is simply that the angular velocity 2 π/Tis constant. Therefore the time /Delta1tto fly through a true anomaly of/Delta1θis (/Delta1θ/2π )T. Because the circle is symmetric about any diameter, the apse line – and therefore the periapsis – can be chosen arbitrarily. 3.4 Elliptical orbits For 0<e<1, we find in integral tables that /integraldisplayθ 0dϑ (1+ecosϑ)2=1 (1−e2)3 2/bracketleftBigg 2 tan−1/parenleftBigg/radicalbigg 1−e 1+etanθ 2/parenrightBigg −e√ 1−e2sinθ 1+ecosθ/bracketrightBigg Therefore, Equation 3.2 in this case becomes µ2 h3t=1 (1−e2)3 2/bracketleftBigg 2 tan−1/parenleftBigg/radicalbigg 1−e 1+etanθ 2/parenrightBigg −e√ 1−e2sinθ 1+ecosθ/bracketrightBigg 110 Chapter 3 Orbital position as a function of time 2π 0 2πe /H11005 0 e /H11005 0.2 e /H11005 0.5 e /H11005 0.8 e /H11005 0.9 True anomaly, uMean anomaly, M e π π Figure 3.2 Mean anomaly versus true anomaly for ellipses of various eccentricities. or Me=2 tan−1/parenleftBigg/radicalbigg 1−e 1+etanθ 2/parenrightBigg −e√ 1−e2sinθ 1+ecosθ(3.3) where Me=µ2 h3(1−e2)3 2t (3.4) Meis called the mean anomaly. Equation 3.3 is plotted in Figure 3.2. Observe that for all values of the eccentricity e,Meis a monotonically increasing function of the true anomaly θ. From Equation 2.72, the formula for the period Tof an elliptical orbit, we haveµ2(1−e2)3 2/h3=2π/T, so that the mean anomaly can be written much more simply as Me=2π Tt (3.5) The angular velocity of the position vecto r of an elliptical orbit is not constant, but since 2 πradians are swept out per period T, the ratio 2 π/Tis the average angular velocity, which is given the symbol nand called the mean motion, n=2π T(3.6) In terms of the mean motion, Equation 3.5 can be written simpler still, Me=nt 3.4 Elliptical orbits 111 F PES r Ab aB DaeuQ O aV Figure 3.3 Ellipse and the circumscribed auxiliary circle. The mean anomaly is the azimuth position (in radians) of a fictitious body moving around the ellipse at the constant angular speed n. For a circular orbit, the mean anomaly Meand the true anomaly θare identical. It is convenient to simplify Equation 3.3 by introducing an auxiliary angle Ecalled the eccentric anomaly, which is shown in Figure 3.3. This is done by circumscribingthe ellipse with a concentric auxiliary circle having a radius equal to the semimajoraxis aof the ellipse. Let Sbe that point on the ellipse whose true anomaly is θ. Through point Swe pass a perpendicular to the apse line, intersecting the auxiliary circle at point Qand the apse line at point V. The angle between the apse line and the radius drawn from the center of the circle to Qon its circumference is the eccentric anomaly E. Observe that Elagsθfrom PtoA, whereas it leads θfrom AtoP. T o find Eas a function of θ, we first observe from Figure 3.3 that, in terms of the eccentric anomaly, OV=acosEwhereas in terms of the true anomaly, OV=ae+rcosθ.T h u s acosE=ae+rcosθ Using Equation 2.62, r=a(1−e2)/(1+ecosθ), we can write this as acosE=ae+a(1−e2)c o sθ 1+ecosθ Simplifying the right-hand side, we get cosE=e+cosθ 1+ecosθ(3.7a) 112 Chapter 3 Orbital position as a function of time π π 23π 2π 2EIVE EI0cos E1 /H110021 Figure 3.4 For 0<cosE<1,Ecan lie in the first or fourth quadrant. For −1<cosE<0,Ecan lie in the second or third quadrant. Solving this for cos θwe obtain the inverse relation, cosθ=e−cosE ecosE−1(3.7b) Substituting Equation 3.7a into the trigonometric identity sin2E+cos2E=1 and solving for sin Eyields sinE=√ 1−e2sinθ 1+ecosθ(3.8) Equation 3.7a would be fine for obtaining Efromθ, except that, given a value of cosEbetween −1 and 1, there are twovalues of Ebetween 0◦and 360◦, as illustrated in Figure 3.4. The same comments hold for Equation 3.8. T o resolve this quadrantambiguity, we use the following trigonometric identity tan 2E 2=1−cosE 1+cosE(3.9) From Equation 3.7a 1−cosE=1−cosθ 1+ecosθ(1−e) and 1 +cosE=1+cosθ 1+ecosθ(1+e) Therefore, tan2E 2=1−e 1+e1−cosθ 1+cosθ=1−e 1+etan2θ 2 where the last step required applying the trig identity in Equation 3.9 to the term (1−cosθ)/(1+cosθ). Finally, therefore, we obtain tanE 2=/radicalbigg 1−e 1+etanθ 2(3.10a) or E=2 tan−1/parenleftBigg/radicalbigg 1−e 1+etanθ 2/parenrightBigg (3.10b) 3.4 Elliptical orbits 113 tanE 2 E 2π 0 π Figure 3.5 T o any value of tan(E /2) there corresponds a unique value of Ein the range 0 to 2 π. 02π 2π Eccentric anomaly, EMean anomaly, M e e = 0e = 0.2e = 0.4e = 0.6e = 0.8e /H11005 1.0 ππ Figure 3.6 Plot of Kepler’s equation for an elliptical orbit. Observe from Figure 3.5 that for any value of tan( E/2), there is only one value of E between 0◦and 360◦. There is no quadrant ambiguity. Substituting Equations 3.8 and 3.10b into Eq uation 3.3 yields Kepler’s equation, Me=E−esinE (3.11) This monotonically increasing relationship between mean anomaly and eccentric anomaly is plotted for several values of eccentricity in Figure 3.6. 114 Chapter 3 Orbital position as a function of time Root xf xi+1 f(xi)xi Slope /H11005df dxx/H11005xi   Figure 3.7 Newton’s method for finding a root of f(x)=0. Given the true anomaly θ, we calculate the eccentric anomaly Eusing Equations 3.10. Substituting Einto Kepler’s formula, Equation 3.11, yields the mean anomaly directly. From the mean anomaly and the period Twe find the time (since periapsis) from Equation 3.5, t=Me 2πT (3.12) On the other hand, if we are given the time, then Equation 3.12 yields the mean anomaly Me. Substituting Meinto Kepler’s equation we get the following expression for the eccentric anomaly, E−esinE=Me We cannot solve this transcendental equation directly for E. A rough value of Emight be read off Figure 3.6. However, an accurate solution requires an iterative, ‘trial anderror’ procedure. Newton’s method, or one of its variants, is one of the more common and efficient ways of finding the root of a well-behaved function. T o find a root of the equationf(x)=0 in Figure 3.7, we estimate it to be x i, and evaluate the function f(x) and its first derivative f/prime(x) at that point. We then extend the tangent to the curve at f(xi) until it intersects the xaxis at xi+1, which becomes our updated estimate of the root. The intercept xi+1is found by setting the slope of the tangent line equal to the slope of the curve at xi, that is, f/prime(xi)=0−f(xi) xi+1−xi from which we obtain xi+1=xi−f(xi) f/prime(xi)(3.13) The process is repeated, using xi+1to estimate xi+2, and so on, until the root has been found to the desired level of precision. 3.4 Elliptical orbits 115 T o apply Newton’s method to the solution of Kepler’s equation, we form the function f(E)=E−esinE−Me and seek the value of eccentric anomaly that makes f(E)=0. Since f/prime(E)=1−ecosE for this problem Equation 3.13 becomes Ei+1=Ei−Ei−esinEi−Me 1−ecosEi(3.14) Algorithm 3.1Solve Kepler’s equation for the eccentric anomaly Egiven the eccentricity eand the mean anomaly Me. See Appendix D.2 for the implementation of this algorithm in MATLAB®. 1. Choose an initial estimate of the root Eas follows (Prussing and Conway, 1993). IfMe<π, then E=Me+e/2. If Me>π, then E=Me−e/2. Remember that the angles Eand Meare in radians. (When using a hand-held calculator, be sure it is in radian mode.) 2. At any given step, having obtained Eifrom the previous step, calculate f(Ei)=Ei−esinEi−Meand f/prime(Ei)=1−ecosEi. 3. Calculate ratio i=f(Ei)/f/prime(Ei). 4. If|ratio i|exceeds the chosen tolerance (e.g., 10−8), then calculate an updated value of E Ei+1=Ei−ratio i Return to step 2. 5. If|ratio i|is less than the tolerance, then accept Eias the solution to within the chosen accuracy. Example 3.1A geocentric elliptical orbit has a perigee radius of 9600 km and an apogee radius of 21 000 km. Calculate the time to fly from perigee Pto a true anomaly of 120◦. The eccentricity is readily obtained from the perigee and apogee radii by means of Equation 2.74, e=ra−rp ra+rp=21 000 −9600 21 000 +9600=0.37255 (a) We find the angular momentum using the orbit equation, 9600=h2 398 6001 1+0.37255 cos(0)⇒ h=72 472 km2/s With hand e, the period of the orbit is obtained from Equation 2.72, T=2π µ2/parenleftbiggh√ 1−e2/parenrightbigg3 =2π 398 6002/parenleftbigg72 472√ 1−0.372552/parenrightbigg3 =18 834 s (b) 116 Chapter 3 Orbital position as a function of time 120/H11034 Earth 21 000 km 9600 kmP AB Figure 3.8 Geocentric elliptical orbit. Equation 3.10a yields the eccentric anomaly from the true anomaly, tanE 2=/radicalbigg 1−e 1+etanθ 2=/radicalbigg 1−0.37255 1+0.37255tan120◦ 2=1.1711 ⇒ E=1.7281 rad Then Kepler’s equation, Equatio n 3.11, is used to find the mean anomaly, Me=1.7281−0.37255 sin 1 .7281=1.3601 rad Finally, the time follows from Equation 3.12, t=Me 2πT=1.3601 2π18 834 =4077 s (1.132 hr)(Example 3.1 continued) Example 3.2In the previous example, find the true anomaly at three hours after perigee passage. Since the time (10 800 seconds) is greater than one-half the period, the true anomaly must be greater than 180◦. First, we use Equation 3.12 to calculate the mean anomaly for t=10 800 s: Me=2πt T=2π10 800 18 830=3.6029 rad (a) Kepler’s equation, E−esin(E)=Me(with all angles in radians), is then employed to find the eccentric anomaly. This transcendental equation will be solved using Algorithm 3.1 with an error tolerance of 10−6. Since Me>π, a good starting value for the iteration is E0=Me−e/2=3.4166. Executing the algorithm yields the following steps: Step 1: E0=3.4166 f(E0)=− 0.085124 3.4 Elliptical orbits 117 f/prime(E0)=1.3585 ratio=− 0.062658 |ratio|>10−6, so repeat. Step 2: E1=3.4166 −(−0.062658) =3.4793 f(E1)=− 0.0002134 f/prime(E1)=1.3515 ratio=− 1.5778 ×10−4 |ratio|>10−6, so repeat. Step 3: E2=3.4793 −(−1.5778 ×10−4)=3.4794 f(E2)=− 1.5366 ×10−9 f/prime(E2)=1.3515 ratio=− 1.137×10−9 |f(E2)|<10−6, so accept E=3.4794 as the solution. Convergence to even more than the desired accuracy occurred after just two iter- ations. With this value of the eccentric anomaly, the true anomaly is found from Equation 3.10a tanθ 2=/radicalbigg 1+e 1−etanE 2=/radicalbigg 1+0.37255 1−0.37255tan3.4794 2=− 8.6721 ⇒θ=193.2◦ Example 3.3Let a satellite be in a 500 km by 5000 km orbit with its apse line parallel to the line from the earth to the sun, as shown below. Find the time that the satellite is in theearth’s shadow if: (a) the apogee is towards the sun; (b) the perigee is towards the sun. (a) If the apogee is towards the sun, as in Figure 3.9, then the satellite is in earth’s shadow between points aand bon its orbit. These are two of the four points of intersection of the orbit with lines parallel to the earth–sun line which are a distance REfrom the center of the earth. The true anomaly of bis therefore given by sin θ=RE/r,w h e r e ris the radial position of the satellite. It follows that the radius of bis r=RE sinθ(a) From Equation 2.62 we also have r=a(1−e2) 1+ecosθ(b) 118 Chapter 3 Orbital position as a function of time (Example 3.3 continued) P ATo the sun ab c dr EarthREu Figure 3.9 Satellite passing in and out of the earth’s shadow. Equating (a) and (b), collecting terms and simplifying yields an equation in θ, ecosθ−(1−e2)a REsinθ+1=0( c ) From the data given in the problem statement, we obtain e=ra−rp ra+rp=(6378 +5000) −(6378 +500) (6378 +5000) +(6378 +500)=0.24649 (d) a=rp+ra 2=(6378 +500)+(6378 +5000) 2=9128 km (e) T=2π√µa3 2=2π√ 398 600(9128)3 2=8679.1s( 2.4109 hr) (f) Substituting (d) and (e) together with RE=6378 km into (c) yields 0.24649 cos θ−1.3442 sin θ=− 1( g ) This equation is of the form acosθ+bsinθ=c (h) It has two roots, which are given by (see Problem 3.9) θ=tan−1b a±cos−1/bracketleftbiggc acos/parenleftbigg tan−1b a/parenrightbigg/bracketrightbigg (i) For the case at hand, θ=tan−1−1.3442 0.24649±cos−1/bracketleftbigg−1 0.24649cos/parenleftbigg tan−1−1.3442 0.24649/parenrightbigg/bracketrightbigg =− 79.607◦±137.03◦ That is θb=57.423◦ θc=− 216.64◦(+143.36◦)(j) 3.4 Elliptical orbits 119 For apogee towards the sun, the flight from perigee to point bwill be in shadow. T o find the time of flight from perigee to point b, we first compute the eccentric anomaly of busing Equation 3.10b: Eb=2 tan−1/parenleftBigg/radicalbigg 1−e 1+etanθb 2/parenrightBigg =2 tan−1/parenleftBigg/radicalbigg 1−0.24649 1+0.24649tan1.0022 2/parenrightBigg =0.80521 rad (k) From this we find the mean anomaly using Kepler’s equation, Me=E−esinE=0.80521 −0.24649 sin 0 .80521 =0.62749 rad (l) Finally, Equation (3.5) yields the time at b, tb=Me 2πT=0.62749 2π8679.1 =866.77 s (m) The total time in shadow, from atob, during which the satellite passes through perigee, is t=2tb=1734 s (28.98 min) (n) (b) If the perigee is towards the sun, then the satellite is in shadow near apogee, from point c(θc=143.36◦)t od on the orbit. Following the same procedure as above we obtain (see Problem 3.12), Ec=2.3364 rad Mc=2.1587 rad (o) tc=2981.8s The total time in shadow, from ctod,i s t=T−2tc=8679.1 −2·2891.8 =2716 s (45.26 min) (p) The time is longer than that given by (n) since the satellite travels slower near apogee. We have observed that there is no closed for m solution for the eccentric anomaly E in Kepler’s equation, E−esinE=Me. However, there exist infinite series solutions. One of these, due to Lagrange (Battin, 1999), is a power series in the eccentricity e, E=Me+∞/summationdisplay n=1anen(3.15) where the coefficients anare given by the somewhat intimidating expression an=1 2n−1floor (n/2)/summationdisplay k=0(−1)k 1 (n−k)!k!(n−2k)n−1sin[(n −2k)M] (3.16) 120 Chapter 3 Orbital position as a function of time 2π Eccentric anomaly, E2π e /H11005 0.65 Exact and N /H11005 10 N /H11005 3 0Mean anomaly, M e π π Figure 3.10 Comparison of the exact solution of Kepler’ s equation with the truncated Lagrange series solution ( N=3 and N=10) for an eccentricity of 0.65. Here, floor (x) means rounded to the next lowest integer [e.g., floor (0.5)=0, floor (π)=3]. If eis sufficiently small, then the Lagrange series converges. That means by including enough terms in the summation, we can obtain Eto any desired degree of precision. Unfortunately, if eexceeds 0.662743419, the series diverges, which means taking more and more terms yields worse and worse results for some values of M. The limiting value for the eccentricity was discovered by the French mathemati- cian Pierre-Simon Laplace (1749–1827) and is called the Laplace limit. In practice, we must truncate the Lagrange series to a finite number of terms N, so that E=Me+N/summationdisplay n=1anen(3.17) For example, setting N=3 and calculating each anby means of Equation 3.16 leads to E=Me+esinMe+e2 2sin 2 Me+e3 8(3 sin 3 Me−sinMe) (3.18) For small values of the eccentricity ethis yields good agreement with the exact solu- tion of Kepler’s equation (plotted in Figure 3.6). However, as we approach the Laplacelimit, the accuracy degrades unless more terms of the series are included. Figure 3.10shows that for an eccentricity of 0.65, just below the Laplace limit, Equation 3.18(N=3) yields a solution which oscillates ar ound the exact solution, but is fairly close to it everywhere. Setting N=10 in Equation 3.17 produces a curve which, at the given scale, is indistinguishable from the exact solution. On the other hand, for aneccentricity of 0.90, far above the Laplace limit, Figure 3.11 reveals that Equation 3.18 3.4 Elliptical orbits 121 π 2π Eccentric anomaly, E2π 0e /H11005 0.9 πMean anomaly, M e ExactN /H11005 10 N /H11005 3 Figure 3.11 Comparison of the exact solution of Kepler’s equation with the truncated Lagrange series solution ( N=3 and N=10) for an eccentricity of 0.90. is a poor approximation to the exact solution, and using N=10 makes matters even worse. Another infinite series for E(Battin, 1999) is given by E=Me+∞/summationdisplay n=12 nJn(ne) sin nM e (3.19) where the coefficients Jnare functions due to the German astronomer and math- ematician Friedrich Bessel (1784–1846). These Bessel functions of the first kind are defined by the infinite series Jn(x)=∞/summationdisplay k=0(−1)k k!(n+k)!/parenleftBigx 2/parenrightBign+2k (3.20) J1through J5are plotted in Figure 3.12. Clearly, they are oscillatory in appearance and tend towards zero with increasing x. It turns out that, unlike the Lagrange series, the Bessel function series solution converges for all values of the eccentricity less than 1. Figure 3.13 shows how thetruncated Bessel series solution E=M e+N/summationdisplay n=12 nJn(ne) sin nM e (3.21) forN=3 and N=10 compares to the exact solution of Kepler’s equation for the very large elliptical eccentricity of e=0.99. It can be seen that the case N=3 yields a poor 122 Chapter 3 Orbital position as a function of time xJn(x)J1 J2 J3J4 J50.5 0.40.2 /H110020.2 /H110020.4 0 5 10 150 Figure 3.12 Bessel functions of the first kind. Eccentric anomaly, Ee /H11005 0.99Mean anomaly, Me2π π 0 π 2πN = 10N /H11005 3Exact Figure 3.13 Comparison of the exact solution of Kepler’s e quation with the truncated Bessel series solution (N=3 and N=10) for an eccentricity of 0.99. approximation for all but a few values of Me. Increasing the number of terms in the series to N=10 obviously improves the approximation, and adding even more terms will make the truncated series solution indistinguishable from the exact solution at the given scale. 3.4 Elliptical orbits 123 Observe that we can combine Equations 3.7 and 2.62 as follows to obtain the orbit equation for the ellipse in terms of the eccentric anomaly: r=a(1−e2) 1+ecosθ=a(1−e2) 1+e/parenleftbigge−cosE ecosE−1/parenrightbigg From this it is easy to see that r=a(1−ecosE) (3.22) In Equation 2.76 we defined the true-anomaly-averaged radius ¯rθof an elliptical orbit. Alternatively, the time-averaged radius ¯rtof an elliptical orbit is defined as ¯rt=1 T/integraldisplayT 0rd t (3.23) According to Equations 3.11 and 3.12, t=T 2π(E−esinE) Therefore, dt=T 2π(1−ecosE)dE Upon using this relationship to change the variable of integration from ttoEand substituting Equation 3.22, Equation 3.23 becomes ¯rt=1 T/integraldisplay2π 0[a(1−ecosE)]/bracketleftbiggT 2π(1−ecosE)/bracketrightbigg dE =a 2π/integraldisplay2π 0(1−ecosE)2dE =a 2π/integraldisplay2π 0(1−2ecosE+e2cos2E)dE =a 2π(2π−0+e2π) so that ¯rt=a/parenleftbigg 1+e2 2/parenrightbigg Time-averaged radius of an elliptical orbit. (3.24) Comparing this result with Equation 2.77 reveals, as we should have expected (Why?), that¯rt>¯rθ. In fact, combining Equations 2.77 and 3.24 yields ¯rθ=a/radicalbigg 3−2¯rt a(3.25) 124 Chapter 3 Orbital position as a function of time 3.5 Parabolic trajectories For the parabola ( e=1), Equation 3.2 becomes µ2 h3t=/integraldisplayθ 0dϑ (1+cosϑ)2(3.26) In integral tables we find that /integraldisplayθ 0dϑ (1+cosϑ)2=1 2tanθ 2+1 6tan3θ 2 Therefore, Equation 3.26 may be written as Mp=1 2tanθ 2+1 6tan3θ 2(3.27) where Mp=µ2t h3(3.28) Mpis dimensionless, and it may be thought of as the ‘mean anomaly’ for the parabola. Equation 3.27 is plotted in Figure 3.14. Equation 3.27 is also known as Barker’sequation. Given the true anomaly θ, we find the time directly from Equations 3.27 and 3.28. If time is the given variable, then we must solve the cubic equation 1 6/parenleftbigg tanθ 2/parenrightbigg3 +1 2tanθ 2−Mp=0Mean anomaly, M p 2π 2πππ True anomaly, u Figure 3.14 Graph of Equation 3.27. 3.6 Hyperbolic trajectories 125 which has but one real root, namely, tanθ 2=/bracketleftbigg 3M p+/radicalBig (3M p)2+1/bracketrightbigg1 3 −/bracketleftbigg (3M p+/radicalBig (3M p)2+1)/bracketrightbigg−1 3 (3.29) Example 3.4A geocentric parabola has a perigee velocity of 10 km/s. How far is the satellite from the center of the earth six hours after perigee passage? Using Equation 2.80, we find the perigee radius, rp=2µ v2 p=2·398 600 102=7972 km so that the angular momentum is h=rpvp=7972·10=79 720 km2/s Now we can calculate the parabolic mean anomaly using Equation 3.28, Mp=µ2t h3=398 6002·(6·3600) 79 7203=6.7737 rad so that 3M p=20.321 rad. Equation 3.29 yields the true anomaly, tanθ 2=/bracketleftBig 20.321 +/radicalbig 20.3212+1/bracketrightBig1 3−/bracketleftBig (20.321 +/radicalbig 20.3212+1)/bracketrightBig−1 3 =3.1481 ⇒θ=144.75◦ Finally, we substitute the true anomaly into the orbit equation to find the radius, r=79 7202 398 6001 1+cos(144.75◦)=86 899 km 3.6 Hyperbolic trajectories For the hyperbola ( e>1), integral tables reveal /integraldisplayθ 0dϑ (1+ecosϑ)2 =1 e2−1/bracketleftbiggesinθ 1+ecosθ−1√ e2−1ln/parenleftbigg√e+1+√e−1 tan(θ/ 2)√e+1−√e−1 tan(θ/ 2)/parenrightbigg/bracketrightbigg so that Equation 3.1 becomes µ2 h3t=1 e2−1esinθ 1+ecosθ−1 (e2−1)3 2ln/parenleftbigg√e+1+√e−1 tan(θ/ 2)√e+1−√e−1 tan(θ/ 2)/parenrightbigg Multiplying both sides by (e2−1)3 2,w eg e t Mh=e√ e2−1 sinθ 1+ecosθ−ln/parenleftbigg√e+1+√e−1 tan(θ/ 2)√e+1−√e−1 tan(θ/ 2)/parenrightbigg (3.30) 126 Chapter 3 Orbital position as a function of time 0.01110010 000 e /H11005 1.1e /H11005 1.5e /H11005 2.0e /H11005 3.0e /H11005 5.0 True anomaly, /H9258Mean anomaly, M h π 2π Figure 3.15 Plots of Equation 3.30 for several different eccentricities. where Mh=µ2 h3(e2−1)3 2t (3.31) Mhis the hyperbolic mean anomaly. Equation 3.30 is plotted in Figure 3.15. Recall that|θ|<cos−1(−1/e). We can simplify Equation 3.30 by introducing an auxiliary angle analogous to the eccentric anomaly Efor the ellipse. Consider a point on a hyperbola whose polar coordinates are randθ. Referring to Figure 3.16, let xbe the distance of the point from the center Cof the hyperbola, and let ybe its distance above the apse line. The ratio y/bdefines the hyperbolic sine of the dimensionless variable Fthat we will use as the hyperbolic eccentric anomaly. That is, we define Fto be such that sinh F=y b(3.32) In view of the equation of a hyperbola x2 a2−y2 b2=1 it is consistent with the definition of sinh Fto define the hyperbolic cosine as cosh F=x a(3.33) (It should be recalled that sinh x=(ex−e−x)/2 and cosh x=(ex+e−x)/2 and, therefore, that cosh2x−sinh2x=1.) 3.6 Hyperbolic trajectories 127 FocusP rp aApse lineCbMx yAsymptote θ Figure 3.16 Hyperbola parameters. From Figure 3.16 we see that y=rsinθ. Substituting this into Equation 3.32, along with r=a(e2−1)/(1+ecosθ) (Equation 2.94) and b=a√ e2−1 (Equation 2.96), we get sinh F=1 brsinθ=1 a√ e2−1a(e2−1) 1+ecosθsinθ so that sinh F=√ e2−1 sinθ 1+ecosθ(3.34) This can be used to solve for Fin terms of the true anomaly, F=sinh−1/parenleftBigg√ e2−1 sinθ 1+ecosθ/parenrightBigg (3.35) Using the formula sinh−1x=ln/parenleftbig x+√ x2+1/parenrightbig , we can, after simplifying the algebra, write Equation 3.35 as F=ln/parenleftBigg sinθ√ e2−1+cosθ+e 1+ecosθ/parenrightBigg Substituting the trigonometric identities sinθ=2 tan(θ/ 2) 1+tan2(θ/2)cosθ=1−tan2(θ/2) 1+tan2(θ/2) 128 Chapter 3 Orbital position as a function of time 0.01110010 000 2 4 1 3 6 5 Eccentric anomaly, Fe /H11005 1.1e /H11005 1.5e /H11005 2.0e /H11005 3.0e /H11005 5.0Mean anomaly, M h Figure 3.17 Plot of Kepler’s equation for the hyperbola. and doing some more algebra yields F=ln/bracketleftBigg 1+e+(e−1) tan2(θ/2)+2 tan(θ/2)√ e2−1 1+e+(1−e) tan2(θ/2)/bracketrightBigg Fortunately, but not too obviously, the numerator and the denominator in the brack- ets have a common factor, so that this expression for the hyperbolic eccentric anomalyreduces to F=ln/bracketleftbigg√ e+1+√e−1 tan(θ/2)√e+1−√e−1 tan(θ/2)/bracketrightbigg (3.36) Substituting Equations 3.34 and 3.36 into Equat ion 3.30 yields Kepler’s equation for the hyperbola, Mh=esinh F−F (3.37) This equation is plotted for several different eccentricities in Figure 3.17. If we substitute the expression for sinh F, Equation 3.34, into the hyperbolic trig identity cosh2F−sinh2F=1, we get cosh2F=1+/parenleftBigg√ e2−1 sinθ 1+ecosθ/parenrightBigg2 A few steps of algebra lead to cosh2F=/parenleftbiggcosθ+e 1+ecosθ/parenrightbigg2 so that cosh F=cosθ+e 1+ecosθ(3.38a) 3.6 Hyperbolic trajectories 129 Solving this for cos θ, we obtain the inverse relation, cosθ=cosh F−e 1−ecosh F(3.38b) The hyperbolic tangent is found in terms of the hyperbolic sine and cosine by the formula tanh F=sinh F cosh F In mathematical handbooks we can find the hyperbolic trig identity, tanhF 2=sinh F 1+cosh F(3.39) Substituting Equations 3.34 and 3.38a into this formula and simplifying yields tanhF 2=/radicalbigg e−1 e+1sinθ 1+cosθ(3.40) Interestingly enough, Equation 3.39 holds for ordinary trig functions, too; that is, tanθ 2=sinθ 1+cosθ Therefore, Equation 3.40 can be written tanhF 2=/radicalbigg e−1 e+1tanθ 2(3.41a) This is a somewhat simpler alternative to E quation 3.36 for computing eccentric anomaly from true anomaly, and it is a whole lot simpler to invert: tanθ 2=/radicalbigg e+1 e−1tanhF 2(3.41b) If time is the given quantity, then Equation 3.3 7–at r anscendental equation – must be solved for Fby an iterative procedure, as was the case for the ellipse. T o apply Newton’s procedure to the solution of Kepler’s equation for the hyperbola, we formthe function f(F)=esinh F−F−M h and seek the value of Fthat makes f(F)=0. Since f/prime(F)=ecosh F−1 Equation 3.13 becomes Fi+1=Fi−esinh Fi−Fi−Mh ecosh Fi−1(3.42) All quantities in this formula are dimensionless (radians, not degrees). 130 Chapter 3 Orbital position as a function of time Algorithm 3.2Solve Kepler’s equation for the hyperbola for the hyperbolic eccentric anomaly F given the eccentricity eand the hyperbolic mean anomaly Mh. See Appendix D.3 for the implementation of this algorithm in MATLAB. 1. Choose an initial estimate of the root F. (a) For hand computations read a rough value of F0(no more than two significant figures) from Figure 3.17 in order to keep the number of iterations to aminimum. (b) In computer software let F0=Mh, an inelegant choice which may result in many iterations but will nevertheless rapidly converge on today’s high speeddesktop and laptop computers. 2. At any given step, having obtained Fifrom the previous step, calculate f(Fi)=esinh Fi−Fi−Mhand f/prime(Fi)=ecosh Fi−1. 3. Calculate ratio i=f(Fi)/f/prime(Fi). 4. If|ratio i|exceeds the chosen tolerance (e.g., 10−8), then calculate an updated value of F, Fi+1=Fi−ratio i Return to step 2. 5. If|ratio i|is less than the tolerance, then accept Fias the solution to within the desired accuracy. Example 3.5A geocentric trajectory has a perigee velocity of 15 km/s and a perigee altitude of 300 km. Find (a) the radius when the true anomaly is 100◦and (b) the position and speed three hours later. (a) The angular momentum is calculated from the given perigee data: h=rpvp=(6378 +300)·15=100 170 km2/s The eccentricity is found by evaluating the orbit equation, r=(h2/µ) [1/(1+ecosθ)], at perigee: 6378+300=100 1702 398 6001 1+e⇒ e=2.7696 (a) Since e>1 the trajectory is a hyperbola. Note that the true anomaly of the asymptote of the hyperbola is, from Equation 2.87, θ∞=cos−1/parenleftbigg −1 2.7696/parenrightbigg =111.17◦ Solving the orbit equation at θ=100◦yields r=100 1702 398 6001 1+2.7696 cos 100◦=48 497 km 3.6 Hyperbolic trajectories 131 (b) The time since perigee passage at θ=100◦must be found next so that we can add the three hour time increment needed to find the final position of the satellite.Using Equation 3.41a to calculate the hyperbolic eccentric anomaly, we find tanhF 2=/radicalbigg 2.7696 −1 2.7696 +1tan100◦ 2=0.81653 ⇒ F=2.2927 rad Kepler’s equation for the hyperbola then yields the mean anomaly, Mh=esinh F−F=2.7696 sinh 2 .2927−2.2927 =11.279 rad Now we can obtain the time since perigee passage by means of Equation 3.31, t=h3 µ21 (e2−1)3 2Mh=100 1703 398 60021 (2.76962−1)3 211.279 =4141 s Three hours later the time since perigee passage is t=4141.4 +3·3600=14 941 s (4 .15 hr) The corresponding mean anomaly, from Equation 3.31, is Mh=398 6002 100 1703(2.76962−1)3 214 941 =40.690 rad (b) We will use Algorithm 3.2 with an error tolerance of 10−6to find the hyperbolic eccen- tric anomaly F. Referring to Figure 3.17, we see that for Mh=40.69 and e=2.7696, Flies between 3 and 4. Let us arbitrarily choose F0=3 as our initial estimate of F. Executing the algorithm yields the following steps: F0=3 Step 1: f(F0)=− 15.944494 f/prime(F0)=26.883397 ratio=− 0.59309818 F1=3−(−0.59309818) =3.5930982 |ratio|>10−6, so repeat. Step 2: f(F1)=6.0114484 f/prime(F1)=49.370747 ratio=− 0.12176134 F2=3.5930982 −(−0.12176134) =3.4713368 |ratio|>10−6, so repeat. 132 Chapter 3 Orbital position as a function of time (Example 3.5 continued)Step 3: f(F2)=0.35812370 f/prime(F2)=43.605527 ratio=8.2128052 ×10−3 F3=3.4713368 −(8.2128052 ×10−3)=3.4631240 |ratio|>10−6, so repeat. Step 4: f(F3)=1.4973128 ×10−3 f/prime(F3)=43.241398 ratio=3.4626836 ×10−5 F4=3.4631240 −(3.4626836 ×10−5)=3.4630894 |ratio|>10−6, so repeat. Step 5: f(F4)=2.6470781 ×10−3 f/prime(F4)=43.239869 ratio=6.1218459 ×10−10 F5=3.4630894 −(6.1218459 ×10−10)=3.4630894 |ratio|<10−6, so accept F=3.4631 as the solution. We substitute this value of Finto Equation 3.41b to find the true anomaly, tanθ 2=/radicalbigg e+1 e−1tanhF 2=/radicalbigg 2.7696+1 2.7696−1tanh3.4631 2=1.3708 ⇒θ=107.78◦ With the true anomaly, the orbital equation yields the radial coordinate at the final time r=h2 µ1 1+ecosθ=100 1702 398 6001 1+2.7696 cos 107 .78=163 180 km The velocity components are obtained from Equation 2.21, v⊥=h r=100 170 163 180=0.61386 km /s and Equation 2.39, vr=µ hesinθ=398 600 100 1702.7696 sin 107 .78◦=10.494 km /s 3.6 Hyperbolic trajectories 133 100°107.78°Initial positionPosition three hours later 163 180 km 48 497 km PerigeeApse lineθ∞ /H11005 117.1° Figure 3.18 Given and computed data for Example 3.5. Therefore, the speed of the spacecraft is v=/radicalBig v2r+v2 ⊥=/radicalbig 10.4942+0.613862=10.51 km /s Note that the hyperbolic excess speed for this orbit is v∞=µ hesinθ∞=398 600 100 170·2.7696 ·sin 111.7◦=10.277 km /s The results of this analysis are shown in Figure 3.18. When determining orbital position as a function of time with the aid of Kepler’s equation, it is convenient to have position ras a function of eccentric anomaly F. This is obtained by substituting Equation 3.38b into Equation 2.94, r=a(e2−1) 1+ecosθ=a(e2−1) 1+e/parenleftbiggcosh F−e 1−ecosF/parenrightbigg This reduces to r=a(ecosh F−1) (3.43) 134 Chapter 3 Orbital position as a function of time 3.7 Universal variables The equations for elliptical and hyperbolic trajectories are very similar, as can be seen from Table 3.1. Observe, for example, that the hyperbolic mean anomaly is obtainedfrom that of the ellipse as follows: M h=µ2 h3(e2−1)3 2t =µ2 h3/bracketleftbig (−1)(1−e2)/bracketrightbig3 2t =µ2 h3(−1)3 2(1−e2)3 2t =µ2 h3(−i)(1−e2)3 2t =− i/bracketleftbiggµ2 h3(1−e2)3 2t/bracketrightbigg =− iMe In fact, the formulas for the hyperbola can all be obtained from those of the ellipse by replacing the variables in the ellipse equations according to the following scheme,wherein ‘ ←’ means ‘replace by’: a←− a b←ib M e←− iMh E←iF(i=√ −1) Note in this regard that sin( iF)=isinh Fand cos( iF)=cosh F. Relations among the circular and hyperbolic trig functions are found in mathematics handbooks, such asBeyer (1991). In the universal variable approach, the semimajor axis of the hyperbola is con- sidered to have a negative value, so that the energy equation (row 5 of Table 3.1) has the same form for any type of orbit, including the parabola, for which a=∞ . In this formulation, the semimajor axis of any orbit is found using (row 3), a=h 2 µ1 1−e2(3.44) If the position rand velocity vare known at a given point on the path, then the energy equation (row 5) is convenient for finding the semimajor axis of any orbit, a=1 2 r−v2 µ(3.45) Kepler’s equation may also be written in terms of a universal variable, or universal ‘anomaly’ χ, that is valid for all orbits. See, for example, Battin (1999), Bond and Allman (1993) and Prussing and Conway (1993). If t0is the time when the universal 3.7 Universal variables 135 T able 3.1 Comparison of some of the orbital formulas for the ellipse and hyperbola Equation Ellipse ( e<1) Hyperbola (e >1) 1. Orbit equation (2.35) r=h2 µ1 1+ecosθsame 2. Conic equation in cartesianx2 a2+y2 b2=1x2 a2−y2 b2=1 coordinates (2.69), (2.99) 3. Semimajor axis (2.61), (2.93) a=h2 µ1 1−e2a=h2 µ1 e2−1 4. Semiminor axis (2.66), (2.96) b=a√ 1−e2 b=a√ e2−1 5. Energy equation (2.71), (2.101)v2 2−µ r=−µ 2av2 2−µ r=µ 2a 6. Mean anomaly (3.4), (3.31) Me=µ2 h3(1−e2)3 2tM h=µ2 h3(e2−1)3 2t 7. Kepler’s equation (3.11), (3.37) Me=E−esinEM h=esinh F−F 8. Orbit equation in terms of eccentric r=a(1−ecosE) r=a(ecosh F−1) anomaly (3.22), (3.43) variable is zero, then the value of χat time t0+/Delta1tis found by iterative solution of the universal Kepler’s equation √µ/Delta1t=r0vr0√µχ2C(αχ2)+(1−αr0)χ3S(αχ2)+r0χ (3.46) in which r0andvr0are the radius and radial velocity at t=t0, andαis the reciprocal of the semimajor axis α=1 a(3.47) α<0,α=0 andα>0 for hyperbolas, parabolas and ellipses, respectively. The units ofχare km1 2(soαχ2is dimensionless). The functions C(z) and S(z)b e l o n gt ot h e class known as Stumpff functions, and they are defined by the infinite series, S(z)=∞/summationdisplay k=0(−1)k zk (2k+3)!=1 6−z 120+z2 5040−z3 362 880+z4 39 916 800 −z5 6 227 020 800+··· (3.48a) C(z)=∞/summationdisplay k=0(−1)k zk (2k+2)!=1 2−z 24+z2 720−z3 40 320+z4 3 628 800 −z5 479 001 600+ ··· (3.48b) 136 Chapter 3 Orbital position as a function of time /H1100230/H1100220/H1100210246810 10 200.10.20.30.4 100 200 300 4000.010.020.03 012 0.5 /H11002500 /H11002400 0 30 0 5000.04 0 zz zS(z)C(z) S(z)C(z) C(z) S(z) Figure 3.19 A plot of the Stumpff functions C(z) and S(z). C(z) and S(z) are related to the circular and hyperbolic trig functions as follows: S(z)=  √ z−sin√z (√z)3(z>0) sinh√−z−√−z (√−z)3(z<0 ) 1 6(z=0)(z=αχ2) (3.49) C(z)=  1−cos√ z z(z>0) cosh√−z−1 −z(z<0 ) 1 2(z=0)(z=αχ2) (3.50) Clearly, z<0,z=0 and z>0 for hyperbolas, parabolas and ellipses, respectively. It should be pointed out that if C(z) and S(z) are computed by the series expansions, Equations 3.48a and 3.48b, then the forms of C(z) and S(z), depending on the sign of z, are selected, so to speak, automatically. C(z) and S(z)b e h a v ea ss h o w n in Figure 3.19. Both C(z) and S(z) are non-negative functions of z. They increase without bound as zapproaches −∞ and tend towards zero for large positive values ofz. As can be seen from Equation 3.50 1, for z>0C(z)=0w h e nc o s√z=1, that is, when z=(2π)2,( 4π)2,( 6π)2,… . The price we pay for using the universal variable formulation is having to deal with the relatively unknown Stumpff functions. However, Equations 3.49 and 3.50are easy to implement in both computer programs and programmable calculators.See Appendix D.4 for the implementation of these expressions in MATLAB. T o gain some insight into how Equation 3.46 represents the Kepler equations for all of the conic sections, let t 0be the time at periapse passage and let us set t0=0, as we have assumed previously. Then /Delta1t=t,vr0=0 and r0equals rp, the periapse radius. In that case Equation 3.46 reduces to √µt=(1−αrp)χ3S(αχ2)+rpχ(t=0 at periapse passage) (3.51) 3.7 Universal variables 137 Consider first the parabola. In that case α=0 and S=S(0)=1/6, so that Equation 3.51 becomes a cubic polynomial in χ, √µt=1 6χ3+rpχ Multiply this equation through by (√µ/h)3to obtain µ2 h3t=1 6/parenleftbiggχ√µ h/parenrightbigg3 +rpχ/parenleftbigg√µ h/parenrightbigg3 Since rp=h2/2µ for a parabola, we can write this as µ2 h3t=1 6/parenleftbigg√µ hχ/parenrightbigg3 +1 2/parenleftbigg√µ hχ/parenrightbigg (3.52) Upon setting χ=htan(θ/ 2)/√µ, Equation 3.52 becomes identical to Equation 3.27, the time versus true anomaly relation for the parabola. Kepler’s equation for the ellipse can be obtained by multiplying Equation 3.51 through by/parenleftBig/radicalbig µ(1−e2)/h/parenrightBig3 : µ2 h3/parenleftbig 1−e2/parenrightbig3 2t=/parenleftbigg χ√µ h/radicalbig 1−e2/parenrightbigg3 (1−αrp)S(z) +rpχ/parenleftbigg√µ h/radicalbig 1−e2/parenrightbigg3 (z=αχ2) (3.53) Recall that for the ellipse, rp=h2/[µ(1 +e)] and α=1/a=µ(1−e2)/h2. Using these two expressions in Equation 3.53, along with S(z)=/bracketleftbig√αχ−sin (√αχ)/bracketrightbig/slashbig α3 2χ3 (from Equation 3.49 1), and working through the algebra ultimately leads to Me=χ√a−esin/parenleftbiggχ√a/parenrightbigg Comparing this with Kepler’s equation for an ellipse (Equation 3.11) reveals that the relationship between the universal variable χand the eccentric anomaly Eisχ=√aE. Similarly, it can be shown for hyperbolic orbits that χ=√−aF . In summary, the universal anomaly χis related to the previously encountered anomalies as follows: χ=  h √µtanθ 2parabola √aE ellipse (t 0=0, at periapsis) √−aF hyperbola(3.54) When t0is the time at a point other than per iapsis, so that Equation 3.46 applies, then Equations 3.54 become χ=  h √µ/parenleftbigg tanθ 2−tanθ0 2/parenrightbigg parabola √a(E−E0) ellipse √−a(F−F0) hyperbola(3.55) 138 Chapter 3 Orbital position as a function of time As before, we can use Newton’s method to solve Equation 3.46 for the universal anomaly χ, given the time interval /Delta1t. T o do so, we form the function f(χ)=r0vr0√µχ2C(z)+(1−αr0)χ3S(z)+r0χ−√µ/Delta1t (3.56) and its derivative df(χ) dχ=2r0vr0√µχC(z)+r0vr0√µχ2dC(z) dzdz dχ +3(1−αr0)χ2S(z)+(1−r0α)χ3dS(z) dzdz dχ+r0 (3.57) where it is to be recalled that z=αχ2(3.58) which means of course that dz dχ=2αχ (3.59) It turns out that dS(z) dz=1 2z[C(z)−3S(z)] (3.60) dC(z) dz=1 2z[1−zS(z)−2C(z)] Substituting Equations 3.58, 3.59 and 3.60 into Equation 3.57 and simplifying the result yields df(χ) dχ=r0vr0√µχ[1−αχ2S(z)]+(1−αr0)χ2C(z)+r0 (3.61) With Equations 3.56 and 3.61, Newton’s algorithm (Equation 3.13) for the universal Kepler equation becomes χi+1=χi−r0vr0√µχ2 iC(zi)+(1−αr0)χ3 iS(zi)+r0χi−√µ/Delta1t r0vr0√µχi[1−αχ2 iS(zi)]+(1−αr0)χ2 iC(zi)+r0(zi=αχ2 i) (3.62) According to Chobotov (2002), a reasonable estimate for the starting value χ0is χ0=√µ|α|/Delta1t (3.63) Algorithm 3.3Solve the universal Kepler’s equation for the universal anomaly χgiven/Delta1t,r0,vr0 andα. See Appendix D.5 for an implementation of this procedure in MATLAB. 1. Use Equation 3.63 for an initial estimate of χ0. 2. At any given step, having obtained χifrom the previous step, calculate f(χi)=r0vr0√µχ2 iC(zi)+(1−αr0)χ3 iS(zi)+r0χi−√µ/Delta1t 3.7 Universal variables 139 and f/prime(χi)=r0vr0√µχi[1−αχ2 iS(zi)]+(1−αr0)χ2 iC(zi)+r0 where zi=αχ2 i. 3. Calculate ratio i=f(χi)/f/prime(χi). 4. If|ratio i|exceeds the chosen tolerance (e.g., 10−8), then calculate an updated value of χ, χi+1=χi−ratio i Return to step 2. 5. If|ratio i|is less than the tolerance, then accept χias the solution to within the desired accuracy. Example 3.6An earth satellite has an initial true anomaly of θ0=30◦, a radius of r0=10 000 km, a n das p e e do f v0=10 km/s. Use the universal Kepler’s equation to find the change in universal anomaly χafter one hour and use that information to determine the true anomaly θat that time. Using the initial conditions, let us first determine the angular momentum and the eccentricity of the trajectory. From the orbit formula, Equation 2.35, we have h=/radicalbig µr0(1+ecosθ0)=/radicalbig 398 600 ·10 000 ·(1+ecos 30◦) =63 135√ 1+0.86602e (a) This, together with the angular momentum formula, Equation 2.21, yields v⊥0=h r0=63 135√1+0.86602e 10 000=6.3135√ 1+0.86602e Using the radial velocity relation, Equation 2.39, we find vr0=µ hesinθ0=398 600 63 135√1+0.86602eesin 30◦=3.1567e√1+0.86602e Sincev2 r0+v2 ⊥0=v2 0, it follows that /parenleftbigg 3.1567e√1+0.86602e/parenrightbigg2 +/parenleftBig 6.3135√ 1+0.86602e/parenrightBig2 =102 which simplifies to become 39 .86e2−17.563e −60.14=0. The only positive root of this quadratic equation is e=1.4682 Substituting this value of the eccentricity back into (a) yields the angular momentum h=95 154 km2/s 140 Chapter 3 Orbital position as a function of time (Example 3.6 continued)The hyperbolic eccentric anomaly F0for the initial conditions may now be found from Equation 3.41a, tanhF0 2=/radicalbigg e−1 e+1tanθ0 2=/radicalbigg 1.4682−1 1.4682+1tan30◦ 2=0.16670 Solving for F0yields F0=0.23448 rad (b) The initial radial speed (required in Equation 3.46) is obtained from Equation 2.39, vr0=µ hesinθ0=398 600 95 154·1.4682·sin 30◦=3.0752 km /s( c ) We calculate the semimajor axis of the orbit by means of Equation 3.44, a=h2 µ1 1−e2=95 1542 398 6001 1−1.46822=− 19 655 km The fact that the semimajor axis is negative means the orbit is a hyperbola. Equation 3.47 implies that α=1 a=1 −19 655=− 5.0878×10−5km−1(d) We will use Algorithm 3.3 with an error tolerance of 10−6to find the universal anomaly. From Equation 3.63, our initial estimate is χ0=√ 398 600 ·| −5.0878×10−6|·3600=115.6 Executing the algorithm yields the following steps: χ0=115.6 Step 1: f(χ0)=− 370 650 .01 f/prime(χ0)=26 956 .300 ratio=− 13.750033 χ1=115.6−(−13.750033) =129.35003 |ratio|>10−6, so repeat. Step 2: f(χ1)=25 729 .002 f/prime(χ1)=30 776 .401 ratio=0.83599669 χ2=129.35003 −0.83599669 =128.51404 |ratio|>10−6, so repeat. 3.7 Universal variables 141 Step 3: f(χ2)=102.83891 f/prime(χ2)=30 530.672 ratio=3.3683800 ×10−3 χ3=128.51404 −3.3683800 ×10−3=128.51067 |ratio|>10−6, so repeat. Step 4: f(χ3)=1.6614116 ×10−3 f/prime(χ3)=30 529.686 ratio=5.4419545 ×10−8 χ4=128.51067 −5.4419545 ×10−8=128.51067 |ratio|<10−6 So we accept χ=128.51 km1 2 as the solution after four iterations. Substituting this value of χtogether with the semimajor axis [Equation (d)] into Equation 3.55 3yields F−F0=χ√−a=128.51√−(−19 655)=0.91664 It follows from (b) that the hyperbolic eccentric anomaly after one hour is F=0.23448 +0.91664 =1.1511 Finally, we calculate the corresponding true anomaly using Equation 3.41b, tanθ 2=/radicalbigg e+1 e−1tanhF 2=/radicalbigg 1.4682 +1 1.4682 −1tanh1.1511 2=1.1926 which means that after one hour θ=100.04◦ Recall from Section 2.11 that the position rand velocity vo nat r a j e c t o r ya ta n yt i m e tcan be found in terms of the position r0and velocity v0at time t0by means of the Lagrange fand gcoefficients and their first derivatives, r=fr0+gv0 (3.64) v=˙fr0+˙gv0 (3.65) 142 Chapter 3 Orbital position as a function of time Equations 2.148 give f,g,˙fand˙gexplicitly in terms of the change in true anomaly /Delta1θover the time interval /Delta1t=t−t0. The Lagrange coefficients can also be derived in terms of changes in the eccentric anomaly /Delta1Efor elliptical orbits, /Delta1Ffor hyperbolas or /Delta1tan(θ/2) for parabolas. However, if we take advantage of the universal variable formulation, we can cover all of these cases with thesame set of Lagrange coefficients. In terms of the universal anomaly χand the Stumpff functions C(z) and S(z), the Lagrange coefficients are (Bond and Allman, 1996) f=1−χ 2 r0C(αχ2) (3.66a) g=/Delta1t−1√µχ3S(αχ2) (3.66b) ˙f=√µ rr0/bracketleftbig αχ3S(αχ2)−χ/bracketrightbig (3.66c) ˙g=1−χ2 rC(αχ2) (3.66d) The implementation of these four functions in MATLAB is found in Appendix D.6. Algorithm 3.4Given r0and v0, find rand vat a time /Delta1tlater. See Appendix D.7 for an implementation of this procedure in MATLAB. 1. Use the initial conditions to find: (a) The magnitude of r0and v0, r0=√r0·r0v0=√v0·v0 (b) The radial component velocity of vr0by projecting v0onto the direction ofr0, vr0=r0·v0 r0 (c) The reciprocal αof the semimajor axis, using Equation 3.45 α=2 r0−v2 0 µ The sign of αdetermines whether the trajectory is an ellipse ( α>0), parabola (α=0) or hyperbola ( α<0). 2. With r0,vr0,αand/Delta1t, use Algorithm 3.3 to find the universal anomaly χ. 3. Substitute α,r0,/Delta1tandχinto Equations 3.66a and 3.66b to obtain f,g. 4. Use Equation 3.64 to compute rand, from that, its magnitude r. 5. Substitute α,r0,randχinto Equations 3.66c and 3.66d to obtain ˙fand˙g. 6. Use Equation 3.65 to compute v. 3.7 Universal variables 143 Example 3.7An earth satellite moves in the xyplane of an inertial frame with origin at the earth’s center. Relative to that frame, the position and velocity of the satellite at time t0are r0=7000.0 ˆi−12 124 ˆj(km) v0=2.6679 ˆi+4.6210 ˆj(km/s) (a) Compute the position and velocity vectors of the satellite 60 minutes later using Algorithm 3.4. Step 1: r0=/radicalbig 7000.02+(−12 124)2=14 000 km v0=/radicalbig 2.66792+4.62102=5.3359 km /s vr0=7000.0 ·2.6679 +(−12 124) ·4.6210 14 000=− 2.6679 km /s α=2 14 000−5.33592 398 600=7.1429 ×10−5km−1 The trajectory is an ellipse, because αis positive. Step 2: Using the results of Step 1, Algorithm 3.3 yields χ=253.53 km1 2 which means z=αχ2=7.1429 ×10−5·253.532=4.5911 Step 3: Substituting the above values of χand zinto Equations 3.66a and 3.66b we find f=1−χ2 r0C(αχ2)=1−253.532 14 0000.3357/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright C(4.5911) =− 0.54123 g=/Delta1t−1√µχ3S(αχ2)=3600−253.532 √ 398 6000.13233/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright S(4.5911) =184.35 s−1 Step 4: r=fr0+gv0 =(−0.54123)(7000.0 ˆi−12.124 ˆj)+184.35(2.6679 ˆi+4.6210 ˆj) =− 3296.8 ˆi+7413.9 ˆj(km) 144 Chapter 3 Orbital position as a function of time (Example 3.7 continued)Therefore, the magnitude of ris r=/radicalbig (−3296.8)2+7413.92=8113.9k m Step 5: ˙f=√µ rr0/bracketleftbig αχ3S(αχ2)−χ/bracketrightbig =√ 398 600 8113.9·14 000 (7.1429×105)·253.532·0.13233/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright S(4.5911) −253.53  =− 0.00055298 s−1 ˙g=1−χ2 rC(αχ2)=1−253.532 8113.90.3357/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright C(4.5911) =− 1.6593 Step 6: v=˙fr0+˙gv0 =(−0.00055298)(7000 .0ˆi−12.124ˆj)+(−1.6593) v0(2.6679ˆi+4.6210ˆj) =− 8.2977ˆi−0.96309ˆj(km/s) The initial and final position and velocity vectors, as well as the trajectory, are accurately illustrated in Figure 3.20. Perigee ˆixˆj y rt /H11005 t0/H110013600 s t /H11005 t0r0Ov v0 Figure 3.20 Initial and final points on a geocentric trajectory. Problems 145 Problems 3.1 Use Newton’s method to find, to eight significant figures, the positive roots of the equation 10 esinx=x2−5x+4. In each case, starting with your initial guess, list each successive approximation until subsequent iterations produce changes only beyond eightsignificant figures. Recall that successive estimates of a root of the equation f(x)=0a r e obtained from the formula x i+1=xi−f(xi)/f/prime(xi). 3.2 Use Newton’s method to find, to eight significant figures, the first four non-negative roots of the equation tan ( x)=tanh (x ). Starting with your initial guess, list each suc- cessive approximation until subsequent iterations produce changes only beyond eightsignificant figures. 3.3 A satellite is in earth orbit for which perigee altitude is 200 km and apogee altitude is 600 km. Find the time interval during which the satellite remains above an altitude of400 km.{Ans.: 47.15 min} 3.4 An earth-orbiting satellite has a perigee radius of 7000 km and an apogee radius of 10 000 km. (a) What true anomaly /Delta1θis swept out between t=0.5 hr and t=1.5 hr after perigee passage? (b) What area is swept out by the position vector during that time interval? {Ans.: (a) 128.7 ◦; (b) 1.03 ×108km2} 3.5 An earth-orbiting satellite has a period of 15.743 hours and a perigee radius of 12 756 km. At time t=10 hours after perigee passage, determine (a) the radius; (b) the speed; (c) the radial component of the velocity. {Ans.: (a) 48 290 km; (b) 2.00 km/s; (c) −0.7210 km/s} 3.6 In terms of the eccentricity eand the period T, calculate (a) the time required to fly from DtoBthrough perigee; (b) the time required to fly from BtoDthrough apogee. {Ans.: (a) tDPB=(1/2−e/π)T; (b) tBAD=(1/2+e/π)T} P AB DF Figure P .3.6 3.7 If the eccentricity of the elliptical orbit is 0.3, calculate, in terms of the period T, the time required to fly from PtoB. {Ans.: 0.157T } 146 Chapter 3 Orbital position as a function of time P AB 90/H11034 F Figure P .3.7 3.8 A satellite in earth orbit has perigee and apogee radii of rp=7000 km and ra=14 000 km, respectively. Find its true anomaly 30 minutes after passing true anomaly of 60◦. {Ans.: 127◦} 3.9 Show that the solution to acosθ+bsinθ=c,w h e r e a,band care given, is θ=φ±cos−1/parenleftBigc acosφ/parenrightBig where tan φ=b/a. 3.10 Calculate the time required to fly from PtoB, in terms of the eccentricity eand the period T.Blies on the minor axis. {Ans.: (0 .25−0.1592 e)T} P AB DF Figure P .3.10 3.11 If the eccentricity of the elliptical orbit is 0.5, calculate, in terms of the period T, the time required to fly from PtoB. {Ans.: 0.170 T} P AB rp2rp F Figure P .3.11 Problems 147 3.12 Verify the results of part (b) of Example 3.3. 3.13 Calculate the time required for a spacecraft launched into a parabolic trajectory at a perigee altitude of 500 km to leave the earth’s sphere of influence (see Table A.2).{Ans.: 7 d 18 hr 34 min} 3.14 A spacecraft on a parabolic trajectory around the earth has a perigee radius of 7500 km. ( a ) H o wl o n gd o e si tt a k et ofl yf r o mθ =− 90 ◦toθ=+ 90◦? (b) How far is the spacecraft from the center of the earth 24 hours after passing through perigee? {Ans.: (a) 1.078 hr; (b) 230 200 km} 3.15 A spacecraft on a hyperbolic trajectory around the earth has a perigee radius of 7500 km and a perigee speed of 1 .1vesc. ( a ) H o wl o n gd o e si tt a k et ofl yf r o mθ =− 90◦toθ=+ 90◦? (b) How far is the spacecraft from the center of the earth 24 hours after passing through perigee? {Ans.: (a) 1.14 hr; (b) 456 000 km} 3.16 A trajectory has a perigee velocity of 11.5 km/s and a perigee altitude of 300 km. If at 6 AM the satellite is traveling towards the earth with a speed of 10 km/s, how far will itbe from the earth’s surface at 11 AM the same day?{Ans.: 88 390 km} 3.17 An incoming object is sighted at an altitude of 37 000 km with a speed of 8 km/s and a flight path angle of −65 ◦. (a) Will it impact the earth or fly by? (b) What is the time to impact or closest passage? {Ans.: (b) 1 hr 24 min} 3.18 At a given instant the radial position of an earth-orbiting satellite is 7200 km, its radial speed is 1 km/s. If the semimajor axis is 10 000 km, use Algorithm 3.3 to find the universalanomaly 60 minutes later. Check your result using Equation 3.55. 3.19 At a given instant a space object has the following position and velocity vectors relative to an earth-centered inertial frame of reference: r 0=20 000 ˆi−105 000 ˆj−19 000 ˆk(km) v0=0.9000 ˆi−3.4000 ˆj−1.5000 ˆk(km/s) Find rand vtwo hours later. {Ans.: r=26 338 ˆi−128 750 ˆj−29 656 ˆk(km); v=0.862800 ˆi−3.2116 ˆj−1.4613 ˆk(km/s)} This page int entionally left blank 4Chapter Orbits in three dimensions Chapter outline 4.1 Introduction 149 4.2 Geocentric right ascension–declination frame 1504.3 State vector and the geocentric equatorial frame 1544.4 Orbital elements and the state vector 158 4.5 Coordinate transformation 164 4.6 Transformation between geocentric equatorial and perifocal frames 172 4.7 Effects of the earth’s oblateness 177 Problems 187 4.1 Introduction The discussion of orbital mechanics up to now has been confined to two dimen- sions, i.e., to the plane of the orbits themselves. This chapter explores the means of describing orbits in three-dimensional space, which, of course, is the setting for realmissions and orbital maneuvers. Our focus will be on the orbits of earth satellites, butthe applications are to any two-body trajectories, including interplanetary missionsto be discussed in Chapter 8. We begin with a discussion of the ancient concept of the celestial sphere and the use of right ascension and declination to d efine the location of stars, planets and other celestial objects on the sphere. This leads to the establishment of the inertial geocentricequatorial frame of reference and the concept of state vector. The six components ofthis vector give the instantaneous position and velocity of an object relative to the 149 150 Chapter 4 Orbits in three dimensions inertial frame and define the characteristics of the orbit. Following that discussion is a presentation of the six classical orbital elements, which also uniquely define the shape and orientation of an orbit and the location of a body on it. We then show howto transform the state vector into orbital elements and vice versa, taking advantage ofthe perifocal frame introduced in Chapter 2. The chapter concludes with a summary of two major perturbations of earth orbits due to the earth’s non-spherical shape. These perturbations are exploited toplace satellites in sun-synchronous and molniya orbits. 4.2 Geocentric right ascension–declination frame The coordinate system used to describe earth orbits in three dimensions is defined in terms of earth’s equatorial plane, the ecliptic plane, and the earth’s axis of rotation.The ecliptic is the plane of the earth’s orbit around the sun, as illustrated in Figure 4.1.The earth’s axis of rotation, which passes through the North and South Poles, is notperpendicular to the ecliptic. It is tilted away by an angle known as the obliquity ofthe ecliptic, ε. For the earth εis approximately 23.4 ◦. Therefore, the earth’s equatorial plane and the ecliptic intersect along a line, which is known as the vernal equinox line.On the calendar,‘vernal equinox’ is the first day of spring in the northern hemisphere,when the noontime sun crosses the equator from south to north. The position of thesun at that instant defines the location of a point in the sky called the vernal equinox,for which the symbol γis used. On the day of the vernal equinox, the number of hours of daylight and darkness is equal; hence, the word equinox. The other equinox occurs N First day of summer ≈ 21 JuneFirst day of winter ≈ 21 DecemberWinter solstice N Summer solsticeγ First day of spring ≈ 21 MarchFirst day of autumn ≈ 21 SeptemberSunVernal equinox NAutumnal equinox N Vernal equinox line Figure 4.1 The earth’s orbit around the sun, viewed from above the ecliptic plane, showing the change of seasons in the northern hemisphere. 4.2 Geocentric right ascension–declination frame 151 precisely one-half year later, when the sun crosses back over the equator from north to south, thereby defining the first day of autumn. The vernal equinox lies today inthe constellation Pisces, which is visible in the night sky during the fall. The directionof the vernal equinox line is from the earth towards γ, as shown in Figure 4.1. For many practical purposes, the vernal equinox line may be considered fixed in space. However, it actually rotates slowly because the earth’s tilted spin axis precesseswestward around the normal to the ecliptic at the rate of about 1.4 ◦per century. This slow precession is due primarily to the action of the sun and the moon on thenon-spherical distribution of mass within the earth. Due to the centrifugal force ofrotation about its own axis, the earth bulges very slightly outward at its equator. Thiseffect is shown highly exaggerated in Figure 4.2. One of the bulging sides is closerto the sun than the other, so the force of the sun’s gravity f 1on its mass is slightly larger than the force f2on the opposite side, farthest from the sun. The forces f1and f2, along with the dominant force Fon the spherical mass, comprise the total force of the sun on the earth, holding in its solar orbit. Taken together, f1and f2produce a net clockwise moment (a vector into the page) about the center of the earth. Thatmoment would rotate the earth’s equator into alignment with the ecliptic if it werenot for the fact that the earth has an angular momentum directed along its south-to-north polar axis due to its spin around that axis at an angular velocity ω Eof 360◦ per day. The effect of the moment is to rotate the angular momentum vector in the direction of the moment (into the page). The result is that the spin axis is forced toprecess in a counterclockwise direction around the normal to the ecliptic, sweepingout a cone as illustrated in the figure. The moon exerts a torque on the earth for thesame reason, and the combined effect of the sun and the moon is a precession of thespin axis, and hence γ, with a period of about 26 000 years. The moon’s action also superimposes a small nutation on the precession. This causes the obliquity εto vary with a maximum amplitude of 0.0025 ◦over a period of 18.6 years. Four thousand years ago, when the first recorded astronomical observations were being made, γwas located in the constellation Aries, the ram. The Greek letter γis a descendent of the ancient Babylonian symbol resembling the head of a ram. N εωE STo the sunEcliptic FC f1 f2ε Figure 4.2 Secondary (perturbing) gravitational forces on the earth. 152 Chapter 4 Orbits in three dimensions 15° 30° 45° 60°75°90°105° 345°315° 330°10°20°30°40°50°60°70° 80° /H1100210° /H1100220° /H1100230° /H1100240° /H1100250°gCelestial equator 0°90° Right ascensionEarth's equatorial planeN S1 hour EastDeclination Figure 4.3 The celestial sphere, with grid lines of right ascension and declination. T o the human eye, objects in the night sky appear as points on a celestial sphere surrounding the earth, as illustrated in Figure 4.3. The north and south poles ofthis fixed sphere correspond to those of the earth rotating within it. Coordinates oflatitude and longitude are used to locate points on the celestial sphere in much thesame way as on the surface of the earth. The projection of the earth’s equatorial planeoutward onto the celestial sphere defines the celestial equator. The vernal equinoxγ, which lies on the celestial equator, is the origin for measurement of longitude, which in astronomical parlance is called right ascension. Right ascension (RA orα) is measured along the celestial equator in degrees east from the vernal equinox. (Astronomers measure right ascension in hours instead of degrees, where 24 hoursequals 360 ◦.) Latitude on the celestial sphere is called declination. Declination (Dec orδ) is measured along a meridian in degrees, positive to the north of the equator and negative to the south. Figure 4.4 is a sky chart showing how the heavenly grid appearsfrom a given point on the earth. Notice that the sun is located at the intersection ofthe equatorial and ecliptic planes, so this must be the first day of spring. Stars are so far away from the earth that their positions relative to each other appear stationary on the celestial sphere. Planets, comets, satellites, etc., move upon the fixed backdrop of the stars. The coordinates of celestial bodies as a functionof time is called an ephemeris, for example, the Astronomical Almanac (US Naval Observatory, 2004). Table 4.1 is an abbreviated ephemeris for the moon and for Venus. An ephemeris depends on the location of the vernal equinox at a given time 4.2 Geocentric right ascension–declination frame 153 MercurySunMoon VenusVernal equinoxCelestial equator10° /H1100220°/H1100230°/H1100240°20°30°40° Ecliptic 15° (1 hr) 30° (2 hr)345° (23 hr) 0° (0 hr) meridian/H1100210°23.5° Figure 4.4 A view of the sky above the eastern horizon from 0◦longitude on the equator at 9 am local time, 20 March, 2004. (Precession epoch 2000.) T able 4.1 Venus and moon ephemeris for 0 hours universal time (Precession epoch: 2000) Ve n u s Mo o n Date RA Dec RA Dec 1 Jan 2004 21 hr 05.0 min −18◦36/prime1 hr 44.9 min +8◦47/prime 1 Feb 2004 23 hr 28.0 min −04◦30/prime4 hr 37.0 min +24◦11/prime 1 Mar 2004 01 hr 30.0 min +10◦26/prime6 hr 04.0 min +08◦32/prime 1 Apr 2004 03 hr 37.6 min +22◦51/prime9 hr 18.7 min +21◦08/prime 1 May 2004 05 hr 20.3 min +27◦44/prime11 hr 28.8 min +07◦53/prime 1 Jun 2004 05 hr 25.9 min +24◦43/prime14 hr 31.3 min −14◦48/prime 1 Jul 2004 04 hr 34.5 min +17◦48/prime17 hr 09.0 min −26◦08/prime 1 Aug 2004 05 hr 37.4 min +19◦04/prime21 hr 05.9 min −21◦49/prime 1 Sep 2004 07 hr 40.9 min +19◦16/prime00 hr 17.0 min −00◦56/prime 1 Oct 2004 09 hr 56.5 min +12◦42/prime02 hr 20.9 min +14◦35/prime 1 Nov 2004 12 hr 15.8 min +00◦01/prime05 hr 26.7 min +27◦18/prime 1 Dec 2004 14 hr 34.3 min −13◦21/prime07 hr 50.3 min +26◦14/prime 1 Jan 2005 17 hr 12.9 min −22◦15/prime10 hr 49.4 min +11◦39/prime or epoch, for we know that even the positions of the stars relative to the equinox change slowly with time. For example, Table 4.2 shows the celestial coordinates of the star Regulus at five epochs since 1700. Currently, the position of the vernal equinox in the year 2000 is used to define the standard grid of the celestial sphere. 154 Chapter 4 Orbits in three dimensions T able 4.2 Variation of the coordinates of the star Regulus due to precession of the equinox Precession epoch RA Dec 1700 9 hr 52.2 min (148.05◦) +13◦25/prime 1800 9 hr 57.6 min (149.40◦) +12◦56/prime 1900 10 hr 3.0 min (150.75◦) +12◦27/prime 1950 10 hr 5.7 min (151.42◦) +12◦13/prime 2000 10 hr 8.4 min (152.10◦) +11◦58/prime In 2025, the position will be updated to that of the year 2050; in 2075 to that of the year 2100; and so on at 50 year intervals. Since observations are made relativeto the actual orientation of the earth, these measurements must be transformed intothe standardized celestial frame of reference. As Table 4.2 suggests, the adjustmentswill be small if the current epoch is within 25 years of the standard precession epoch. 4.3 State vector and the geocentric equatorial frame At any given time, the state vector of a satellite comprises its velocity vand acceleration a. Orbital mechanics is concerned with spe cifying or predicting state vectors over intervals of time. From Chapter 2, we know that the equation governing the statevector of a satellite traveling around the earth is, under the familiar assumptions, ¨r=−µ r3r (4.1) ris the position vector of the satellite relative to the center of the earth. The com- ponents of rand, especially, those of its time derivatives ˙r=vand¨r=a, must be measured in a non-rotating frame attached to the earth. A commonly used non-rotating right-handed cartesian coordinate system is the geocentric equatorial frame shown in Figure 4.5. The Xaxis points in the vernal equinox direction. The XYplane is the earth’s equatorial plane, and the Zaxis coincides with the earth’s axis of rotation and points northward. The unit vectors ˆI,ˆJandˆKform a right-handed triad. The non-rotating geocentric equatorial frame serves as an inertial frame for the two-bodyearth satellite problem, as embodied in Equation 4.1. It is not truly an inertial frame,however, since the center of the earth is always accelerating towards a third body, thesun (to say nothing of the moon), a fact which we ignore in the two-body formulation. In the geocentric equatorial frame the state vector is given in component form by r=XˆI+YˆJ+ZˆK (4.2) v=v XˆI+vYˆJ+vZˆK (4.3) Ifris the magnitude of the position vector, then r=rˆur (4.4) 4.3 State vector and the geocentric equatorial frame 155 Y XZ Celestial north pole Right ascension, ar Celestial equatorDeclination, d Vernal equinox, gIntersection of equatorial and ecliptic planesEarth's equatorial planeCelestial sphereSatellitev ˆIˆK ˆJ Figure 4.5 Geocentric equatorial frame. From Figure 4.5 we see that the components of ˆur(the direction cosines of r)a r e found in terms of the right ascension αand declination δas follows, ˆur=cosδcosαˆI+cosδsinαˆJ+sinδˆK (4.5) Therefore, given the state vector, we can then compute the right ascension and decli- nation. However, the right ascension and declination alone do not furnish r. For that we need the distance rto obtain rfrom Equation 4.4. Example 4.1If the position vector of the International Space Station is r=− 5368ˆI−1784ˆJ+3691ˆK(km) what are its right ascension and declination? The magnitude of ris r=/radicalbig (−5368)2+(−1784)2+36912=6754 km Hence, ˆur=r r=− 0.7947 ˆI−0.2642 ˆJ+0.5464 ˆK (a) From this and Equation 4.5 we see that sin δ=0.5464 which means δ=sin−10.5464 =33.12◦ There is no quadrant ambiguity since, by definition, the declination lies between −90◦ and+90◦, which is precisely the range of the principal values of the arcsin function. It also follows that cos δcannot be negative. 156 Chapter 4 Orbits in three dimensions (Example 4.1 continued)From Equation 4.5 and Equation (a) just above we have cosδcosα=− 0.7947 (b) cosδsinα=− 0.2642 (c) Therefore cosα=−0.7947 cos 33.12◦=− 0.9489 which implies α=cos−1(−0.9489) =161.6◦(second quadrant) or 198 .4◦(third quadrant) From (c) we observe that sin αis negative, which means αlies in the third quadrant, α=198.4◦ If we are provided with the state vector r0,v0at a given instant, then we can determine the state vector at any other time in terms of the initial vector by means of theexpressions r=fr 0+gv0 v=˙fr0+˙gv0(4.6) where the Lagrange coefficients fand gand their time derivatives are given in Equa- tion 3.66. Specifying the total of six components of r0and v0therefore completely determines the size, shape and orientation of the orbit. Example 4.2At time t0the state vector of an earth satellite is r0=1600ˆI+5310ˆJ+3800ˆK(km) (a) v0=− 7.350ˆI+0.4600ˆJ+2.470ˆK(km/s) (b) Determine the position and velocity 3200 seconds later and plot the orbit in three dimensions. We will use the universal variable formulation and Algorithm 3.4, which was illus- trated in detail in Example 3.7. Therefore, only the results of each step are presentedhere. Step 1: α=1.4613×10 −4km−1. Since this is positive, the orbit is an ellipse. Step 2: χ=294.42 km1 2. Step 3: f=− 0.94843 and g=− 354.89 s−1. 4.3 State vector and the geocentric equatorial frame 157 Step 4: r=1090.9 ˆI−5199.4 ˆJ−4480.6 ˆK(km), r=6949.8k m . Step 5: ˙f=0.00045324 s−1,˙g=− 0.88479. Step 6: v=7.2284 ˆI+1.9997 ˆJ−0.46311 ˆK(km/s) T o plot the orbit, we observe that one complete revolution means a change in the eccentric anomaly Eof 2πradians. According to Equation 3.54 2, the corresponding change in the universal anomaly is χ=√aE=/radicalbigg 1 αE=/radicalbigg 1 0.00014613·2π=519.77 km1 2 Letting χvary from 0 to 519.77 in small increments, we employ the Lagrange coefficient formulation (Equatio n 3.64 plus 3.66a and 3.66b) to compute r=/bracketleftbigg 1−χ2 r0C(αχ2)/bracketrightbigg r0+/bracketleftbigg /Delta1t−1√µχ3S(αχ2)/bracketrightbigg v0 where /Delta1tf o rag i v e nv a l u eo f χis given by Equation 3.45. Using a computer to plot the points obtained in this fashion yields Figure 4.6, which also shows the state vectorsatt 0and t0+3200 s. XYZ r0v0 Equatorial plane vt /H11005 t0 t /H11005 t0/H110013200 sDescending node Ascending node r Figure 4.6 The orbit corresponding to the initial conditions given in Equations (a) and (b) of Example 4.2. 158 Chapter 4 Orbits in three dimensions The previous example illustrates the fact that the six quantities or orbital elements comprising the state vector rand vcompletely determine the orbit. Other elements may be chosen. The classical orbital elements are introduced and related to the statevector in the next section. 4.4 Orbital elements and the state vector T o define an orbit in the plane requires two parameters: eccentricity and angularmomentum. Other parameters, such as the semimajor axis, the specific energy, and (for an ellipse) the period, are obtained from these two. T o locate a point on the orbitrequires a third parameter, the true anomaly, which leads us to the time since perigee.Describing the orientation of an orbit in three dimensions requires three additional parameters, called the Euler angles, which are illustrated in Figure 4.7. First, we locate the intersection of the orbital plane with the equatorial ( XY) plane. That line is called the node line. The point on the node line where the orbitpasses above the equatorial plane from below it is called the ascending node. Thenode line vector Nextends outward from the origin through the ascending node. At the other end of the node line, where the orbit dives below the equatorial plane, is thedescending node. The angle between the positive Xaxis and the node line is the first Euler angle /Omega1, the right ascension of the ascending node. Recall from Section 4.2 that right ascension is a positive number lying between 0 ◦and 360◦. The dihedral angle between the orbital plane and the equatorial plane is the inclination i, measured according to the right-hand rule, that is, counterclockwise around the node line vector from the equator to the orbit. The inclination is also theangle between the positive Zaxis and the normal to the plane of the orbit. The two Y Xii /H9275/H9258PerigeeZEarth /H11032s north polar axis Satellite v Nhe Ascending node ˆˆ IˆK r /H9253Earth /H11032s equatorial plane J Node lineΩ Figure 4.7 Geocentric equatorial frame and the orbital elements. 4.4 Orbital elements and the state vector 159 equivalent means of measuring iare indicated in Figure 4.7. Recall from Chapter 2 that the angular momentum vector his normal to the plane of the orbit. Therefore, the inclination iis the angle between the positive Zaxis and h. The inclination is a positive number between 0◦and 180◦. It remains to locate the perigee of the orbit. Recall that perigee lies at the inter- section of the eccentricity vector ewith the orbital path. The third Euler angle ω, the argument of perigee, is the angle between the node line vector Nand the eccentricity vector e, measured in the plane of the orbit. The argument of perigee is a positive number between 0◦and 360◦. In summary, the six orbital elements are hspecific angular momentum iinclination /Omega1right ascension ( RA) of the ascending node eeccentricity ωargument of perigee θtrue anomaly The angular momentum hand true anomaly θare frequently replaced by the semimajor axis aand the mean anomaly M, respectively. Given the position rand velocity vof a satellite in the geocentric equatorial frame, how do we obtain the orbital elements? The step-by-step procedure is outlined in Algorithm 4.1. Note that each step incorporates results obtained in the previous steps. Algorithm 4.1Obtain orbital elements from the state vector. A MATLAB version of this procedure appears in Appendix D.8. Applying this algorithm to orbits around other planets or the sun amounts to defining the frame of reference and substituting the appropriategravitational parameter µ. 1. Calculate the distance, r=√r·r=/radicalbig X2+Y2+Z2 2. Calculate the speed, v=√v·v=/radicalBig v2 X+v2 Y+v2 Z 3. Calculate the radial velocity, vr=r·v/r=(XvX+YvY+ZvZ)/r Note that if vr>0, the satellite is flying away from perigee. If vr<0, it is flying towards perigee. 4. Calculate the specific angular momentum, h=r×v=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆIˆJˆK XYZ v XvYvZ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle 160 Chapter 4 Orbits in three dimensions (Algorithm 4.1 continued)5. Calculate the magnitude of the specific angular momentum, h=√ h·h the first orbital element. 6. Calculate the inclination, i=cos−1/parenleftbigghZ h/parenrightbigg (4.7) This is the second orbital element. Recall that imust lie between 0◦and 180◦,s o there is no quadrant ambiguity. If 90◦<i≤180◦, the orbit is retrograde. 7. Calculate N=ˆK×h=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆIˆJˆK 001 h XhYhZ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle(4.8) This vector defines the node line. 8. Calculate the magnitude of N, N=√ N·N 9. Calculate the RAof the ascending node, /Omega1=cos−1(NX/N) the third orbital element. If ( NX/N)>0, then /Omega1lies in either the first or fourth quadrant. If ( NX/N)<0, then /Omega1lies in either the second or third quadrant. T o place /Omega1in the proper quadrant, observe that the ascending node lies on the positive side of the vertical XZplane (0 ≤/Omega1< 180◦)i fNY>0. On the other hand, the ascending node lies on the negative side of the XZplane (180◦≤/Omega1< 360◦) ifNY<0. Therefore, NY>0 implies that 0 </Omega1< 180◦, whereas NY<0 implies that 180◦</Omega1< 360◦. In summary, /Omega1=  cos−1/parenleftbiggNX N/parenrightbigg (NY≥0) 360◦−cos−1/parenleftbiggNX N/parenrightbigg (NY<0)(4.9) 10. Calculate the eccentricity vector. Starting with Equation 2.30, e=1 µ/bracketleftBig v×h−µr r/bracketrightBig =1 µ/bracketleftBig v×(r×v)−µr r/bracketrightBig =1 µ bac−cabrule/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright rv2−v(r·v)−µr r  so that e=1 µ/bracketleftBig/parenleftBig v2−µ r/parenrightBig r−rvrv/bracketrightBig (4.10) 11. Calculate the eccentricity, e=√e·e 4.4 Orbital elements and the state vector 161 the fourth orbital element. Substituting Equation 4.10 leads to a form depending only on the scalars obtained thus far, e=1 µ/radicalBig (2µ−rv2)rv2r+(µ−rv2)2 (4.11) 12. Calculate the argument of perigee, ω=cos−1(N·e/Ne) the fifth orbital element. If N·e>0, then ωlies in either the first or fourth quadrant. If N·e<0, then ωlies in either the second or third quadrant. T o placeωin the proper quadrant, observe that perigee lies above the equatorial plane (0 ≤ω< 180◦)i fepoints up (in the positive Zdirection), and perigee lies below the plane (180◦≤ω< 360◦)i f epoints down. Therefore, eZ≥0 implies that 0<ω<180◦, whereas eZ<0 implies that 180◦<ω<360◦. T o summarize, ω=  cos−1/parenleftbiggN·e Ne/parenrightbigg (eZ≥0) 360◦−cos−1/parenleftbiggN·e Ne/parenrightbigg (eZ<0)(4.12) 13. Calculate the true anomaly, θ=cos−1/parenleftBige·r er/parenrightBig the sixth and final orbital element. If e·r>0, then θlies in the first or fourth quadrant. If e·r<0, then θlies in the second or third quadrant. T o place θin the proper quadrant, note that if the satellite is flying away from perigee ( r·v≥0), then 0 ≤θ<180◦, whereas if the satellite is flying towards perigee ( r·v<0), then 180◦≤θ<360◦. Therefore, using the results of step 3 above θ=  cos−1/parenleftBige·r er/parenrightBig (vr≥0) 360◦−cos−1/parenleftBige·r er/parenrightBig (vr<0)(4.13a) Substituting Equation 4.10 yields an alternative form of this expression, θ=  cos−1/bracketleftbigg1 e/parenleftbiggh2 µr−1/parenrightbigg/bracketrightbigg (vr≥0) 360◦−cos−1/bracketleftbigg1 e/parenleftbiggh2 µr−1/parenrightbigg/bracketrightbigg (vr<0)(4.13b) The procedure described above for calculating the orbital elements is not unique. Example 4.3Given the state vector, r=− 6045ˆI−3490ˆJ+2500ˆK(km) v=− 3.457ˆI+6.618ˆJ+2.533ˆK(km/s) find the orbital elements h,i,/Omega1,e,ωandθusing Algorithm 4.1. 162 Chapter 4 Orbits in three dimensions (Example 4.3 continued)Step 1: r=√r·r=/radicalbig (−6045)2+(−3490)2+25002=7414 km (a) Step 2: v=√v·v=/radicalbig (−3.457)2+6.6182+2.5332=7.884 km/s (b) Step 3: vr=v·r r=(−3.457)·(−6045) +6.618·(−3490) +2.533·2500 7414 =0.5575 km/s (c) Sincevr>0, the satellite is flying away from perigee. Step 4: h=r×v=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆI ˆJ ˆK −6045 −3490 2500 −3.457 6 .618 2 .533/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=− 25 380 ˆI+6670ˆJ−52 070 ˆK(km 2/s) (d) Step 5: h=√ h·h=/radicalbig (−25 380)2+66702+(−52 070)2=58 310 km2/s (e) Step 6: i=cos−1hZ h=cos−1/parenleftbigg−52 070 58 310/parenrightbigg =153.2◦(f) Since iis greater than 90◦, this is a retrograde orbit. Step 7: N=ˆK×h=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆI ˆJ ˆK 001 −25 380 6670 −52 070/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=− 6670ˆI−25 380 ˆJ (g) Step 8: N=√ N·N=/radicalbig (−6670)2+(−25 380)2=26 250 (h) Using (g) and (h), we compute the right ascension of the node. Step 9: /Omega1=cos−1NX N=cos−1/parenleftbigg−6670 26 250/parenrightbigg =104.7◦or 255 .3◦ From (g) we know that NY<0; therefore, /Omega1must lie in the third quadrant, /Omega1=255.3◦(i) 4.4 Orbital elements and the state vector 163 Step 10: e=1 µ/bracketleftBig/parenleftBig v2−µ r/parenrightBig r−(r·v)v/bracketrightBig =1 398 600/bracketleftbigg/parenleftbigg 7.8842−398 600 7414/parenrightbigg (−6045 ˆI−3490ˆJ+2500ˆK) −4133(−3.457 ˆI+6.618ˆJ+2.533ˆK)/bracketrightbigg =− 0.09160 ˆI−0.1422 ˆJ+0.02644 ˆK (j) Step 11: e=√e·e=/radicalbig (−0.09160)2+(−0.1422)2+(0.02644)2=0.1712 (k) Clearly, the orbit is an ellipse. Step 12: ω=cos−1N·e Ne =cos−1/bracketleftbigg(−6670)(−0.09160) +(−25 380)(−0.1422) +(0)(0.02644) (26 250)(0.1712)/bracketrightbigg =20.07◦or 339 .9◦ ωlies in the first quadrant if eZ>0, which is true in this case, as we see from (j). Therefore, ω=20.07◦(l) Step 13: θ=cos−1/parenleftBige·r er/parenrightBig =cos−1/bracketleftbigg(−0.09160)(−6045) +(−0.1422) ·(−3490) +(0.02644)(2500) (0.1712)(7414)/bracketrightbigg =28.45◦or 331 .6◦ From (c) we know that vr>0, which means 0 ≤θ<180◦. Therefore, θ=28.45◦ Having found the orbital elements, we can go on to compute other parameters. The perigee and apogee radii are rp=h2 µ1 1+ecos(0)=58 3102 398 6001 1+0.1712=7284 km ra=h2 µ1 1+ecos(180◦)=58 3102 398 6001 1−0.1712=10 290 km 164 Chapter 4 Orbits in three dimensions (Example 4.3 continued)From these it follows that the semimajor axis of the ellipse is a=1 2(rp+ra)=8788 km This leads to the period, T=2π√µa3 2=2.278 hr The orbit is illustrated in Figure 4.8. XYZ Perigee Apogeerv Ascending node Descending nodeNode lineΩ /H11005255° Equatorial planeu/H1100528.45 ° (Retrograde orbit)Initial state Apse linev/H1100520.07 ° Figure 4.8 A plot of the orbit identified in Example 4.3. We have seen how to obtain the orbital elements from the state vector. T o arrive at the state vector, given the orbital elements, requires performing coordinatetransformations, which are discussed in the next section. 4.5 Coordinate transformation Figure 4.9 shows two cartesian coordinate systems: the unprimed system with axesxyz, and the primed system with axes x /primey/primez/prime. The orthogonal unit basis vectors for the unprimed system are ˆi,ˆjandˆk. The fact they are unit vectors means ˆi·ˆi=ˆj·ˆj=ˆk·ˆk=1 (4.14) Since they are orthogonal, ˆi·ˆj=ˆi·ˆk=ˆj·ˆk=0 (4.15) 4.5 Coordinate transformation 165 Oxy zQ33Q32Q31ij ki′j′ k′x′ y′ z′ˆ ˆ ˆˆ ˆ Figure 4.9 Two sets of cartesian reference axes, xyzand x/primey/primez/prime. The orthonormal basis vectors ˆi/prime,ˆj/primeandˆk/primeof the primed system share these same properties. That is, ˆi/prime·ˆi/prime=ˆj/prime·ˆj/prime=ˆk/prime·ˆk/prime=1 (4.16) and ˆi/prime·ˆj/prime=ˆi/prime·ˆk/prime=ˆj/prime·ˆk/prime=0 (4.17) We can express the unit vectors of the primed system in terms of their components in the unprimed system as follows ˆi/prime=Q11ˆi+Q12ˆj+Q13ˆk ˆj/prime=Q21ˆi+Q22ˆj+Q23ˆk (4.18) ˆk/prime=Q31ˆi+Q32ˆj+Q33ˆk The Qs in these expressions are just the direction cosines of ˆi/prime,ˆj/primeandˆk/prime. Figure 4.9 illustrates the components of ˆk/prime, which are, of course, the projections of ˆk/primeonto the x,yand zaxes. The unprimed unit vectors may be resolved into components along the primed system to obtain a set of equations similar to Equations 4.18: ˆi=Q/prime 11ˆi/prime+Q/prime 12ˆj/prime+Q/prime 13ˆk/prime ˆj=Q/prime 21ˆi/prime+Q/prime 22ˆj/prime+Q/prime 23ˆk/prime(4.19) ˆk=Q/prime 31ˆi/prime+Q/prime 32ˆj/prime+Q/prime 33ˆk/prime 166 Chapter 4 Orbits in three dimensions However, ˆi/prime·ˆi=ˆi·ˆi/prime, so that, from Equations 4.18 1and 4.19 1, we find Q11=Q/prime 11. Likewise, ˆi/prime·ˆj=ˆj·ˆi/prime, which, according to Equations 4.18 1and 4.19 2, means Q12=Q/prime 21. Proceeding in this fashion, it is clear that the direction cosines in Equations 4.18 maybe expressed in terms of those in Equations 4.19. That is, Equations 4.19 may bewritten ˆi=Q 11ˆi/prime+Q21ˆj/prime+Q31ˆk/prime ˆj=Q12ˆi/prime+Q22ˆj/prime+Q32ˆk/prime(4.20) ˆk=Q13ˆi/prime+Q23ˆj/prime+Q33ˆk/prime Substituting Equations 4.20 into Eq uations 4.14 and making use of Equations 4.16 and 4.17, we get the three relations ˆi·ˆi=1⇒Q2 11+Q2 21+Q2 31=1 ˆj·ˆj=1⇒Q2 12+Q2 22+Q2 32=1 (4.21) ˆk·ˆk=1⇒Q2 13+Q2 23+Q2 33=1 Substituting Equations 4.20 into Equat ions 4.15 and, again, making use of Equations 4.16 and 4.17, we obtain the three equations ˆi·ˆj=0⇒Q11Q12+Q21Q22+Q31Q32=0 ˆi·ˆk=0⇒Q11Q13+Q21Q23+Q31Q33=0 (4.22) ˆj·ˆk=0⇒Q12Q13+Q22Q23+Q32Q33=0 Let [ Q] represent the matrix of direction cosines of ˆi/prime,ˆj/primeandˆk/primerelative to ˆi,ˆjandˆk, as given by Equations 4.19. Then [Q]= Q11 Q12 Q13 Q21 Q22 Q23 Q31 Q32 Q33 = ˆi/prime·ˆiˆi/prime·ˆjˆi/prime·ˆk ˆj/prime·ˆiˆj/prime·ˆjˆj/prime·ˆk ˆk/prime·ˆiˆk/prime·ˆjˆk/prime·ˆk  (4.23) The transpose of the matrix [ Q], denoted [ Q]T, is obtained by interchanging the rows and columns of [ Q]. Thus, [Q]T= Q11 Q21 Q31 Q12 Q22 Q32 Q13 Q23 Q33 = ˆi·ˆi/primeˆi·ˆj/primeˆi·ˆk/prime ˆj·ˆi/primeˆj·ˆj/primeˆj·ˆk/prime ˆk·ˆi/primeˆk·ˆj/primeˆk·ˆk/prime  (4.24) Forming the product [ Q]T[Q]w eg e t [Q]T[Q]= Q11 Q21 Q31 Q12 Q22 Q32 Q13 Q23 Q33  Q11 Q12 Q13 Q21 Q22 Q23 Q31 Q32 Q33  = Q2 11+Q2 21+Q2 31 Q11Q12+Q21Q22+Q31Q32 Q11Q13+Q21Q23+Q31Q33 Q12Q11+Q22Q21+Q32Q31 Q2 12+Q2 22+Q2 32 Q12Q13+Q22Q23+Q32Q33 Q13Q11+Q23Q21+Q33Q31 Q13Q12+Q23Q22+Q33Q32 Q2 13+Q2 23+Q2 33  4.5 Coordinate transformation 167 From this we obtain, with the aid of Equations 4.21 and 4.22, [Q]T[Q]=[1] (4.25) where [1]= 100 010001  [1] stands for the identity matrix or unit matrix. In a similar fashion, we can substitute Equations 4.18 into Equations 4.16 and 4.17 and make use of Equations 4.14 and 4.15 to finally obtain [Q][Q] T=[1] (4.26) Since [Q ] satisfies Equations 4.25 and 4.26, it is called an orthogonal matrix. Let vbe a vector. It can be expressed in terms of its components along the unprimed system, v=vxˆi+vyˆj+vzˆk or along the primed system, v=v/prime xˆi/prime+v/prime yˆj/prime+v/prime zˆk/prime These two expressions for vare equivalent ( v=v) since a vector is independent of the coordinate system used to describe it. Thus, v/prime xˆi/prime+v/prime yˆj/prime+v/prime zˆk/prime=vxˆi+vyˆj+vzˆk (4.27) Substituting Equations 4.20 into the right-hand side of Equation 4.27 yields v/prime xˆi/prime+v/prime yˆj/prime+v/prime zˆk/prime=vx(Q11ˆi/prime+Q21ˆj/prime+Q31ˆk/prime)+vy(Q12ˆi/prime+Q22ˆj/prime+Q32ˆk/prime) +vz(Q13ˆi/prime+Q23ˆj/prime+Q33ˆk/prime) Upon collecting terms on the right, we get v/prime xˆi/prime+v/prime yˆj/prime+v/prime zˆk/prime=(Q11vx+Q12vy+Q13vz)ˆi/prime+(Q21vx+Q22vy+Q23vz)ˆj/prime +(Q31vx+Q32vy+Q33vz)ˆk/prime Equating the components of like unit vectors on each side of the equals sign yields v/prime x=Q11vx+Q12vy+Q13vz v/prime y=Q21vx+Q22vy+Q23vz (4.28) v/prime z=Q31vx+Q32vy+Q33vz In matrix notation, this may be written {v/prime}=[Q ]{v} (4.29) 168 Chapter 4 Orbits in three dimensions where {v/prime}=  v/prime x v/primey v/primez  {v}=  vx vy vz  (4.30) and [ Q] is given by Equation 4.23. Equation 4.28 (or Equation 4.29) shows how to transform the components of the vector vin the unprimed system into its components in the primed system. The inverse transformation, from primed to unprimed, is foundby multiplying Equation 4.29 through by [ Q] T: [Q]T{v/prime}=[Q]T[Q]{v} But, according to Equation 4.25, [ Q][Q]T=[1], so that [Q]T{v/prime}=[1]{v} Since [ 1]{v}={v}, we obtain {v}=[Q]T{v/prime} (4.31) Therefore, to go from the primed system to the unprimed system use [ Q], and in the reverse direction – from primed to unprimed – use [ Q]T. Example 4.4In Figure 4.10, the x/primeaxis is defined by the line segment O/primeP.T h e x/primey/primeplane is defined by the intersecting line segments O/primePand O/primeQ.T h e z/primeaxis is normal to the plane of O/primePand O/primeQand obtained by rotating O/primePtowards O/primeQand using the right-hand rule. (a) Find the transformation matrix [ Q]. (b) If {v}=⌊ 246 ⌋T, find{v/prime}. (c) If{v/prime}=⌊ 240 ⌋T, find{v}. (a) Resolve the directed line segments→ O/primePand→ O/primeQinto components along the unprimed system: → O/primeP=(−5−3)ˆi+(5−1)ˆj+(4−2)ˆk=− 8ˆi+4ˆj+2ˆk → O/primeQ=(−6−3)ˆi+(3−1)ˆj+(5−2)ˆk=− 9ˆi+2ˆj+3ˆk O′(3, 1, 2)P (/H110025, 5, 4)Q (/H110026, 3, 5) xyz Oˆj′ ˆ i′ˆk′ Figure 4.10 Defining a unit triad from the coordinates of three non-collinear points, O/prime,Pand Q. 4.5 Coordinate transformation 169 Taking the cross product of→ O/primePinto→ O/primeQy i e l d sav e c t o r Z/primewhich lies in the direction of the desired positive z/primeaxis: Z/prime=→ O/primeP×→ O/primeQ=8ˆi+6ˆj+20ˆk Taking the cross product of Z/primeinto→ O/primePthen yields a vector Y/primewhich points in the positive y/primedirection: Y/prime=Z×→ O/primeP=− 68ˆi−176ˆj+80ˆk Normalizing the vectors→ O/primeP,Y/primeand Z/primeproduces the ˆi/prime,ˆj/primeandˆk/primeunit vectors, respectively. Thus ˆi/prime=→ O/primeP /bardbl→ O/primeP/bardbl=− 0.8729 ˆi+0.4364 ˆj+0.2182 ˆk ˆj/prime=Y/prime /bardblY/prime/bardbl=− 0.3318 ˆi−0.8588 ˆj+0.3904 ˆk and ˆk/prime=Z/prime /bardblZ/prime/bardbl=0.3578 ˆi+0.2683 ˆj+0.8944 ˆk The components of ˆi/prime,ˆj/primeandˆk/primeare the rows of the orthogonal transformation matrix [Q ]. Thus, [Q]= −0.8729 0.4364 0.2182 −0.3318 −0.8588 0.3904 0.3578 0.2683 0.8944  (b) {v/prime}=[Q ]{v}= −0.8729 0.4364 0.2182 −0.3318 −0.8588 0.3904 0.3578 0.2683 0.8944   2 46  =  1.309 −1.756 7.155   (c) {v}=[Q ]T{v/prime}= −0.8729 −0.3318 0.3578 0.4364 −0.8588 0.2683 0.2182 0.3904 0.8944   2 40  =  −3.073 −2.562 1.998   Let us consider the special case in which the coordinate transformation involves a rotation about only one of the coordinate axes, as shown in Figure 4.11. If therotation is about the xaxis, then according to Equations 4.18 and 4.23, ˆi /prime=ˆi ˆj/prime=(ˆj/prime·ˆi)ˆi+(ˆj/prime·ˆj)ˆj+(ˆj/prime·ˆk)ˆk=cosφˆj+cos (90 −φ)ˆk=cosφˆj+sin (φ )ˆk ˆk/prime=(ˆk/prime·ˆj)ˆj+(ˆk/prime·ˆk)ˆk=cos (90◦+φ)ˆj+cosφˆk=− sinφˆj+cosφˆk 170 Chapter 4 Orbits in three dimensions φφ i, i′jk j′k′ ˆ ˆ ˆ ˆ ˆ ˆ Figure 4.11 Rotation about the xaxis. or   ˆi/prime ˆj/prime ˆk/prime  = 10 0 0c o s φ sinφ 0−sinφcosφ   ˆi ˆj ˆk   The transformation from the xyzcoordinate system to the xy/primez/primesystem having a common xaxis is given by the matrix coefficient of the unit vectors on the right. Since this is a rotation through the angle φabout the xaxis, we denote this matrix by [R1(φ)], in which the subscript 1 stands for axis 1 (the xaxis). Thus, [R1(φ)]= 10 0 0c o s φ sinφ 0−sinφcosφ  (4.32) If the rotation is about the yaxis, as shown in Figure 4.12, then Equation 4.18 yields ˆi/prime=(ˆi/prime·ˆi)ˆi+(ˆi/prime·ˆk)ˆk=cosφˆi+cos (φ+90◦)ˆk=cosφˆi−sinφˆk ˆj/prime=ˆj ˆk/prime=(ˆk/prime·ˆi)ˆi+(ˆk/prime·ˆk)ˆk=cos (90◦−φ)ˆi+cosφˆk=sinφˆi+cosφˆk or, more compactly,   ˆi/prime ˆj/prime ˆk/prime  = cosφ0−sinφ 01 0 sinφ0c o s φ   ˆi ˆj ˆk   4.5 Coordinate transformation 171 ik k′ j, j′φ φ i′ˆ ˆ ˆ ˆ ˆ ˆ Figure 4.12 Rotation about the yaxis. φ φ ijk, k′ j′ i′ ˆ ˆ ˆ ˆ ˆ ˆ Figure 4.13 Rotation about the zaxis. We represent this transformation between two cartesian coordinate systems having a common yaxis (axis 2) as [R 2(φ)]. Therefore, [R2(φ)]= cosφ0−sinφ 01 0 sinφ0c o s φ  (4.33) Finally, if the rotation is about the zaxis, as shown in Figure 4.13, then we have from Equation 4.18 that ˆi/prime=(ˆi/prime·ˆi)ˆi+(ˆi/prime·ˆj)ˆj=cosφˆi+cos (90◦−φ)ˆj=cosφˆi+sinφˆj ˆj/prime=(ˆj/prime·ˆi)ˆi+(ˆj/prime·ˆj)ˆj=cos (90◦+φ)ˆi+cosφˆj=− sinφˆi+cosφˆj ˆk/prime=ˆk 172 Chapter 4 Orbits in three dimensions or  ˆi/prime ˆj/prime ˆk/prime  = cosφsinφ0 −sinφ cosφ0 00 1   ˆi ˆj ˆk   In this case the rotation is around axis 3, the zaxis, so [R3(φ)]= cosφsinφ0 −sinφcosφ0 00 1  (4.34) A transformation between two cartesian coordinate systems can be broken down into a sequence of two-dimensional rotations using the matrices [ Ri(φ)],i=1, 2, 3. We will use this to great advantage in the following sections. 4.6 Transformation between geocentric equatorial and perifocal frames The perifocal frame of reference for a given orbit was introduced in Section 2.10. Figure 4.14 illustrates the relationship between the perifocal and geocentric equatorialframes. Since the orbit lies in the ¯x¯yplane, the components of the state vector of a body relative to its perifocal reference are, according to Equations 2.109 and 2.115, r=¯xˆp+¯yˆq=h 2 µ1 1+ecosθ(cosθˆp+sinθˆq) (4.35) v=˙¯xˆp+˙¯yˆq=µ h[−sinθˆp+(e+cosθ)ˆq] (4.36) ˆp ˆIˆJˆqˆK ˆw FocusPeriapseSemilatus rectumz XYZ g Axes of the geocentric equatorial framey x Figure 4.14 Perifocal ( ¯x¯y¯z) and geocentric equatorial ( XYZ ) frames. 4.6 Transformation between geocentric equatorial and perifocal frames 173 In matrix notation these may be written {r}¯x=h2 µ1 1+ecosθ  cosθ sinθ 0  (4.37) {v}¯x=µ h  −sinθ e+cosθ 0  (4.38) The subscript ¯xis shorthand for ‘the ¯x¯y¯zcoordinate system’ and is used to indicate that the components of these vectors are given in the perifocal frame, as opposed to, say, the geocentric equatorial frame (Equations 4.2 and 4.3). The transformation from the geocentric equatorial frame into the perifocal frame may be accomplished by the sequence of three rotations illustrated in Figure 4.15. Thefirst rotation, ①, is around the ˆKaxis, through the right ascension /Omega1. It rotates the i 2 133 /H92752 1 23 2i i/H9275 i/H9275/H9275 ˆwˆw ˆw ˆKˆpˆp ˆJˆJ ˆKˆK11 3ˆqˆq ˆI′ˆI′ ˆIˆI ˆJ′ˆJ′ ˆJ′ ˆI′ˆI′ ˆJ′′ˆJ′′ˆJ′′ΩΩ Ω Ω Figure 4.15 Sequence of three rotations transforming ˆIˆJˆKintoˆpˆqˆw. The ‘eye’ viewing down an axis sees the illustrated rotation about that axis. 174 Chapter 4 Orbits in three dimensions ˆI,ˆJdirections into the ˆI/prime,ˆJ/primedirections. Viewed down the Zaxis, this rotation appears as shown in the insert at the top of the figure. The orthogonal transformation matrixassociated with this rotation is [R 3(/Omega1)]= cos/Omega1 sin/Omega1 0 −sin/Omega1 cos/Omega1 0 00 1  (4.39) Recall that the subscript on Rmeans that the rotation is around the ‘3’ direction, in this case the ˆKaxis. The second rotation, ②, is around the node line ( ˆI/prime), through the angle irequired to bring the XYplane parallel to the orbital plane. In other words, it rotates ˆKinto alignment with ˆw, andˆJ/primesimultaneously rotates into ˆJ/prime/prime. The insert in the lower right of Figure 4.15 shows how this rotation appears when viewed from the ˆI/primedirection. The orthogonal transformation matrix for this rotation is [R1(i)]= 10 0 0c o s isini 0−sin icosi  (4.40) The third and final rotation, ③, is in the orbital plane and rotates the unit vectors ˆI/prime andˆJ/prime/primethrough the angle ωaround the ˆwaxis so that they become aligned with ˆpand ˆq, respectively. This rotation appears from the ˆwdirection as shown in the insert on the left of Figure 4.15. The orthogonal transformation matrix is seen to be [R3(ω)]= cosωsinω0 −sinω cosω0 00 1  (4.41) Finally, let us note that the transformation matrix [ Q]X¯xfrom the geocentric equa- torial frame into the perifocal frame is just the product of the three rotation matricesgiven by Equations 4.39, 4.40 and 4.41; i.e., [Q] X¯x=[R3(ω)][R 1(i)][R 3(/Omega1)] (4.42) Substituting the three matrices on the right and carrying out the matrix multiplica- tions yields [Q]X¯x = cos/Omega1cosω−sin/Omega1sinωcosi sin/Omega1cosω+cos/Omega1cosisinωsinisinω −cos/Omega1sinω−sin/Omega1cosicosω−sin/Omega1sinω+cos/Omega1cosicosω sinicosω sin/Omega1sini −cos/Omega1sini cosi  (4.43) Remember, this is an orthogonal matrix, so that for the inverse transformation, from ¯x¯y¯ztoXYZ we have [Q ]¯xX=([Q]X¯x)T,o r [Q]¯xX = cos/Omega1cosω−sin/Omega1sinωcosi−cos/Omega1sinω−sin/Omega1cosicosω sin/Omega1sini sin/Omega1cosω+cos/Omega1cosisinω−sin/Omega1sinω+cos/Omega1cosicosω−cos/Omega1sini sinisinω sinicosω cosi  (4.44) 4.6 Transformation between geocentric equatorial and perifocal frames 175 If the components of the state vector are given in the geocentric equatorial frame r={r}X=  X Y Z  v={v}X=  vX vY vZ   the components in the perifocal frame are found by carrying out the matrix multiplications {r}¯x=  ¯x ¯y 0  =[Q]X¯x{r}X {v}¯x=  ˙¯x ˙¯y 0  =[Q]X¯x{v}X (4.45) Likewise, the transformation from perifocal to geocentric equatorial components is {r}X=[Q]¯xX{r}¯x{v}X=[Q]¯xX{v}¯x (4.46) Algorithm 4.2Given the orbital elements h,e,i,/Omega1,ωandθ, compute the position vectors rand v in the geocentric equatorial frame of reference. A MATLAB implementation of this procedure is listed in Appendix D.9. This algorithm can be applied to orbits aroundother planets or the sun. 1. Calculate position vector {r}¯xin perifocal coordinates using Equation 4.37. 2. Calculate velocity vector {v}¯xin perifocal coordinates using Equation 4.38. 3. Calculate the matrix [ Q]¯xXof the transformation from perifocal to geocentric equatorial coordinates using Equation 4.44. 4. Transform {r}¯xand{v}¯xinto the geocentric frame by means of Equations 4.46. Example 4.5For a given earth orbit, the elements are h=80 000 km2/s, e=1.4, i=30◦, /Omega1=40◦,ω=60◦andθ=30◦. Using Algorithm 4.2 find the state vectors rand v in the geocentric equatorial frame. Step 1: {r}¯x=h2 µ1 1+ecosθ  cosθ sinθ 0  =80 0002 398 6001 1+1.4 cos 30◦  cos 30◦ sin 30◦ 0  =  6285.0 3628.6 0  km Step 2: {v}¯x=µ h  −sinθ e+cosθ 0  =398 600 80 000  −sin 30◦ 1.4+cos 30◦ 0  =  −2.4913 11.290 0  km/s Step 3: [Q]X¯x= cosωsinω0 −sinω cosω0 00 1  10 0 0c o s isini 0−sin icosi  cos/Omega1 sin/Omega1 0 −sin/Omega1 cos/Omega1 0 00 1  176 Chapter 4 Orbits in three dimensions (Example 4.5 continued) = cos 60◦sin 60◦0 −sin 60◦cos 60◦0 00 1  10 0 0 cos 30◦sin 30◦ 0−sin 30◦cos 30◦  cos 40◦sin 40◦0 −sin 40◦cos 40◦0 00 1  = −0.099068 0 .89593 0 .43301 −0.94175 −0.22496 0 .25 0.32139 −0.38302 0 .86603  This is the transformation matrix for XYZ→¯x¯y¯z. The transformation matrix for ¯x¯y¯z→XYZ is the transpose, [Q]¯xX= −0.099068 −0.94175 0 .32139 0.89593 −0.22496 −0.38302 0.43301 0 .25 0 .86603  Step 4:The geocentric equatorial position vector is {r} X=[Q]¯xX{r}¯x = −0.099068 −0.94175 0 .32139 0.89593 −0.22496 −0.38302 0.43301 0 .25 0 .86603   6285.0 3628.6 0  =  −4040 48153629  (km) (a) whereas the geocentric equatorial velocity vector is {v}X=[Q]¯xX{v}¯x = −0.099068 −0.94175 0 .32139 0.89593 −0.22496 −0.38302 0.43301 0 .25 0 .86603   −2.4913 11.290 0  =  −10.39 −4.772 1.744  (km/s) The state vectors rand vare shown in Figure 4.16. By holding all of the orbital parameters except the true anomaly fixed and allowing θto take on a range of values, we generate a sequence of position vectors r¯xfrom Equations 4.37. Each of these is projected into the geocentric equatorial frame as in (a), using repeatedly the same transformation matrix [ Q]¯xX. By connecting the end points of all of the position vectors rX, we trace out the trajectory illustrated in Figure 4.16. XZ v Descending node Ω /H11005 40°u /H11005 30° i /H11005 30° Ascending nodePerigee Yv /H11005 60°r Figure 4.16 A portion of the hyperbolic trajectory of Example 4.5. 4.7 Effects of the earth’s oblateness 177 4.7 Effects of the earth’s oblateness The earth, like all of the planets with comparable or higher rotational rates, bulges out at the equator because of centrifugal force. The earth’s equatorial radius is 21 km(13 miles) larger than the polar radius. This flattening at the poles is called oblateness,which is defined as follows oblateness =equatorial radius −polar radius equatorial radius The earth is an oblate spheroid, lacking the pe rfect symmetry of a sphere. (A basketball can be made an oblate spheroid by sitting on it.) This lack of symmetry means that the force of gravity on an orbiting body is not directed towards the center of theearth. Whereas the gravitational field of a per fectly spherical planet depends only on the distance from its center, oblateness causes a variation also with latitude, that is,the angular distance from the equator (or pole). This is called a zonal variation. Thedimensionless parameter which quantifies the major effects of oblateness on orbits isJ 2, the second zonal harmonic. J2is not a universal constant. Each planet has its own value, as illustrated in Table 4.3, which lists variations of J2as well as oblateness. The gravitational acceleration (force per unit mass) arising from an oblate planet is given by ¨r=−µ r2ˆur+p The first term on the right is the familiar one (Equation 2.15) due to a spherical planet. The second term, p, which is several orders of magnitude smaller than µ/r2, is a perturbing acceleration due to the oblateness. This perturbing acceleration canbe resolved into components, p=p rˆur+p⊥ˆu⊥+phˆh where ˆur,ˆu⊥andˆhare the radial, transverse and normal unit vectors attached to the satellite, as illustrated in Figure 4.17. ˆurpoints in the direction of the radial position T able 4.3 Oblateness and second zonal harmonics Planet Oblateness J 2 Mercury 0.000 60×10−6 Venus 0.000 4.458×10−6 Earth 0.003353 1.08263 ×10−3 Mars 0.00648 1 .96045 ×10−3 Jupiter 0.06487 14 .736×10−3 Saturn 0.09796 16 .298×10−3 Uranus 0.02293 3.34343 ×10−3 Neptune 0.01708 3 .411×10−3 Pluto 0.000 – (Moon) 0.0012 202.7×10−6 178 Chapter 4 Orbits in three dimensions ˆh r XYZ ˆ u⊥ urˆ Figure 4.17 Unit vectors attached to an orbiting body. vector r,ˆhis the unit vector normal to the plane of the orbit and ˆu⊥is perpendicular tor, lying in the orbital plane and pointing in the direction of the motion. The perturbation components pr,p⊥andphare all directly proportional to J2and are functions of otherwise familiar orbital parameters as well as the planet radius R, pr=−µ r23 2J2/parenleftbiggR r/parenrightbigg2/bracketleftbig 1−3 sin2isin2(ω+θ)/bracketrightbig p⊥=−µ r23 2J2/parenleftbiggR r/parenrightbigg2 sin2isin[2(ω+θ)] ph=−µ r23 2J2/parenleftbiggR r/parenrightbigg2 sin 2 isin(ω+θ) These relations are derived by Prussing and Conway (1993), who also show how pr, p⊥and phinduce time rates of change in all of the orbital parameters. For example, ˙/Omega1=h µsin(ω+θ) sini(1+ecosθ)ph ˙ω=−rcosθ ehpr+(2+ecosθ)sinθ ehp⊥−rsin(ω+θ) htaniph Clearly, the time variation of the right ascension /Omega1depends only on the component of the perturbing force normal to the (instantaneous) orbital plane, whereas the rate ofchange of the argument of perigee is influenced by all three perturbation components. Integrating ˙/Omega1over one complete orbit yields the average rate of change, ˙/Omega1 avg=1 T/integraldisplayT 0˙/Omega1dt 4.7 Effects of the earth’s oblateness 179 20 40 60 80/H110028/H110026/H110024/H1100220 100 0/H110022 /H1100210 i, degrees300 km500 km700 km900 km1100 kme /H11005 0.001 90Ω, degrees per day 20 40 60 80051015 0 100/H11002563.4 i, degrees20 300 km 500 km 700 km 900 km 1100 kme /H11005 0.001v, degrees per day Figure 4.18 Regression of the node and advance of perigee for nearly circular orbits of altitudes 300 to 1100 km. where Tis the period. Carrying out the mathematical details leads to an expression for the average rate of precession of the node line, and hence, the orbital plane, ˙/Omega1=−/bracketleftBigg 3 2√µJ2R2 /parenleftbig 1−e2/parenrightbig2a7 2/bracketrightBigg cosi (4.47) where we have dropped the subscript avg.Randµare the radius and gravitational parameter of the planet, aand eare the semimajor axis and eccentricity of the orbit, and iis the orbit’s inclination. Observe that if 0 ≤i<90◦, then ˙/Omega1<0. That is, for posigrade orbits, the node line drifts westward. Since the right ascension ofthe node continuously decreases, this phenomenon is called regression of the nodes.If 90 ◦<i≤180◦, we see that ˙/Omega1>0. The node line of retrograde orbits therefore advances eastward. For polar orbits ( i=90◦), the node line is stationary. In a similar fashion the time rate of change of the argument of perigee is found to be ˙ω=−/bracketleftBigg 3 2√µJ2R2 /parenleftbig 1−e2/parenrightbig2a7 2/bracketrightBigg/parenleftbigg5 2sin2i−2/parenrightbigg (4.48) This expression shows that if 0◦≤i<63.4◦or 116 .6◦<i≤180◦then˙ωis positive, which means the perigee advances in the direction of the motion of the satellite (hence,the name advance of perigee for this phenomenon). If 63 .4 ◦<i≤116.6◦, the perigee regresses, moving opposite to the direction of motion. i=63.4◦and i=116.6◦are the critical inclinations at which the apse line does not move. Observe that the coefficient of the trigonometric terms in Equations 4.47 and 4.48 are identical. Figure 4.18 is a plot of Equations 4.47 and 4.48 for several low-earth orbits. The effect of oblateness on both ˙/Omega1and˙ωis greatest at low inclinations, for which the orbit is near the equatorial bulge for longer portions of each revolution. The effectdecreases with increasing semimajor axis because the satellite becomes further fromthe bulge and its gravitational influence. Obviously, ˙/Omega1=˙ω=0i fJ 2=0 (no equatorial bulge). 180 Chapter 4 Orbits in three dimensions The time averaged rates of change for the inclination, eccentricity and semimajor axis are zero. Example 4.6The space shuttle is in a 280 km by 400 km orbit with an inclination of 51.43◦. Find the rates of node regression and perigee advance. The perigee and apogee radii are rp=6378+280=6658 km ra=6378+400=6778 km Therefore the eccentricity and semimajor axis are e=ra−rp ra+rp=0.008931 a=1 2(ra+rp)=6718 km From Equation 4.47 we obtain the rate of node line regression: ˙/Omega1=−/bracketleftBigg 3 2√ 398 600 ·0.0010826 ·63782 /parenleftbig 1−0.00893122/parenrightbig2·67187 2/bracketrightBigg cos 51.43◦ =− 1.6786×10−6·cos 51.43◦ =− 1.0465×10−6rad/s or ˙/Omega1=5.181◦per day to the west From Equation 4.48, ˙ω=s a m ea si n ˙/Omega1/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright −1.6786×10−6·/parenleftbigg5 2sin251.43◦−2/parenrightbigg =+ 7.9193×10−7rad/s or ˙ω=3.920◦per day in the flight direction The effect of orbit inclination on node regression and advance of perigee is taken advantage of for two very important types of orbits. Sun-synchronous orbits arethose whose orbital plane makes a constant angle αwith the radial from the sun, as illustrated in Figure 4.19. For that to occur, the orbital plane must rotate in inertialspace with the angular velocity of the earth in its orbit around the sun, which is 360 ◦ per 365.26 days, or 0.9856◦per day. With the orbital plane precessing eastward at this rate, the ascending node will lie at a fixed local time. In the illustration it hap-pens to be 3 pm. During every orbit, the satellite sees any given swath of the planetunder nearly the same conditions of daylight or darkness day after day. The satel-lite also has a constant perspective on the sun. Sun-synchronous satellites, like theNOAA Polar-orbiting Operational Environmental Satellites (NOAA/POES) and those 4.7 Effects of the earth’s oblateness 181 N N Na a a gggAscending node (a.n.) a.n. a.n.Sun Sun-synchronous orbitEarth's orbit0.9856° 24 hr 0.9856° 24 hr12 noon 12 noon 12 noon3 PM 3 PM 3PMΩ Ω Ω Figure 4.19 Sun-synchronous orbit. of the Defense Meteorological Satellite Program (DMSP) are used for global weather coverage, while Landsat and the French SPOT series are intended for high-resolutionearth observation. Example 4.7A satellite is to be launched into a sun-synchronous circular orbit with period of 100 minutes. Determine the required altitude and inclination of its orbit. We find the altitude zfrom the period relation for a circular orbit, Equation 2.54: T=2π√µ(RE+z)3 2⇒100·60=2π√ 398 600(6378 +z)3 2⇒z=758.63 km For a sun-synchronous orbit, the ascending node must advance at the rate ˙/Omega1=2πrad 365.26 ·24·3600 s=1.991×10−7rad/s Substituting this and the altitude into Equation 4.47, we obtain, 1.991×10−7=−/bracketleftBigg 3 2√ 398 600 ·0.00108263 ·63782 (1−02)2(6378+758.63 )7 2/bracketrightBigg cosi⇒cosi=− 0.14658 Thus, the inclination of the orbit is i=cos−1(−0.14658) =98.43◦ This illustrates the fact that sun-synchronous orbits are very nearly polar orbits (i=90◦). 182 Chapter 4 Orbits in three dimensions N g PerigeeApogee Figure 4.20 A typical Molniya orbit (to scale). Moscow Molniya is visible from Moscow when the track is north of this curve.180W 90N 60N30N 0 30S 60S90S90N 60N30N 0 30S 60S 90S 180W 150W 120W 90W 60W 30W 0 30E 60E 90E 120E 150E 180E150W 120W 90W 60W 30W 0 30E 60E 90E 120E 150E 180E Figure 4.21 Ground track of a Molniya satellite. Tick marks are one hour apart. 4.7 Effects of the earth’s oblateness 183 If a satellite is launched into an orbit with an inclination of 63.4◦(prograde) or 116.6◦(retrograde), then Equation 4.48 shows that the apse line will remain stationary. The Russian space program made this a key element in the design of thesystem of Molniya (‘lightning’) communications satellites. All of the Russian launchsites are above 45 ◦latitude, the northernmost, Plesetsk, being located at 62.8◦N. As we shall see in Chapter 6, launching a satellite into a geostationary orbit would involve a costly plane change maneuver. Furthermore, recall from Example 2.4 thata geostationary satellite cannot view effectively the far northern latitudes into whichRussian territory extends. The Molniya telecommunications satellites are launched from Plesetsk into 63 ◦ inclination orbits having a period of 12 hours. From Equation 2.73 we conclude thatthe apse line of these orbits is 53 000 km long. Perigee (typically 500 km altitude) lies in the southern hemisphere, while apogee is at an altitude of 40 000 km (25 000 miles)above the northern latitudes, farther out than the geostationary satellites. Figure 4.20illustrates a typical Molniya orbit, and Figure 4.21 shows a ground track. A Molniya ‘constellation’ consists of eight satellites in planes separated by 45 ◦. Each satellite is above 30◦north latitude for over eight hours, coasting towards and away from apogee. Example 4.8Determine the perigee and apogee for an earth satellite whose orbit satisfies all of the following conditions: it is sun-synchronous, its argument of perigee is constant, andits period is three hours. The period determines the semimajor axis, T=2π √µa3 2⇒3·3600=2π√ 398 600a3 2⇒a=10 560 km For the apse line to be stationary we know from Equation 4.48 that i=64.435◦ ori=116.57◦. But an inclination of less than 90◦causes a westward regression of the node, whereas a sun-synchonous orbit requires an eastward advance, which i=116.57◦provides. Substituting this, the semimajor axis and the ˙/Omega1in radians per second for a sun-synchronous orbit (cf. Example 4.7) into Equation 4.47, we get 1.991×10−7=−3 2√ 398 600 ·0.0010826 ·63782 /parenleftbig 1−e2/parenrightbig2·10 5607 2cos 116.57◦⇒e=0.3466 Now we can find the angular momentum from the period expression (Equation 2.72) T=2π µ2/parenleftbiggh√ 1−e2/parenrightbigg3 ⇒3·3600=2π 398 6002/parenleftbiggh√ 1−0.346552/parenrightbigg3 ⇒h=60 850 km2/s Finally, to obtain the perigee and apogee radii, we use the orbit formula: zp+6378=h2 µ1 1+e=60 8602 398 6001 1+0.34655⇒zp=522.6k m za+6378=h2 µ1 1−e⇒za=7842 km 184 Chapter 4 Orbits in three dimensions Example 4.9Given the following state vector of a satellite in geocentric equatorial coordinates, r=− 3670ˆI−3870ˆJ+4400ˆKkm v=4.7ˆI−7.4ˆJ+1ˆKkm/s find the state vector four days (96 hours) later, assuming that there are no perturbations other than the influence of the earth’s oblateness on /Omega1andω. Four days is a long enough time interval that we need to take into consideration not only the change in true anomaly but also the regression of the ascending node and the advance of perigee. First we must determine the orbital elements at the initial time using Algorithm 4.1, which yields h=58 930 km2/s i=39.687◦ e=0.42607 (the orbit is an ellipse) /Omega10=130.32◦ ω0=42.373◦ θ0=52.404◦ We use Equation 2.61 to determine the semimajor axis, a=h2 µ1 1−e2=58 9302 398 6001 1−0.42612=10 640 km so that, according to Equation 2.73, the period is T=2π√µa3 2=10 928 s From this we obtain the mean motion n=2π T=0.00057495 rad/s The initial value E0of eccentric anomaly is found from the true anomaly θ0using Equation 3.10a, tanE0 2=/radicalbigg 1−e 1+etanθ0 2=/radicalbigg 1−0.42607 1+0.42607tan52.404◦ 2⇒E0=0.60520 rad With E0, we use Kepler’s equation to calculate the time t0since perigee at the initial epoch, nt0=E0−esinE0⇒0.00057495 t0=0.60520 −0.42607 sin 0 .60520 ⇒t1=631.00 s Now we advance the time to tf, that of the final epoch, given as 96 hours later. That is, /Delta1t=345 600 s, so that tf=t1+/Delta1t=631.00+345 600 =346 230 s 4.7 Effects of the earth’s oblateness 185 The number of periods nPsince passing perigee in the first orbit is nP=tf T=346 230 10 928=31.682 From this we see that the final epoch occurs in the 32nd orbit, whereas t0was in orbit 1. Time since passing perigee in the 32nd orbit, which we will denote t32,i s t32=(31.682 −31)T⇒t32=7455.7s The mean anomaly corresponding to that time in the 32nd orbit is M32=nt32=0.00057495 ·7455.7 =4.2866 rad Kepler’s equation yields the eccentric anomaly E32−esinE32=M32⇒E32−0.42607 sin E32=4.2866 ⇒E32=3.9721 rad (Algorithm 3.1) The true anomaly follows in the usual way, tanθ32 2=/radicalbigg 1+e 1−etanE32 2⇒θ32=211.25◦ At this point, we use the newly found true anomaly to calculate the state vector of the satellite in perifocal coordinates. Thus, from Equation 4.35 r¯x=rcosθ32ˆp+rsinθ32ˆq=− 11 714 ˆp−7108.8 ˆq(km) or, in matrix notation, {r}¯x=  −11 714 −7108.8 0  (km) Likewise, from Equation 4.36, v¯x=−µ hsinθ32ˆp+µ h(e+cosθ32)ˆq=3.5093 ˆp−2.9007 ˆq(km/s) or {v}¯x=  3.5093 −2.9007 0  (km/s) Before we can project r¯xand v¯xinto the geocentric equatorial frame, we must update the right ascension of the node and the argument of perigee. The regression rate of the ascending node is ˙/Omega1=−/bracketleftBigg 3 2√µJ2R2 /parenleftbig 1−e2/parenrightbig2a7 2/bracketrightBigg cosi=−3 2√ 398 600 ·00108263 ·63782 /parenleftbig 1−0.426072/parenrightbig2·10 6447 2cos 39.69◦ =− 3.8514 ×10−7(rad/s) =− 2.2067 ×10−5◦/s Therefore, right ascension at epoch in the 32nd orbit is /Omega132=/Omega10+˙/Omega1/Delta1t=130.32 +(−2.2067 ×10−5)·345 600 =122.70◦ 186 Chapter 4 Orbits in three dimensions (Example 4.9 continued)Likewise, the perigee advance rate is ˙ω=− 3 2√µJ2R2 /parenleftbig 1−e2/parenrightbig2a7 2 /parenleftbigg5 2sin2i−2/parenrightbigg =4.9072 ×10−7rad/s=2.8116 ×10−5◦/s which means the argument of perigee at epoch in the 32nd orbit is ω32=ω0+˙ω/Delta1t=42.373 +2.8116 ×10−5·345 600 =52.090◦ Substituting the updated values of /Omega1andω, together with the inclination i, into Equation 4.43 yields the updated transformation matrix from geocentric equatorial to the perifocal frame, [Q]X¯x= cosω32 sinω32 0 −sinω32 cosω32 0 00 1  10 0 0c o s isini 0−sin icosi  cos/Omega132 sin/Omega132 0 −sin/Omega132 cos/Omega132 0 00 1  = cos 52.09◦sin 52.09◦0 −sin 52.09◦cos 52.09◦0 00 1  10 0 0 cos 39 .687◦sin 39.687◦ 0−sin 39.687◦cos 39.687◦  × cos 122.70◦sin 122.70◦0 −sin 122.70◦cos 122.70◦0 00 1  or [Q]X¯x= −0.84285 0.18910 0.50383 0.028276 −0.91937 0.39237 0.53741 0.34495 0.76955  For the inverse transformation, from perifocal to geocentric equatorial, we need the transpose of this matrix, [Q]¯xX= −0.84285 0.18910 0.50383 0.028276 −0.91937 0.39237 0.53741 0.34495 0.76955 T = −0.84285 0.028276 0.53741 0.18910 −0.91937 0.34495 0.50383 0.39237 0.76955  Thus, according to Equations 4.46, the final state vector in the geocentric equatorial frame is {r}X=[Q]¯xX{r}¯x = −0.84285 0.028276 0.53741 0.18910 −0.91937 0.34495 0.50383 0.39237 0.76955   −11 714 −7108.8 0  =  9672 4320 −8691  (km) {v}X=[Q]¯xX{v}¯x = −0.84285 0.028276 0.53741 0.18910 −0.91937 0.34495 0.50383 0.39237 0.76955   3.5093 −2.9007 0  =  −3.040 3.330 0.6299  (km/s) Problems 187 or, in vector notation, rX=9672ˆI+4320ˆJ−8691ˆK(km) vX=− 3.040ˆI+3.330ˆJ+0.6299 ˆK(km/s) The two orbits are plotted in Figure 4.22. X YZ g 130.3°122.7°52.09° 42.37° Orbit 1 Orbit 32rt /H11005 0 rt /H11005 96 hrOrbit 1 Orbit 32Perigees Node lines Figure 4.22 The initial and final position vectors. Problems 4.1 Find the orbital elements of a geocentric satellite whose inertial position and velocity vectors in a geocentric equatorial frame are r=2615ˆI+15 881 ˆJ+3980ˆK(km) v=− 2.767ˆI−0.7905 ˆJ+4.980ˆK(km/s) {Ans.: e=0.3760, h=95 360 km2/s,i=63.95◦,/Omega1=73.71◦,ω=15.43◦,θ=0.06764◦} 4.2 At a given instant the position rand velocity vof a satellite in the geocentric equatorial frame are r=12 670 ˆK(km) and v=− 3.874ˆJ−0.7905 ˆK(km/s). Find the orbital elements.{Ans.: h=49 080 km 2/s,e=0.5319, /Omega1=90◦,ω=259.5◦,θ=190.5◦,i=90◦} 4.3 At time tothe position rand velocity vof a satellite in the geocentric equatorial frame are r=6472.7 ˆI−7470.8 ˆJ−2469.8 ˆK(km) and v=3.9914 ˆI+2.7916 ˆJ−3.2948 ˆK(km/s). Find the orbital elements. {Ans.: h=58 461 km2/s,e=0.2465, /Omega1=110◦,ω=75◦,θ=130◦,i=35◦} 4.4 Given that, with respect to the geocentric equatorial frame, r=− 6634.2 ˆI−1261.8 ˆJ−5230.9 ˆK(km), v=5.7644 ˆI−7.2005 ˆJ−1.8106 ˆK(km/s) 188 Chapter 4 Orbits in three dimensions and the eccentricity vector is e=− 0.40907ˆI−0.48751ˆJ−0.63640 ˆK(dimensionless) calculate the true anomaly θof the earth-orbiting satellite. {Ans.: 330◦} 4.5 Given that, relative to the geocentric equatorial frame, r=− 6634.2ˆI−1261.8ˆJ−5230.9ˆK(km) the eccentricity vector is e=− 0.40907ˆI−0.48751ˆJ−0.63640 ˆK(dimensionless) and the satellite is flying towards perigee, calculate the inclination of the orbit. {Ans.: 69.3◦} 4.6 The right-handed, primed xyzsystem is defined by the three points A,BandC.T h e x/primey/prime plane is defined by the plane ABC .T h e x/primeaxis runs from Athrough B.T h e z/primeaxis is defined by the cross product of→ ABinto→ AC, so that the +y/primeaxis lies on the same side of thex/primeaxis as point C. (a) Find the orthogonal transformation matrix [Q]relating the two coordinate bases. (b) If the components of a vector vin the primed system are ⌊2−13⌋T, find the components of vin the unprimed system. {Ans.: ⌊−1.307 2 .390 2 .565⌋T} xy zx'y'z' C (3, 9, /H110022) B (4, 6, 5) A (1, 2, 3) Figure P .4.6 4.7 The unit vectors in a uvwcartesian coordinate frame have the following components in thexyzframe ˆu=0.26726ˆi+0.53452ˆj+0.80178 ˆk ˆv=− 0.44376ˆi+0.80684ˆj−0.38997 ˆk ˆw=− 0.85536ˆi−0.25158ˆj+0.45284 ˆk If, in the xyzframe, V=− 50ˆi+100ˆj+75ˆk, find the components of the vector Vin the uvwframe. {Ans.: V=100.2ˆu+73.62ˆv+51.57ˆw} Problems 189 4.8 Calculate the transformation matrix [Q] for the sequence of two rotations: α=40◦ about the positive Xaxis, followed by β=25◦about the positive y/primeaxis. The result is that the XYZ axes are rotated into the x/prime/primey/primez/prime/primeaxes. {Partial ans.: Q11=0.9063 Q12=0.2716 Q13=− 0.3237} XYZ y'z' /H9251/H9252z'' x''/H9252 /H9251/H9251 /H9252 Figure P .4.8 4.9 At time tothe position rand velocity vof a satellite in the geocentric equatorial frame are r=− 5102ˆI−8228ˆJ−2105ˆK(km) v=− 4.348ˆI+3.478ˆJ−2.846ˆK(km/s) Find rand vat time to+50 minutes. ( to/negationslash=0!) {Ans.: r=− 4198ˆI+7856ˆJ−3199ˆK(km); v=4.952ˆI+3.482ˆJ+2.495ˆK(km/s)} 4.10 For a spacecraft, the following orbital parameters are given: e=1.5; perigee altitude =300 km; i=35◦;/Omega1=130◦;ω=115◦. Calculate rand vat perigee relative to (a) the perifocal reference frame, and (b) the geocentric equatorial frame. {Ans.: (a) r=6678ˆp(km), v=12.22ˆq(km/s) (b)r=− 1984ˆI−5348ˆJ+3471ˆK(km), v=10.36ˆI−5.763ˆJ−2.961ˆK(km/s)} 4.11 For the spacecraft of Problem 4.10 calculate rand vat two hours past perigee relative to (a) the perifocal reference frame, and (b) the geocentric equatorial frame. {Ans.: (a) r=− 25 010 ˆp+48 090 ˆq(km), v=− 4.335ˆp+5.075ˆq(km/s) (b)r=48 200 ˆI−2658ˆJ−24 660 ˆK(km), v=5.590ˆI+1.078ˆJ−3.484ˆK(km/s)} 4.12 Calculate rand vfor the satellite in Problem 4.3 at time t0+50 minutes. ( to/negationslash=0!) {Ans.: r=6864ˆI+5916ˆJ−5933ˆK(km), v=− 3.564ˆI+3.905ˆJ+1.410ˆK(km/s)} 4.13 For a spacecraft, the following orbital parameters are given: e=1.2; perigee altitude =200 km; i=50◦;/Omega1=75◦;ω=80◦. Calculate rand vat perigee relative to (a) the perifocal reference frame, and (b) the geocentric equatorial frame. {Ans.: (a) r=6578ˆp(km); v=11.55ˆq(km/s) (b)r=− 3726ˆI+2181ˆJ+4962ˆK(km), v=− 4.188ˆI−10.65ˆJ+1.536ˆK(km/s)} 190 Chapter 4 Orbits in three dimensions 4.14 For the spacecraft of Exercise 4.13 calculate rand vat two hours past perigee relative to (a) the perifocal reference frame, and (b) the geocentric equatorial frame. {Ans.: (a) r=− 26 340 ˆp+37 810 ˆq(km), v=− 4.306ˆp+3.298ˆq(km/s) (b)r=1207ˆI−43 600 ˆJ−14 840 ˆK(km), v=1.243ˆI−4.4700ˆJ−2.810ˆK(km/s)} 4.15 Given that e=0.7,h=75 000 km2/s, and θ=25◦, calculate the components of velocity in the geocentric equatorial frame if [ Q]X¯x= −0.83204 −0.13114 0 .53899 0.02741 −0.98019 −0.19617 0.55403 −0.14845 0 .81915 . {Ans.: v=2.103ˆI−8.073ˆJ−2.885ˆK(km/s)} 4.16 The apse line of the elliptical orbit lies in the XYplane of the geocentric equatorial frame, whose Zaxis lies in the plane of the orbit. At B(for which θ=140◦) the perifo- cal velocity vector is {v}¯x=⌊ − 3.208−0.8288 0 ⌋T(km/s). Calculate the geocentric equatorial components of the velocity at B. {Ans.: {v}X=⌊ − 1.604−2.778−0.8288⌋T(km/s)} 60° XYZ B PerigeeApogee Figure P .4.16 4.17 A satellite in earth orbit has the following orbital parameters: a=7016 km, e=0.05, i=45◦,/Omega1=0◦,ω=20◦andθ=10◦. Find the position vector in the geocentric- equatorial frame.{Ans.: r=5776.4ˆI+2358.2ˆJ+2358.2ˆK(km)} 4.18 Calculate the orbital inclination required to place an earth satellite in a 500 km by 1000 km sun-synchronous orbit.{Ans.: 98.37 ◦} 4.19 The space shuttle is in a circular orbit of 180 km altitude and inclination 30◦. What is the spacing, in kilometers, between successive ground tracks at the equator, includingthe effect of earth’s oblateness?{Ans.: 2511 km} Problems 191 4.20 A satellite in a circular, sun-synchronous low earth orbit passes over the same point on the equator once each day, at 12 o’clock noon. Calculate the inclination, altitude andperiod of the orbit.{This problem has more than one solution.} 4.21 The orbit of a satellite around an unspecified planet has an inclination of 40 ◦, and its perigee advances at the rate of 7◦per day. At what rate does the node line regress? {Ans.: ˙/Omega1=5.545◦/day} 4.22 At a given time, the position and velocity of an earth satellite in the geocentric equatorial frame are r=− 2429.1 ˆI+4555.1 ˆJ+4577.0 ˆK(km) and v=− 4.7689 ˆI− 5.6113 ˆJ+3.0535 ˆK(km/s). Find rand vprecisely 72 hours later, taking into consideration the node line regression and the advance of perigee.{Ans.: r=4596ˆI+5759ˆJ−1266ˆKkm, v=− 3.601ˆI+3.179ˆJ+5.617ˆKkm/s} This page int entionally left blank 5Chapter Preliminary orbit determination Chapter outline 5.1 Introduction 193 5.2 Gibbs’ method of orbit determination from three position vectors 194 5.3 Lambert’s problem 202 5.4 Sidereal time 213 5.5 Topocentric coordinate system 218 5.6 Topocentric equatorial coordinate system 221 5.7 Topocentric horizon coordinate system 223 5.8 Orbit determination from angle and range measurements 228 5.9 Angles-only preliminary orbit determination 2355.10 Gauss’s method of preliminary orbit determination 236 Problems 250 5.1 Introduction In this chapter we will consider some (by no means all) of the classical ways in which the orbit of a satellite can be determined from earth-bound observations. All of the methods presented here are based on the two-body equations of motion. As such, they must be considered preliminary orbit determination techniques becausethe actual orbit is influenced over time by other phenomena (perturbations), such asthe gravitational force of the moon and sun, atmospheric drag, solar wind and thenon-spherical shape and non-uniform mass distribution of the earth. We took a brief 193 194 Chapter 5 Preliminary orbit determination look at the dominant effects of the earth’s oblateness in Section 4.7. T o accurately propagate an orbit into the future from a set of initial observations requires takingthe various perturbations, as well as instrumentation errors themselves, into account.More detailed considerations, including the means of updating the orbit on the basisof additional observations, are beyond our scope. Introductory discussions may befound elsewhere – see Bate, Mueller and White (1971), Boulet (1991), Prussing andConway (1993) and Wiesel (1997), to name but a few. We begin with the Gibbs method of predicting an orbit using three geocentric position vectors. This is followed by a presentation of Lambert’s problem, in whichan orbit is determined from two position vectors and the time between them. Boththe Gibbs and Lambert procedures are based on the fact that two-body orbits lie ina plane. The Lambert problem is more complex and requires using the Lagrange f andgfunctions introduced in Chapter 2 as well as the universal variable formulation introduced in Chapter 3. The Lambert algorithm is employed in Chapter 8 to analyzeinterplanetary missions. In preparation for explaining how satellites are tracked, the Julian day numbering scheme is introduced along with the notion of sidereal time. This is followed bya description of the topocentric coordinate systems and the relationships amongtopocentric right ascension/declension angles and azimuth/elevation angles. We then describe how orbits are determined from measuring the range and angular orientationof the line of sight together with their rates. The chapter concludes with a presentationof the Gauss method of angles-only orbit determination. 5.2 Gibbs’ method of orbit determination from three position vectors Suppose that from observations of a space object at the three successive times t1,t2 and t3(t1<t2<t3) we have obtained the geocentric position vectors r1,r2and r3. The problem is to determine the velocities v1,v2and v3att1,t2and t3assuming that the object is in a two-body orbit. The solution using purely vector analysis is dueto J. W. Gibbs (1839–1903), an American scholar who is known primarily for hiscontributions to thermodynamics. Our explanation is based on that in Bate, Mueller and White (1971). We know that the conservation of angular momentum requires that the position vectors of an orbiting body must lie in the same plane. In other words, the unitvector normal to the plane of r 2and r3must be perpendicular to the unit vector in the direction of r1. Thus, if ˆur1=r1/r1andˆC23=(r2×r3)//bardblr2×r3/bardbl, then the dot product of these two unit vectors must vanish, ˆur1·ˆC23=0 Furthermore, as illustrated in Figure 5.1, the fact that r1,r2and r3lie in the same plane means we can apply scalar factors c1and c3tor1and r3so that r2is the vector sum of c1r1and c3r3 r2=c1r1+c3r3 (5.1) The coefficients c1and c3are readily obtained from r1,r2and r3as we shall see in Section 5.10 (Equations 5.89 and 5.90). 5.2 Gibbs’ method of orbit determinat ion from three position vectors 195 r1r3r2 c3r3 c1r1 Figure 5.1 Any one of a set of three coplanar vectors ( r1,r2,r3) can be expressed as the vector sum of the other two. T o find the velocity vcorresponding to any of the three given position vectors r, we start with Equation 2.30, which may be written v×h=µ/parenleftBigr r+e/parenrightBig where his the angular momentum and eis the eccentricity vector. T o isolate the velocity, take the cross product of this equation with the angular momentum, h×(v×h)=µ/parenleftbiggh×r r+h×e/parenrightbigg (5.2) By means of the bac−cabrule (Equation 2.23), the left side becomes h×(v×h)=v(h·h)−h(h·v) But h·h=h2and v×h=0, since vis perpendicular to h. Therefore h×(v×h)=h2v which means Equation 5.2 may be written v=µ h2/parenleftbiggh×r r+h×e/parenrightbigg (5.3) In Section 2.10 we introduced the perifocal coordinate system, in which the unit vector ˆplies in the direction of the eccentricity vector eandˆwis the unit vector normal to the orbital plane, in the direction of the angular momentum vector h. Thus, we can write e=eˆp (5.4a) h=hˆw (5.4b) 196 Chapter 5 Preliminary orbit determination so that Equation 5.3 becomes v=µ h2/parenleftbigghˆw×r r+hˆw×eˆp/parenrightbigg =µ h/bracketleftbiggˆw×r r+e(ˆw׈p)/bracketrightbigg (5.5) Sinceˆp,ˆqandˆwform a right-handed triad of unit vectors, it follows that ˆp׈q=ˆw, ˆq׈w=ˆpand ˆw׈p=ˆq (5.6) Therefore, Equation 5.5 reduces to v=µ h/parenleftbiggˆw×r r+eˆq/parenrightbigg (5.7) This is an important result, because if we can somehow use the position vectors r1,r2and r3to calculate ˆq,ˆw,hand e, then the velocities v1,v2and v3will each be determined by this formula. So far the only condition we have imposed on the three position vectors is that they are coplanar (Equation 5.1). T o bring in the fact that they describe an orbit, letus take the dot product of Equation 5.1 with the eccentricity vector eto obtain the scalar equation r 2·e=c1r1·e+c3r3·e (5.8) According to Equation 2.34 – the orbit equation – we have the following relation among h,eand each of the position vectors, r1·e=h2 µ−r1 r2·e=h2 µ−r2 r3·e=h2 µ−r3 (5.9) Substituting these relations into Equation 5.8 yields h2 µ−r2=c1/parenleftbiggh2 µ−r1/parenrightbigg +c3/parenleftbiggh2 µ−r3/parenrightbigg (5.10) T o eliminate the unknown coefficients c1and c2from this expression, let us take the cross product of Equation 5.1 first with r1and then r3. This results in two equations, both having r3×r1on the right, r2×r1=c3(r3×r1) r2×r3=− c1(r3×r1) (5.11) Now multiply Equation 5.10 through by the vector r3×r1to obtain h2 µ(r3×r1)−r2(r3×r1)=c1(r3×r1)/parenleftbiggh2 µ−r1/parenrightbigg +c3(r3×r1)/parenleftbiggh2 µ−r3/parenrightbigg Using Equations 5.11, this becomes h2 µ(r3×r1)−r2(r3×r1)=− (r2×r3)/parenleftbiggh2 µ−r1/parenrightbigg +(r2×r1)/parenleftbiggh2 µ−r3/parenrightbigg 5.2 Gibbs’ method of orbit determinat ion from three position vectors 197 Observe that c1and c2have been eliminated. Rearranging terms we get h2 µ(r1×r2+r2×r3+r3×r1)=r1(r2×r3)+r2(r3×r1)+r3(r1×r2) (5.12) This is an equation involving the given position vectors and the unknown angular momentum h. Let us introduce the following notation for the vectors on each side of Equation 5.12, N=r1(r2×r3)+r2(r3×r1)+r3(r1×r2) (5.13) and D=r1×r2+r2×r3+r3×r1 (5.14) Then Equation 5.12 may be written more simply as N=h2 µD from which we obtain N=h2 µD (5.15) where N=/bardblN/bardbland D=/bardblD/bardbl. It follows from Equation 5.15 that the angular momentum his determined from r1,r2and r3by the formula h=/radicalbigg µN D(5.16) Since r1,r2and r3are coplanar, all of the cross products r1×r2,r2×r3and r3×r1 lie in the same direction, namely, normal to the orbital plane. Therefore, it is clear from Equation 5.14 that Dmust be normal to the orbital plane. In the context of the perifocal frame, we use ˆwto denote the orbit unit normal. Therefore, ˆw=D D(5.17) So far we have found handˆwin terms of r1,r2and r3. We need likewise to find an expression for ˆqto use in Equation 5.7. From Equations 5.4a, 5.6, and 5.17 it follows that ˆq=ˆw׈p=1 De(D×e) (5.18) Substituting Equation 5.14 we get ˆq=1 De[(r1×r2)×e+(r2×r3)×e+(r3×r1)×e] (5.19) We can apply the bac−cabrule to the right side by noting (A×B)×C=− C×(A×B)=B(A·C)−A(B·C) 198 Chapter 5 Preliminary orbit determination Using this vector identity we obtain (r2×r3)×e=r3(r2·e)−r2(r3·e) (r3×r1)×e=r1(r3·e)−r3(r1·e) (r1×r2)×e=r2(r1·e)−r1(r2·e) Once again employing Equations 5.9, these become (r2×r3)×e=r3/parenleftbiggh2 µ−r2/parenrightbigg −r2/parenleftbiggh2 µ−r3/parenrightbigg =h2 µ(r3−r2)+r3r2−r2r3 (r3×r1)×e=r1/parenleftbiggh2 µ−r3/parenrightbigg −r3/parenleftbiggh2 µ−r1/parenrightbigg =h2 µ(r1−r3)+r1r3−r3r1 (r1×r2)×e=r2/parenleftbiggh2 µ−r1/parenrightbigg −r1/parenleftbiggh2 µ−r2/parenrightbigg =h2 µ(r2−r1)+r2r1−r1r2 Summing these three equations, collecting terms and substituting the result into Equation 5.19 yields ˆq=1 DeS (5.20) where S=r1(r2−r3)+r2(r3−r1)+r3(r1−r2) (5.21) Finally, we substitute Equations 5.16, 5.17 and 5.20 into Equation 5.7 to obtain v=µ h/parenleftbiggˆw×r r+eˆq/parenrightbigg =µ/radicalBig µN D/bracketleftBiggD D×r r+e/parenleftbigg1 DeS/parenrightbigg/bracketrightBigg Simplifying this expression for the velocity yields v=/radicalbiggµ ND/parenleftbiggD×r r+S/parenrightbigg (5.22) All of the terms on the right depend only on the given position vectors r1,r2and r3. The Gibbs procedure may be summarized in the following algorithm. Algorithm 5.1Gibbs’ method of preliminary orbit determination. A MATLAB implementation of this procedure is found in Appendix D.10. Given r1,r2and r3, the steps are as follows. 1. Calculate r1,r2and r3. 2. Calculate C12=r1×r2,C23=r2×r3and C31=r3×r1. 3. Verify that ˆur1·ˆC23=0. 4. Calculate N,Dand Susing Equations 5.13, 5.14 and 5.21, respectively. 5. Calculate v2using Equation 5.22. 6. Use r2and v2to compute the orbital elements by means of Algorithm 4.1. 5.2 Gibbs’ method of orbit determinat ion from three position vectors 199 Example 5.1The geocentric position vectors of a space object at three successive times are r1=− 294.32 ˆI+4265.1 ˆJ+5986.7 ˆK(km) r2=− 1365.5 ˆI+3637.6 ˆJ+6346.8 ˆK(km) r3=− 2940.3 ˆI+2473.7 ˆJ+6555.8 ˆK(km) Determine the classical orbital elements using Gibbs’ procedure. Step 1: r1=/radicalbig (−294.32)2+4265.12+5986.72=7356.5k m r2=/radicalbig (−1365.5)2+3637.62+6346.82=7441.7k m r3=/radicalbig (−2940.3)2+2473.72+6555.82=7598.9k m Step 2: C12=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆI ˆJ ˆK −294.32 4265.1 5986.7−1365.5 3637.6 6346.8/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =(5.292 ˆI−6.3066 ˆJ+4.7531 ˆK)×10 6(km2) C23=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆI ˆJ ˆK −1365.5 3637.6 6346.8−2940.3 2473.7 6555.8/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =(8.1473 ˆI−9.7095 ˆJ+7.3179 ˆK)×10 6(km2) C31=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆI ˆJ ˆK −2940.3 2473.7 6555.8−294.32 4265.1 5986.7/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =(−1.3151 ˆI+1.5673 ˆJ−1.1812 ˆK)×10 6(km2) Step 3: ˆC23=C23 /bardblC23/bardbl=8.1473 ˆI−9.7095 ˆJ+7.3179 ˆK/radicalbig 8.14732+(−9.7095 )2+7.31792 =0.55667 ˆI−0.66341 ˆJ+0.5000 ˆK Therefore ˆur1·ˆC23=−294.32 ˆI+4265.1 ˆJ+5986.7 ˆK 7356.5·(0.55667 ˆI−0.66341 ˆJ+0.5000 ˆK) =6.9200 ×10−20 This certainly is close enough to zero for our purposes. The three vectors r1,r2and r3are coplanar. 200 Chapter 5 Preliminary orbit determination (Example 5.1 continued)Step 4: N=r1C23+r2C31+r3C12 =7356.5[(8.1473ˆI−9.7095ˆJ+7.3179ˆK)×106] +7441.7[(−1.3151ˆI+1.5673ˆJ−1.1812ˆK)×106] +7598.9[(5.292ˆI−6.3066ˆJ+4.7531ˆK)×106] or N=(2.2807ˆI−2.7181ˆJ+2.0486ˆK)×109(km3) so that N=/radicalbig [2.28072+(−2.7181)2+2.04862]×1018 =4.0971×109(km3) D=C12+C23+C31 =[(5.292ˆI−6.3066ˆJ+4.7531ˆK)×106]+[(8.1473ˆI−9.7095ˆJ +7.3179ˆK)×106]+[(−1.3151ˆI+1.5673ˆJ−1.1812ˆK)×106] or D=(2.8797ˆI−3.4319ˆJ+2.5866ˆK)×106(km2) so that D=/radicalbig [2.87972+(−3.4319)2+2.58662]×1012 =5.1731×105(km2) Lastly, S=r1(r2−r3)+r2(r3−r1)+r3(r1−r2) =(−294.32ˆI+4265.1ˆJ+5986.7ˆK)(7441 .7−7598.9) +(−1365.5ˆI+3637.6ˆJ+6346.8ˆK)(7598 .9−7356.5) +(−2940.3ˆI+2473.7ˆJ+6555.8ˆK)(7356 .5−7441.7) or S=− 34 213 ˆI+533.51ˆJ+38 798 ˆK(km2) Step 5: v2=/radicalbiggµ ND/parenleftbiggD×r2 r2+S/parenrightbigg 5.2 Gibbs’ method of orbit determinat ion from three position vectors 201 =/radicalBigg 398 600 (4.0971 ×109)(5.1731 ×103) × /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆI ˆJ ˆK 2.8797 ×10 6−3.4319 ×1062.5866 ×106 −1365.5 3637.6 6346.8/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle 7441.7+/parenleftbigg−34 213 ˆI+533.51 ˆJ +38 798 ˆK/parenrightbigg  or v2=− 6.2171 ˆI−4.0117 ˆJ+1.5989 ˆK(km/s) Step 6: Using r2and v2, Algorithm 4.1 yields the orbital elements: a=8000 km e=0.1 i=60◦ /Omega1=40◦ ω=30◦ θ=50◦(for position vector r2) The orbit is sketched in Figure 5.2. XYZ r1r2r3 Ascending nodePerigee50° Figure 5.2 Sketch of the orbit of Example 5.1. 202 Chapter 5 Preliminary orbit determination 5.3 Lambert’s problem Suppose we know the position vectors r1and r2of two points P1and P2on the path of mass maround mass M, as illustrated in Figure 5.3. r1and r2determine the change in the true anomaly /Delta1θ, since cos/Delta1θ=r1·r2 r1r2(5.23) where r1=√r1·r1 r2=√r2·r2 (5.24) However, if cos /Delta1θ > 0, then /Delta1θlies in either the first or fourth quadrant, whereas if cos/Delta1θ < 0, then /Delta1θlies in the second or third quadrant. (Recall Figure 3.4.) The first step in resolving this quadrant ambiguity is to calculate the Zcomponent of r1×r2, (r1×r2)Z=ˆK·(r1×r2)=ˆK·(r1r2sin/Delta1θˆw)=r1r2sin/Delta1θ(ˆK·ˆw) where ˆwis the unit normal to the orbital plane. Therefore, ˆK·ˆw=cosi,w h e r e iis the inclination of the orbit, so that (r1×r2)Z=r1r2sin/Delta1θcosi (5.25) We use the sign of the scalar ( r1×r2)Zto determine the correct quadrant for /Delta1θ. Y XZ Trajectory Ascending node IK w Jr1Fundamental planer2/H9004/H9258i i MP1P2m cˆ ˆ ˆˆˆ Figure 5.3 Lambert’s problem. 5.3 Lambert’s problem 203 There are two cases to consider: prograde trajectories (0 <i<90◦), and retrograde trajectories (90◦<i<180◦). For prograde trajectories (like the one illustrated in Figure 5.3), cos i>0, so that if ( r1×r2)Z>0, then Equation 5.25 implies that sin /Delta1θ > 0, which means 0◦</Delta1 θ<180◦. Since /Delta1θtherefore lies in the first or second quadrant, it follows that/Delta1θi sg i v e nb yc o s−1(r1·r2/r1r2). On the other hand, if ( r1×r2)Z<0, Equa- tion 5.25 implies that sin /Delta1θ < 0, which means 180◦</Delta1 θ<360◦. In this case /Delta1θ lies in the third or fourth quadrant and is given by 360◦−cos−1(r1·r2/r1r2). For retrograde trajectories, cos i<0. Thus, if (r 1×r2)Z>0 then sin /Delta1θ < 0, which places /Delta1θin the third or fourth quadrant. Similarly, if ( r1×r2)Z>0,/Delta1θmust lie in the first or second quadrant. This logic can be expressed more concisely as follows: /Delta1θ=  cos −1/parenleftbiggr1·r2 r1r2/parenrightbigg if(r1×r2)Z≥0 360◦−cos−1/parenleftbiggr1·r2 r1r2/parenrightbigg if(r1×r2)Z<0prograde trajectory ............................................................. cos−1/parenleftbiggr1·r2 r1r2/parenrightbigg if(r1×r2)Z<0 360◦−cos−1/parenleftbiggr1·r2 r1r2/parenrightbigg if(r1×r2)Z≥0retrograde trajectory (5.26) J. H. Lambert (1728–1777) was a French-born German astronomer, physicist and mathematician. According to a theorem of Lambert, the transfer time /Delta1tfrom P1 toP2is independent of the orbit’s eccentricity and depends only on the sum r1+r2 of the magnitudes of the position vectors, the semimajor axis aand the length cof the chord joining P1and P2. It is noteworthy that the period (of an ellipse) and the specific mechanical energy are also independent of the eccentricity (Equations 2.73,2.70 and 2.100). If we know the time of flight /Delta1tfrom P 1toP2, then Lambert’s problem is to find the trajectory joining P1and P2. The trajectory is determined once we find v1, because, according to Equations 2.125 and 2.126, the position and velocity of anypoint on the path are determined by r 1and v1. That is, in terms of the notation in Figure 5.3, r2=fr1+gv1 (5.27a) v2=˙fr1+˙gv1 (5.27b) Solving the first of these for v1yields v1=1 g(r2−fr1) (5.28) Substitute this result into Equation 5.27b to get v2=˙fr1+˙g g(r2−fr1)=˙g gr2−f˙g−˙fg gr1 204 Chapter 5 Preliminary orbit determination But, according to Equation 2.129, f˙g−˙fg=1. Hence, v2=1 g(˙gr2−r1) (5.29) By means of Algorithm 4.1 we can find the orbital elements from either r1and v1or r2and v2. Clearly, Lambert’s problem is solved once we determine the Lagrange coefficients f,gand˙g. We will follow the procedure presented by Bate, Mueller and White (1971) and Bond and Allman (1996). The Lagrange fandgcoefficients and their time derivatives are listed as functions of the change in true anomaly /Delta1θin Equations 2.148, f=1−µr2 h2(1−cos/Delta1θ) g=r1r2 hsin/Delta1θ (5.30a) ˙f=µ h1−cos/Delta1θ sin/Delta1θ/bracketleftbiggµ h2(1−cos/Delta1θ)−1 r1−1 r2/bracketrightbigg ˙g=1−µr1 h2(1−cos/Delta1θ) (5.30b) Equations 3.66 express these quantities in terms of the universal anomaly χ, f=1−χ2 r1C(z) g=/Delta1t−1√µχ3S(z) (5.31a) ˙f=√µ r1r2χ[zS(z)−1] ˙g=1−χ2 r2C(z) (5.31b) where z=αχ2.T h e fand gfunctions do not depend on the eccentricity, which would seem to make them an obvious choice for the solution of Lambert’s problem.Ignoring for the time being that z=αχ 2, the unknowns on the right of the above sets of equations are h,χand z, whereas /Delta1θ,/Delta1t,rand r0are given. While /Delta1θappears throughout Equations 5.30, the time interval /Delta1tdoes not. However, /Delta1tdoes appear in Equation 5.31a. A relationship between /Delta1θand/Delta1tcan therefore be found by equating the two expressions for g, r1r2 hsin/Delta1θ=/Delta1t−1√µχ3S(z) (5.32) T o eliminate the unknown angular momentum h, equate the expressions for fin Equations 5.30a and 5.31a, 1−µr2 h2(1−cos/Delta1θ)=1−χ2 r1C(z) Upon solving this for hwe obtain h=/radicalBigg µr1r2(1−cos/Delta1θ) χ2C(z)(5.33) (Equating the two expressions for ˙gleads to the same result.) Substituting Equation 5.33 into 5.32, simplifying and rearranging terms yields √µ/Delta1t=χ3S(z)+χ/radicalbig C(z)/parenleftbigg sin/Delta1θ/radicalbiggr1r2 1−cos/Delta1θ/parenrightbigg (5.34) 5.3 Lambert’s problem 205 The term in parentheses on the right is a constant comprised solely of the given data. Let us assign it the symbol A, A=sin/Delta1θ/radicalbiggr1r2 1−cos/Delta1θ(5.35) Then Equation 5.34 assumes the simpler form √µ/Delta1t=χ3S(z)+Aχ/radicalbig C(z) (5.36) The right side of this equation contains both of the unknown variables χand z. We cannot use the fact that z=αχ2to reduce the unknowns to one since αis the reciprocal of the semimajor axis of the unknown orbit. In order to find a relationship between zandχwhich does not involve orbital parameters, we equate the expressions for ˙f(Equations 5.30b and 5.31b) to obtain µ h1−cos/Delta1θ sin/Delta1θ/bracketleftbiggµ h2(1−cos/Delta1θ)−1 r1−1 r2/bracketrightbigg =√µ r1r2χ[zS(z)−1] Multiplying through by r1r2and substituting for the angular momentum using Equation 5.33 yields µ/radicalBigg µr1r2(1−cos/Delta1θ) χ2C(z)1−cos/Delta1θ sin/Delta1θ µ µr1r2(1−cos/Delta1θ) χ2C(z)(1−cos/Delta1θ)−r1−r2  =√µχ[zS(z)−1] Simplifying and dividing out common factors leads to √1−cos/Delta1θ√r1r2sin/Delta1θ/radicalbig C(z)[χ2C(z)−r1−r2]=zS(z)−1 We recognize the reciprocal of Aon the left, so we can rearrange this expression to read as follows, χ2C(z)=r1+r2+AzS(z)−1√C(z) The right-hand side depends exclusively on z. Let us call that function y(z), so that χ=/radicalBigg y(z) C(z)(5.37) where y(z)=r1+r2+AzS(z)−1√C(z)(5.38) Equation 5.37 is the relation between χand zthat we were seeking. Substituting it back into Equation 5.36 yields √µ/Delta1t=/bracketleftbiggy(z) C(z)/bracketrightbigg3 2 S(z)+A/radicalbig y(z) (5.39) 206 Chapter 5 Preliminary orbit determination We can use this equation to solve for z, given the time interval /Delta1t. It must be done iteratively. Using Newton’s method, we form the function F(z)=/bracketleftbiggy(z) C(z)/bracketrightbigg3 2 S(z)+A/radicalbig y(z)−√µ/Delta1t (5.40) and its derivative F/prime(z)=1 2/radicalbig y(z)C(z)5/braceleftBig [2C(z)S/prime(z)−3C/prime(z)S(z)]y2(z) +/bracketleftBig AC(z)5 2+3C(z)S(z)y(z)/bracketrightBig y/prime(z)/bracerightBig (5.41) in which C/prime(z) and S/prime(z) are the derivatives of the Stumpff functions, which are given by Equations 3.60. y/prime(z) is obtained by differentiating y(z) in Equation 5.38, y/prime(z)=A 2C(z)3 2{[1−zS(z)]C/prime(z)+2[S(z)+zS/prime(z)]C(z)} If we substitute Equations 3.60 into this expression a much simpler form is obtained, namely y/prime(z)=A 4/radicalbig C(z) (5.42) This result can be worked out by using Equations 3.49 and 3.50 to express C(z) and S(z) in terms of the more familiar trig functions. Substituting Equation 5.42 along with Equations 3.60 into Equation 5.41 yields F/prime(z)=  /bracketleftbiggy(z) C(z)/bracketrightbigg3 2/braceleftbigg1 2z/bracketleftbigg C(z)−3 2S(z) C(z)/bracketrightbigg +3 4S(z)2 C(z)/bracerightbigg +A 8/bracketleftBigg 3S(z) C(z)/radicalbig y(z)+A/radicalBigg C(z) y(z)/bracketrightBigg (z/negationslash=0) √ 2 40y(0)3 2+A 8/bracketleftbigg/radicalbig y(0)+A/radicalbigg1 2y(0)/bracketrightbigg (z=0)(5.43) Evaluating F/prime(z)a tz=0 must be done carefully (and is therefore shown as a special case), because of the zin the denominator within the curly brackets. T o handle z=0, we assume that zis very small (almost, but not quite zero) so that we can retain just the first two terms in the series expansions of C(z) and S(z) (Equations 3.47 and 3.48), C(z)=1 2−z 24+... S(z)=1 6−z 120+... 5.3 Lambert’s problem 207 Then we evaluate the term within the curly brackets as follows: 1 2z/bracketleftbigg C(z)−3 2S(z) C(z)/bracketrightbigg ≈1 2z /parenleftbigg1 2−z 24/parenrightbigg −3 2/parenleftbigg1 6−z 120/parenrightbigg /parenleftbigg1 2−z 24/parenrightbigg  =1 2z/bracketleftBigg/parenleftbigg1 2−z 24/parenrightbigg −3/parenleftbigg1 6−z 120/parenrightbigg/parenleftbigg 1−z 12/parenrightbigg−1/bracketrightBigg ≈1 2z/bracketleftbigg/parenleftbigg1 2−z 24/parenrightbigg −3/parenleftbigg1 6−z 120/parenrightbigg/parenleftbigg 1+z 12/parenrightbigg/bracketrightbigg =1 2z/parenleftbigg −7z 120+z2 480/parenrightbigg =−7 240+z 960 In the third step we used the familiar binomial expansion theorem, (a+b)n=an+nan−1b+n(n−1) 2!an−2b2+n(n−1)(n−2) 3!an−3b3+... (5.44) to set (1 −z/12)−1≈1+z/12, which is true if zis close to zero. Thus, when zis actually zero, 1 2z/bracketleftbigg C(z)−3 2S(z) C(z)/bracketrightbigg =−7 240 Evaluating the other terms in F/prime(z) presents no difficulties. F(z) in Equation 5.40 and F/prime(z) in Equation 5.43 are used in Newton’s formula, Equation 3.13, for the iterative procedure, zi+1=zi−F(zi) F/prime(zi)(5.45) For choice of a starting value for z, recall that z=(1/a )χ2. According to Equation 3.54, z=E2for an ellipse and z=− F2for a hyperbola. Since we do not know what the orbit is, setting z0=0 seems a reasonable, simple choice. Alternatively, one can plot or tabulate F(z) and choose z0to be a point near where F(z) changes sign. Substituting Equations 5.37 and 5.39 into Equations 5.31 yields the Lagrange coefficients as functions of zalone: f=1−/bracketleftbigg/radicalbiggy(z) C(z)/bracketrightbigg2 r1C(z)=1−y(z) r1(5.46a) g=1√µ/braceleftBigg/bracketleftbiggy(z) C(z)/bracketrightbigg3 2 S(z)+A/radicalbig y(z)/bracerightBigg −1√µ/bracketleftbiggy(z) C(z)/bracketrightbigg3 2 S(z)=A/radicalBigg y(z) µ(5.46b) 208 Chapter 5 Preliminary orbit determination ˙f=√µ r1r2/radicalBigg y(z) C(z)[zS(z)−1] (5.46c) ˙g=1−/bracketleftbigg/radicalbiggy(z) C(z)/bracketrightbigg2 r2C(z)=1−y(z) r2(5.46d) We are now in a position to present the solution of Lambert’s problem in universal variables, following Bond and Allman (1996). Algorithm 5.2Solve Lambert’s problem. A MATLAB implementation appears in Appendix D.12. Given r1,r2and/Delta1t, the steps are as follows. 1. Calculate r1and r2using Equation 5.24. 2. Choose either a prograde or retrograde trajectory and calculate /Delta1θ using Equation 5.26. 3. Calculate Ain Equation 5.35. 4. By iteration, using Equations 5.40, 5.43 and 5.45, solve Equation 5.39 for z.T h e sign of ztells us whether the orbit is a hyperbola ( z<0), parabola ( z=0) or ellipse ( z>0). 5. Calculate yusing Equation 5.38. 6. Calculate the Lagrange f,gand˙gfunctions using Equations 5.46. 7. Calculate v1and v2from Equations 5.28 and 5.29. 8. Use r1and v1(orr2and v2) in Algorithm 4.1 to obtain the orbital elements. Example 5.2The position of an earth satellite is first determined to be r1=5000ˆI+10 000 ˆJ+ 2100ˆK(km). After one hour the position vector is r2=− 14 600 ˆI+2500ˆJ+ 7000ˆK(km). Determine the orbital elements and find the perigee altitude and the time since perigee passage of the first sighting. We first must execute the steps of Algorithm 5.2 in order to find v1and v2. Step 1: r1=/radicalbig 50002+10 0002+21002=11 375 km r2=/radicalbig (−14 600)2+25002+70002=16 383 km Step 2: assume a prograde trajectory: r1×r2=(64.75ˆI−65.66ˆJ+158.5ˆK)×106 cos−1r1·r2 r1r2=100.29◦ Since the trajectory is prograde and the zcomponent of r1×r2is positive, it follows from Equation 5.26 that /Delta1θ=100.29◦ 5.3 Lambert’s problem 209 Step 3: A=sin/Delta1θ/radicalbiggr1r2 1−cos/Delta1θ=sin 100.29◦/radicalbigg 11 375 ·16 383 1−cos 100.29◦=12 372 km Step 4: Using this value of Aand/Delta1t=3600 s, we can evaluate the functions F(z) and F/prime(z) given by Equations 5.40 and 5.43, respectively. Let us first plot F(z) to get at least a rough idea of where it crosses the zaxis. As can be seen from Figure 5.4, F(z)=0 near z=1.5. With z0=1.5 as our initial estimate, we execute Newton’s procedure, Equation 5.45, zi+1=zi−F(zi) F/prime(zi) z1=1.5−−14 476.4 362 642=1.53991 z2=1.53991 −23.6274 363 828=1.53985 z3=1.53985 −6.29457 ×10−5 363 826=1.53985 Thus, to five significant figures z=1.5398. The fact that zis positive means the orbit is an ellipse. Step 5: y=r1+r2+AzS(z)−1√C(z)=11 375 +16 383 +12 3721.5398S (1.5398)√C(1.5398)=13 523 km Step 6: Equations 5.46 yield the Lagrange functions f=1−y r1=1−13 523 11 375=− 0.18877 1 20F(z) z /H110025/H110031055/H11003105 Figure 5.4 Graph of F(z). 210 Chapter 5 Preliminary orbit determination (Example 5.2 continued) g=A/radicalbiggy µ=12 372/radicalbigg 13 523 398 600=2278.9s ˙g=1−y r2=1−13 523 16 383=0.17457 Step 7: v1=1 g(r2−fr1)=1 2278.9[(−14 600 ˆI+2500ˆJ+7000ˆK) −(−0.18877)(5000 ˆI+10 000 ˆJ+2100ˆK)] v1=− 5.9925ˆI+1.9254ˆJ+3.2456ˆK(km) v2=1 g(˙gr2−r1)=1 2278.9[(0.17457)( −14 600 ˆI+2500ˆJ+7000ˆK) −(5000ˆI+10 000 ˆJ+2100ˆK)] v2=− 3.3125ˆI−4.1966ˆJ−0.38529 ˆK(km) Step 8: Using r1and v1, Algorithm 4.1 yields the orbital elements: h=80 470 km2/s a=20 000 km e=0.4335 /Omega1=44.60◦ i=30.19◦ ω=30.71◦ θ1=350.8◦ This elliptical orbit is plotted in Figure 5.5. The perigee of the orbit is rp=h2 µ1 1+ecos (0)=80 4702 398 6001 1+0.4335=11 330 km Therefore the perigee altitude is 11 330 −6378=4952 km . T o find the time of the first sighting, we first calculate the eccentric anomaly by means of Equation 3.10b, E1=2 tan−1/parenleftBigg/radicalbigg 1−e 1+etanθ 2/parenrightBigg =2 tan−1/parenleftBigg/radicalbigg 1−0.4335 1+0.4335tan350.8◦ 2/parenrightBigg =2 tan−1(−0.05041) =− 0.1007 rad Then using Kepler’s equation for the el lipse (Equation 3.11), the mean anomaly is found to be Me1=E1−esinE1=− 0.1007−0.4335 sin( −0.1007) =− 0.05715 rad 5.3 Lambert’s problem 211 so that from Equation 3.4, the time since perigee passage is t1=h3 µ21 /parenleftbig 1−e2/parenrightbig3 2Me1=80 4703 398 60021 /parenleftbig 1−0.43352/parenrightbig3 2(−0.05715) =− 256.1s The minus sign means there are 256.1 seconds until perigee encounter after the initial sighting. XYZ /H9253r1r2 Perigee ApogeeAscending nodeDescending node Equatorial plane 44.6°EarthP1P2 Figure 5.5 The solution of Lambert’s problem. Example 5.3A meteoroid is sighted at an altitude of 267 000 km. 13.5 hours later, after a change in true anomaly of 5◦, the altitude is observed to be 140 000 km. Calculate the perigee altitude and the time to perigee after the second sighting. We have P1:r1=6378+267 000 =273 378 km P2:r2=6378+140 000 =146 378 km /Delta1t=13.5·3600=48 600 s /Delta1θ=5◦ Since r1,r2and/Delta1θare given, we can skip to step 3 of Algorithm 5.2 and compute A=2.8263 ×105km 212 Chapter 5 Preliminary orbit determination (Example 5.3 continued)Then, solving for zas in the previous example, we obtain z=− 0.17344 Since zis negative, the path of the meteoroid is a hyperbola. With zavailable, we evaluate the Lagrange functions, f=0.95846 g=47 708 s (a) ˙g=0.92241 Step 7 requires the initial and final position vectors. Therefore, for the purposes of this problem let us define a geocentric coordinate system with the xaxis aligned with r1and the yaxis at 90◦thereto in the direction of the motion (see Figure 5.6). The zaxis is therefore normal to the plane of the orbit. Then r1=r1ˆi=273 378 ˆi(km) r2=r2cos/Delta1θˆi+r2sin/Delta1θˆj=145 820 ˆi+12 758 ˆj(km) (b) With (a) and (b) we obtain the velocity at P1, v1=1 g(r2−fr1) =1 47 708[(145 820 ˆi+12 758 ˆj)−0.95846(273 378 ˆi)] =− 2.4356ˆi−0.26741ˆj(km/s) Using r1and v1, Algorithm 4.1 yields h=73 105 km2/s e=1.0506 θ1=205.16◦ The orbit is now determined except for its orientation in space, for which no information was provided. In the plane of the orbit, the trajectory is as shown inFigure 5.6. The perigee radius is rp=h2 µ1 1+ecos(0)=6538.2k m which means the perigee altitude is dangerously low for a large meteoroid, zp=6538.2−6378=160.2k m (100 miles) T o find the time of flight from P2to perigee, we note that the true anomaly of P2is θ2=θ1+5◦=210.16◦ 5.4 Sidereal time 213 The hyperbolic eccentric anomaly F2follows from Equation 3.42, F2=2 tanh−1/parenleftBigg/radicalbigg e−1 e+1tanθ2 2/parenrightBigg =− 1.3347 rad From this we appeal to Kepler’s equat ion (Equation 3.37) for the mean anomaly Mh, Mh2=esinh (F 2)−F2=− 0.52265 rad Finally, Equation 3.31 yields the time t2=Mh2h3 µ2/parenleftbig e2−1/parenrightbig3 2=− 38 396 s The minus sign means that 38 396 seconds (a scant 10.6 hours) remain until the meteoroid passes through perigee. P1P2210.16°205.16° 273 378 km 146 378 km r1r2 xy Figure 5.6 Solution of Lambert’s problem for the incoming meteoroid. 5.4 Sidereal time T o deduce the orbit of a satellite or celestial body from observations requires, among other things, recording the time of each observation. The time we use in every daylife, the time we set our clocks by, is solar time. It is reckoned by the motion of thesun across the sky. A solar day is the time required for the sun to return to the sameposition overhead, that is, to lie on the same meridian. A solar day – from high noonto high noon – comprises 24 hours. Universal time (UT) is determined by the sun’spassage across the Greenwich meridian, which is zero degrees terrestrial longitude. 214 Chapter 5 Preliminary orbit determination See Figure 1.9. At noon UT the sun lies on the Greenwich meridian. Local standard time, or civil time, is obtained from universal time by adding one hour for each timezone between Greenwich and the site, measured westward. Sidereal time is measured by the rotation of the earth relative to the fixed stars (i.e., the celestial sphere, Figure 4.3). The time it takes for a distant star to return toits same position overhead, i.e., to lie on the same meridian, is one sidereal day (24sidereal hours). As illustrated in Figure 4.19, the earth’s orbit around the sun resultsin the sidereal day being slightly shorter than the solar day. One sidereal day is 23hours and 56 minutes. T o put it another way, the earth rotates 360 ◦in one sidereal day whereas it rotates 360.986◦in a solar day. Local sidereal time θof a site is the time elapsed since the local meridian of the site passed through the vernal equinox. The number of degrees (measured eastward) between the vernal equinox and the local meridian is the sidereal time multipliedby 15. T o know the location of a point on the earth at any given instant relative tothe geocentric equatorial frame requires knowing its local sidereal time. The localsidereal time of a site is found by first determining the Greenwich sidereal time θ G (the sidereal time of the Greenwich meridian), and then adding the east longitude (or subtracting the west longitude) of the site. Algorithms for determining sidereal timerely on the notion of the Julian day (JD). The Julian day number is the number of days since noon UT on 1 January 4713 BC. The origin of this time scale is placed in antiquity so that, except for pre- historic events, we do not have to deal with positive and negative dates. The Julianday count is uniform and continuous and does not involve leap years or differ-ent numbers of days in different months. The number of days between two eventsis found by simply subtracting the Julian day of one from that of the other. TheJulian day begins at noon rather than at midnight so that astronomers observingthe heavens at night would not have to deal with a change of date during theirwatch. The Julian day numbering system is not to be confused with the Julian calendar which the Roman emperor Julius Caesar introduced in 46 BC. The Gregorian calendar, introduced in 1583, has largely supplanted the Julian calender and is in common civiluse today throughout much of the world. J 0is the symbol for the Julian day number at 0 hr UT (which is half way into the Julian day). At any other UT, the Julian day is given by JD=J0+UT 24(5.47) Algorithms and tables for obtaining J0from the ordinary year ( y), month ( m) and day (d) exist in the literature and on the World Wide Web. One of the simplest formulas is found in Boulet (1991), J0=367y−INT  7/bracketleftbigg y+INT/parenleftbiggm+9 12/parenrightbigg/bracketrightbigg 4  +INT/parenleftbigg275m 9/parenrightbigg +d+1 721 013 .5 (5.48) 5.4 Sidereal time 215 where y,mand dare integers lying in the following ranges 1901≤y≤2099 1≤m≤12 1≤d≤31 INT(x ) means to retain only the integer portion of x, without rounding (or, in other words, round towards zero); that is, INT( −3.9) =− 3 and INT(3.9) =3. Appendix D.12 lists a MATLAB implementation of Equation 5.48. Example 5.4What is the Julian day number for 12 May 2004 at 14:45:30 UT? In this case y=2004, m=5 and d=12. Therefore, Equation 5.48 yields the Julian day number at 0 hr UT, J0=367·2004−INT  7/bracketleftbigg 2004+INT/parenleftbigg5+9 12/parenrightbigg/bracketrightbigg 4  +INT/parenleftbigg275·5 9/parenrightbigg +12+1 721 013 .5 =735 468 −INT/braceleftbigg7[2004+1] 4/bracerightbigg +152+12+1 721 013 .5 =735 468 −3508+152+12+1 721 013 .5 or J0=2 453 137 .5d a y s The universal time, in hours, is UT=14+45 60+30 3600=14.758 hr Therefore, from Equation 5.47 we obtain the Julian day number at the desired UT, JD=2 453 137 .5+14.758 24=2 453 138 .115 days Example 5.5Find the elapsed time between 4 October 1957 UT 19:26:24 and the date of the previous example. Proceeding as in Example 5.4 we find that the Julian day number of the given event (the launch of the first man-made satellite, Sputnik I) is JD1=2 436 116 .3100 days The Julian day of the previous example is JD2=2 453 138 .1149 days Hence, the elapsed time is /Delta1JD=2 453 138 .1149−2 436 116 .3100=17 021.805 days (46 years, 220 days) 216 Chapter 5 Preliminary orbit determination The current Julian epoch is defined to have been noon on 1 January 2000. This epoch is denoted J2000 and has the exact Julian day number 2 451 545.0. Since there are365.25 days in a Julian year, a Julian century has 36 525 days. It follows that the timeT 0in Julian centuries between the Julian day J0and J2000 is T0=J0−2 451 545 36 525(5.49) The Greenwich sidereal time θG0at 0 hr UT may be found in terms of this dimen- sionless time (Seidelmann, 1992, Section 2.24). θG0in degrees is given by the series θG0=100.4606184 +36 000 .77004 T0+0.000387933 T2 0−2.583(10−8)T3 0(degrees) (5.50) This formula can yield a value outside of the range 0 ≤θG0≤360◦. If so, then the appropriate integer multiple of 360◦m u s tb ea d d e do rs u b t r a c t e dt ob r i n g θG0into that range. Once θG0has been determined, the Greenwich sidereal time θGat any other universal time are found using the relation θG=θG0+360.98564724UT 24(5.51) where UTis in hours. The coefficient of the second term on the right is the number of degrees the earth rotates in 24 hours (solar time). Finally, the local sidereal time θof a site is obtained by adding its east longitude /Lambda1to the Greenwich sidereal time, θ=θG+/Lambda1 (5.52) /H9253 North poleSiteGreenwich/H9258G/H9258 Greenwichat 0 hr UTΛ θG0 Figure 5.7 Schematic of the relationship among θG0,θG,/Lambda1andθ. 5.4 Sidereal time 217 Here again it is possible for the computed value of θto exceed 360◦. If so, it must be reduced to within that limit by subtracting the appropriate integer multiple of 360◦. Figure 5.7 illustrates the relationship among θG0,θG,/Lambda1andθ. Algorithm 5.3Calculate the local sidereal time, given the date, the local time and the east longitude of the site. This is implemented in MATLAB in Appendix D.13. 1. Using the year, month and day, calculate J0using Equation 5.48. 2. Calculate T0by means of Equation 5.49. 3. Compute θG0from Equation 5.50. If θG0lies outside the range 0◦≤θG0≤360◦, then subtract the multiple of 360◦required to place θG0in that range. 4. Calculate θGusing Equation 5.51. 5. Calculate the local sidereal time θby means of Equation 5.52, adjusting the final value so it lies between 0 and 360◦. Example 5.6Use Algorithm 5.3 to find the local sidereal time (in degrees) of T okyo, Japan, on 3 March 2004 at 4:30:00 UT. The east longitude of T okyo is 139.80◦. (This places T okyo nine time zones ahead of Greenwich, so the local time is 1:30 in theafternoon.) Step 1: J 0=367·2004−INT  7/bracketleftbigg 2004+INT/parenleftbigg3+9 12/parenrightbigg/bracketrightbigg 4  +INT/parenleftbigg275·3 9/parenrightbigg +3+1 721 013 .5 =2 453 067 .5d a y s Recall that the .5 means that we are half way into the Julian day, which began at noon UT of the previous day. Step 2: T0=2 453 067 .5−2 451 545 36 525=0.041683778 Step 3: θG0=100.4606184 +36 000.77004(0.041683778) +0.000387933(0.041683778)2−2.583(10−8)(0.041683778)3 =1601.1087◦ The right-hand side is too large. We must reduce θG0to an angle which does not exceed 360◦. T o that end observe that INT(1601.1087/360) =4 218 Chapter 5 Preliminary orbit determination (Example 5.6 continued)Hence, θG0=1601.1087−4·360=161.10873◦(a) Step 4: The universal time of interest in this problem is UT=4+30 60+0 3600=4.5h r Substitute this and (a) into Equation 5.51 to get the Greenwich sidereal time: θG=161.10873 +360.985647244.5 24=228.79354◦ Step 5: Add the east longitude of T okyo to this value to obtain the local sidereal time, θ=228.79354 +139.80=368.59◦ T o reduce this result into the range 0 ≤θ≤360◦we must subtract 360◦to get θ=368.59−360=8.59◦(0.573 hr) Observe that the right ascension of a celestial body lying on T okyo’s meridian is 8.59◦. 5.5 Topocentric coordinate system A topocentric coordinate system is one which is centered at the observer’s location on the surface of the earth. Consider an object B– a satellite or celestial body – and an observer Oon the earth’s surface, as illustrated in Figure 5.8. ris the position of the body Brelative to the center of attraction C;Ris the position vector of the observer relative to C; and/rho1is the position of the body Brelative to the observer. r,R and/rho1comprise the fundamental vector triangle. The relationship among these three vectors is r=R+/rho1 (5.53) As we know, the earth is not a sphere, but a slightly oblate spheroid. This ellipsoidal shape is exaggerated in Figure 5.8. The location of the observation site Ois determined by specifying its east longitude /Lambda1and latitude φ. East longitude /Lambda1is measured positive eastward from the Greenwich meridian to the meridian through O. The angle between the vernal equinox direction (XZ plane) and the meridian of Ois the local sidereal timeθ. Likewise, θGis the Greenwich sidereal time. Once we know θG, then the local sidereal time is given by Equation 5.52. Latitude φis the angle between the equator and the normal ˆnto the earth’s surface atO. Since the earth is not a perfect sphere, the position vector R, directed from the center Cof the earth to O, does not point in the direction of the normal except at the equator and the poles. 5.5 T opocentric coordinate system 219 X γˆ I ˆ J ˆ K Local meridianPolar axis Z, z/H11032 Y x/H11032 θ(East longitude)CO ReRRp C/H11032φ (latitude)Rφ ˆ n B (tracked object) r θGGreenwich meridian Equator Λ/rho1 Figure 5.8 Oblate spheroidal earth (exaggerated). The oblateness, or flattening f, was defined in Section 4.7, f=Re−Rp Re where Reis the equatorial radius and Rpis the polar radius. (Review from Table 4.3 that f=0.00335 for the earth.) Figure 5.9 shows the ellipse of the meridian through O. Obviously, Reand Rpare, respectively, the semimajor and semiminor axes of the ellipse. According to Equation 2.66, Rp=Ra/parenleftbig 1−e2/parenrightbig It is easy to show from the above two relations that flattening and eccentricity are related as follows e=/radicalBig 2f−f2 f=1−/radicalbig 1−e2 As illustrated in Figure 5.8 and again in Figure 5.9, the normal to the earth’s surface at Ointersects the polar axis at a point C/primewhich lies below the center Cof the earth (if Ois in the northern hemisphere). The angle φbetween the normal and the equator is called the geodetic latitude, as opposed to geocentric latitude φ/prime, which is the angle between the equatorial plane and line joining Oto the center of the earth. The distance from CtoC/primeisRφe2sin2φ,w h e r e Rφ, the distance from C/primetoO, is a function of latitude (Seidelmann, 1991, Section 4.22) Rφ=Re/radicalbig 1−e2sin2φ=Re/radicalBig 1−/parenleftbig 2f−f2/parenrightbig sin2φ(5.54) 220 Chapter 5 Preliminary orbit determination Thus, the meridional coordinates of Oare x/prime O=Rφcosφ z/prime O=/parenleftbig 1−e2/parenrightbig Rφsinφ=(1−f)2Rφsinφ If the observation point Ois at an elevation Habove the ellipsoidal surface, then we must add Hcosφtox/prime Oand Hsinφtoz/prime Oto obtain x/prime O=Rccosφ z/prime O=Rssinφ (5.55a) where Rc=Rφ+HR s=(1−f)2Rφ+H (5.55b) Observe that whereas Rcis the distance of Ofrom point C/primeon the earth’s axis, Rsis the distance from Oto the intersection of the line OC/primewith the equatorial plane. The geocentric equatorial coordinates of Oare X=x/prime Ocosθ Y=x/prime Osinθ Z=z/prime O where θis the local sidereal time given in Equation 5.52. Hence, the position vector Rshown in Figure 5.8 is R=RccosφcosθˆI+RccosφsinθˆJ+RssinφˆK C Equator ReONorth pole RpTangent Rfe2 sin fx'z'k' R C'f'f Rfi'n z'O x'Oˆ ˆ ˆ Figure 5.9 The relationship between geocentric latitude ( φ/prime) and geodetic latitude ( φ). 5.6 T opocentric equatorial coordinate system 221 Substituting Equation 5.54 and Equations 5.55b yields R= Re/radicalBig 1−/parenleftbig 2f−f2/parenrightbig sin2φ+H cosφ(cosθˆI+sinθˆJ) + Re/parenleftbig 1−f/parenrightbig2 /radicalBig 1−/parenleftbig 2f−f2/parenrightbig sin2φ+H sinφˆK (5.56) In terms of the geocentric latitude φ/prime R=Recosφ/primecosθˆI+Recosφ/primesinθˆJ+Resinφ/primeˆK By equating these two expressions for Rand setting H=0 it is easy to show that at sea level geodetic latitude is related to geocentric latitude φ/primeas follows, tanφ/prime=(1−f)2tanφ 5.6 Topocentric equatorial coordinate system The topocentric equatorial coordinate system with origin at point Oon the surface of the earth uses a non-rotating set of xyzaxes through Owhich coincide with the XYZ axes of the geocentric equatorial frame, as illustrated in Figure 5.10. As can be X /H9253IˆJˆK Y θCO R RerZ /H9253xyz /H9251/H9254B /H9278/H11032 Equatoriˆjˆk /rho1 ˆˆ Figure 5.10 T opocentric equatorial coordinate system. 222 Chapter 5 Preliminary orbit determination inferred from the figure, the relative position vector /rho1in terms of the topocentric right ascension and declination is /rho1=/rho1cosδcosαˆI+/rho1cosδsinαˆJ+/rho1sinδˆK since at all times, ˆi=ˆI,ˆj=ˆJandˆk=ˆKfor this frame of reference. We can write /rho1as /rho1=/rho1ˆ/rho1 where /rho1is the slant range and ˆ/rho1is the unit vector in the direction of /rho1, ˆ/rho1=cosδcosαˆI+cosδsinαˆJ+sinδˆK (5.57) Since the origins of the geocentric and topocentric systems do not coincide, the direction cosines of the position vectors rand/rho1will in general differ. In particular the topocentric right ascension and declination of an earth-orbiting body Bwill not be the same as the geocentric right ascension and declination. This is an example ofparallax. On the other hand, if /bardblr/bardbl/greatermuch/bardbl R/bardblthen the difference between the geocentric and topocentric position vectors, and hence the right ascension and declination, isnegligible. This is true for the distant planets and stars. Example 5.7At the instant when the Greenwich sidereal time is θG=126.7◦, the geocentric equatorial position vector of the International Space Station is r=− 5368ˆI−1784ˆJ+3691ˆK(km) Find the topocentric right ascension and declination at sea level ( H=0), latitude φ=20◦and east longitude /Lambda1=60◦. According to Equation 5.52, the local sidereal time at the observation site is θ=θG+/Lambda1=126.7+60=186.7◦ Substituting Re=6378 km, f=0.003353 (Table 4.3), θ=189.7◦andφ=20◦into Equation 5.56 yields the geocentric position vector of the site: R=− 5955ˆI−699.5ˆJ+2168ˆK(km) Having found R, we obtain the position vector of the space station relative to the site from Equation 5.53: /rho1=r−R =(−5368ˆI−1784ˆJ+3691ˆK)−(−5955ˆI−699.5ˆJ+2168ˆK) =586.8ˆI−1084ˆJ+1523ˆK(km) The magnitude of this vector is /rho1=1960 km, so that ˆ/rho1=/rho1 /rho1=0.2994ˆI−0.5533ˆJ+0.7773ˆK 5.7 T opocentric horizon coordinate system 223 Comparing this equation with Equation 5.57 we see that cosδcosα=0.2997 cosδsinα=− 0.5524 sinδ=0.7778 From these we obtain the topocentric declension, δ=sin−10.7773 =51.01◦(a) as well as sinα=−0.5533 cosδ=− 0.8795 cosα=0.2994 cosδ=0.4759 Thus α=cos−1(0.4759) =61.58◦(first quadrant) or 298 .4◦(fourth quadrant) Since sin αis negative, αmust lie in the fourth quadrant, so that the right ascension is α=298.4◦(b) Compare (a) and (b) with the geocentric right ascension α0and declination δ0, which were computed in Example 4.2, α0=198.4◦δ0=33.12◦ 5.7 Topocentric horizon coordinate system The topocentric horizon system was introduced in Section 1.6 and is illustrated again in Figure 5.11. It is centered at the observation point Owhose position vector is R. The xyplane is the local horizon, which is the plane tangent to the ellipsoid at point O.T h e zaxis is normal to this plane directed outward towards the zenith. The x axis is directed eastward and the yaxis points north. Because the xaxis points east, this may be referred to as an ENZ (East-North-Zenith) frame. In the SEZ topocentric reference frame the xaxis points towards the south and the yaxis towards the east. The SEZ frame is obtained from ENZ b ya9 0◦clockwise rotation around the zenith. Therefore, the matrix of the transformation from NEZ toSEZ is[R3(−90◦)],w h e r e [R3(φ)]is found in Equation 4.33. The position vector /rho1of a body Brelative to the topocentric horizon system in Figure 5.11 is /rho1=/rho1cosasinAˆi+/rho1cosacosAˆj+/rho1sinaˆk 224 Chapter 5 Preliminary orbit determination X /H9253JK Y θCO RZ φx (East)z (Zenith) ˆ iˆkBA aj y (North) C/H11032Equator I i/H11032ˆˆˆ /rho1 ˆ ˆ Figure 5.11 T opocentric horizon ( xyz) coordinate system on the surface of the oblate earth. in which /rho1is the range; Ais the azimuth measured positive clockwise from due north (0≤A≤360◦); and ais the elevation angle or altitude measured from the horizontal to the line of sight of the body B(−90◦≤a≤90). The unit vector ˆ/rho1in the line of sight direction is ˆ/rho1=cosasinAˆi+cosacosAˆj+sinaˆk (5.58) The transformation between geocentric equatorial and topocentric horizon systems is found by first determining the projections of the topocentric base vectors ˆiˆjˆkonto those of the geocentric equatorial frame. From Figure 5.11 it is apparent that ˆk=cosφˆi/prime+sinφˆK and ˆi/prime=cosθˆI+sinθˆJ where ˆi/primelies in the local meridional plane and is normal to the Zaxis. Hence ˆk=cosφcosθˆI+cosφsinθˆJ+sinφˆK (5.59) The eastward-directed unit vector ˆimay be found by taking the cross product of ˆK into the unit normal ˆk, ˆi=ˆK׈k/vextenddouble/vextenddouble/vextenddoubleˆK׈k/vextenddouble/vextenddouble/vextenddouble=−cosφsinθˆI+cosφcosθˆJ /radicalBig cos2φ/parenleftbig sin2θ+cos2θ/parenrightbig=− sinθˆI+cosθˆJ (5.60) 5.7 T opocentric horizon coordinate system 225 Finally, crossing ˆkintoˆiyieldsˆj, ˆj=ˆk׈i=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆI ˆJ ˆK cosφcosθcosφsinθsinφ −sinθ cosθ 0/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=− sinφcosθˆI−sinφsinθˆJ+cosφˆK (5.61) Let us denote the matrix of the transformation from geocentric equatorial to topocen- tric horizon as [ Q] Xx. Recall from Section 4.5 that the rows of this matrix comprise the direction cosines of ˆi,ˆjandˆk, respectively. It follows from Equations 5.59 through 5.61 that [Q]Xx= −sinθ cosθ 0 −sinφcosθ−sinφsinθcosφ cosφcosθ cosφsinθ sinφ  (5.62a) The reverse transformation, from topocentric horizon to geocentric equatorial, is represented by the transpose of this matrix, [Q]xX= −sinθ−sinφcosθcosφcosθ cosθ−sinφsinθcosφsinθ 0c o s φ sinφ  (5.62b) Observe that these matrices also represent the transformation between topocentric horizontal and topocentric equatorial frames because the unit basis vectors of thelatter coincide with those of the geocentric equatorial coordinate system. Example 5.8The east longitude and latitude of an observer near San Francisco are /Lambda1=238◦and φ=38◦, respectively. The local si dereal time, in degrees, is θ=215.1◦(12 hr 42 min). At that time the planet Jupiter is observed by means of a telescope to be located at azimuth A=214.3◦and angular elevation a=43◦. What are Jupiter’s right ascension and declination in the topocentric equatorial system? The given information allows us to formulate the matrix of the transformation from topocentric horizon to topocentric equatorial using Equation 5.62b, [Q]xX= −sin 215.1◦−sin 38◦cos 215.1◦cos 38◦cos 215.1◦ cos 215.1◦−sin 38◦sin 215.1◦cos 38◦sin 215.1◦ 0 cos 38◦sin 38◦  = 0.5750 0.5037 −0.6447 −0.8182 0.3540 −0.4531 00 .7880 0.6157  From Equation 5.58 we have ˆ/rho1=cosasinAˆi+cosacosAˆj+sinaˆk =cos 43◦sin 214.3◦ˆi+cos 43◦cos 214.3◦ˆj+sin 43◦ˆk =− 0.4121 ˆi−0.6042 ˆj+0.6820 ˆk 226 Chapter 5 Preliminary orbit determination (Example 5.8 continued)Therefore, in matrix notation the topocentric horizon components of ˆ/rho1are {ˆ/rho1}x=  −0.4121 −0.6042 0.6820   We obtain the topocentric equatorial components {ˆ/rho1}Xby the matrix operation {ˆ/rho1}X=[Q]xX{ˆ/rho1}x= 0.5750 0 .5037 −0.6447 −0.8182 0 .3540 −0.4531 00 .7880 0 .6157   −0.4121 −0.6042 0.6820   =  −0.9810 −0.1857 −0.05621   so that ˆ/rho1=− 0.9810ˆI−0.1857ˆJ−0.05621 ˆK Recall Equation 5.57, ˆ/rho1=cosδcosαˆI+cosδsinαˆJ+sinδˆK Comparing the Zcomponents of these two expressions, we see that sinδ=− 0.0562 which means the topocentric equatorial declension is δ=sin−1(−0.0562) =− 3.222◦ Equating the Xand Ycomponents yields sinα=−0.1857 cosδ=− 0.1860 cosα=−0.9810 cosδ=− 0.9825 Therefore, α=cos−1(−0.9825) =169.3◦(second quadrant) or 190 .7◦(fourth quadrant) Since sin αis negative, αis in the fourth quadrant, which means the topocentric equatorial right ascension is α=190.7◦ Jupiter is sufficiently far away that we can ignore the radius of the earth in Equa- tion 5.53. That is, to our level of precision, there is no distinction between the topocentric equatorial and geocentric equatorial systems: r≈/rho1 Therefore the topocentric right ascension and declination computed above are the same as the geocentric equatorial values. 5.7 T opocentric horizon coordinate system 227 Example 5.9At a given time, the geocentric equatorial p osition vector of the International Space Station is r=− 2032.4 ˆI+4591.2 ˆJ−4544.8 ˆK(km) Determine the azimuth and elevation angle relative to a sea-level ( H=0) observer whose latitude is φ=− 40◦and local sidereal time is θ=110◦. Using Equation 5.56 we find the position vector of the observer to be R=− 1673ˆI+4598ˆJ−4078ˆK(km) For the position vector of the space station relative to the observer we have (Equation 5.53) /rho1=r−R =(−2032 ˆI+4591ˆJ−4545ˆK)−(−1673 ˆI+4598ˆJ−4078ˆK) =− 359.0ˆI−6.342ˆJ−466.9ˆK(km) or, in matrix notation, {/rho1}X=  −359.0 −6.342−466.9  (km) T o transform these geocentric equatorial components into the topocentric horizon system we need the transformation matrix [ Q]Xx, which is given by Equation 5.62a, [Q]Xx= −sinθ cosθ 0 −sinφcosθ−sinφsinθcosφ cosφcosθ cosφsinθ sinφ  = −sin 110◦cos 110◦0 −sin(−40◦)cos 110◦−sin(−40◦)sin 110◦cos(−40◦) cos(−40◦)cos 110◦cos(−40◦)sin 110◦sin(−40◦)  Thus, {/rho1}x=[Q]Xx{/rho1}X= −0.9397 −0.3420 0 −0.2198 0.6040 0.7660−0.2620 0.7198 −0.6428   −359.0 −6.342−466.9   =  339.5 −282.6 389.6  (km) or, reverting to vector notation, /rho1=339.5ˆi−282.6ˆj+389.6ˆk(km) The magnitude of this vector is /rho1=589.0 km. Hence, the unit vector in the direction of/rho1is ˆ/rho1=/rho1 /rho1=0.5765 ˆi−0.4787 ˆj+0.6615 ˆk 228 Chapter 5 Preliminary orbit determination (Example 5.9 continued)Comparing this with Equation 5.58 we see that sin a=0.6615, so that the angular elevation is a=sin−10.6615=41.41◦ Furthermore sinA=0.5765 cosa=0.7687 cosA=−0.4787 cosa=− 0.6397 It follows that A=cos−1(−0.6397) =129.8◦(second quadrant) or 230 .2◦(third quadrant) Amust lie in the second quadrant because sin A>0. Thus the azimuth is A=129.8◦ 5.8 Orbit determination from angle and range measurements We know that an orbit around the earth is determined once the state vectors rand vin the inertial geocentric equatorial frame are provided at a given instant of time (epoch). Satellites are of course observed from the earth’s surface and not from itscenter. Let us briefly consider how the state vector is determined from measurementsby an earth-based tracking station. The fundamental vector triangle formed by the topocentric position vector /rho1of a satellite relative to a trac king station, the position vector Rof the station relative to the center of attraction Cand the geocentric position vector rwas illustrated in Figure 5.8 and is shown again schematically in Figure 5.12. The relationship amongthese three vectors is given by Equation 5.53, which can be written r=R+/rho1ˆ/rho1 (5.63) where the range /rho1is the distance of the body Bfrom the tracking site and ˆ/rho1is the unit vector containing the directional information about B. By differentiating Equation 5.63 with respect to time we obtain the velocity vand acceleration a, v=˙r=˙R+˙/rho1ˆ/rho1+/rho1˙ˆ/rho1 (5.64) a=¨r=¨R+¨/rho1ˆ/rho1+2˙/rho1˙ˆ/rho1+/rho1¨ˆ/rho1 (5.65) The vectors in these equations must all be expressed in the common basis ( ˆIˆJˆK)o f the inertial (non-rotating) geocentric equatorial frame. Since Ris a vector fixed in the earth, whose constant angular velocity is /Omega1=ω EˆK (see Equation 2.57), it follows from Equations 1.24 and 1.25 that ˙R=/Omega1×R (5.66) 5.8 Orbit determination from angle and range measurements 229 r RO CB /H9253/rho1 Figure 5.12 Earth-orbiting body Bt r a c k e db ya no b s e r v e rO . ¨R=/Omega1×(/Omega1×R) (5.67) IfLX,LYand LZare the topocentric equatorial direction cosines, then the direction cosine vector ˆ/rho1is ˆ/rho1=LXˆI+LYˆJ+LZˆK (5.68) and its first and second derivatives are ˙ˆ/rho1=˙LXˆI+˙LYˆJ+˙LZˆK (5.69) and ¨ˆ/rho1=¨LXˆI+¨LYˆJ+¨LZˆK (5.70) Comparing Equations 5.57 and 5.68 reveals that the topocentric equatorial direction cosines in terms of the topocentric right ascension αand declension δare   LX LY LZ  =  cosαcosδ sinαcosδ sinδ  (5.71) Differentiating this equation twice yields   ˙LX ˙LY ˙LZ  =  −˙αsinαcosδ−˙δcosαsinδ ˙αcosαcosδ−˙δsinαsinδ ˙δcosδ  (5.72) and   ¨LX ¨LY ¨LZ  =  −¨αsinαcosδ−¨δcosαsinδ−/parenleftbig ˙α2+˙δ2/parenrightbig cosαcosδ+2˙α˙δsinαsinδ ¨αcosαcosδ−¨δsinαsinδ−/parenleftbig ˙α2+˙δ2/parenrightbig sinαcosδ−2˙α˙δcosαsinδ ¨δcosδ−˙δ2sinδ   (5.73) 230 Chapter 5 Preliminary orbit determination Equations 5.71 through 5.73 show how the direction cosines and their rates are obtained from the right ascension and declination and their rates. In the topocentric horizon system, the relative position vector is written ˆ/rho1=lxˆi+lyˆj+lzˆk (5.74) where, according to Equation 5.58, the direction cosines lx,lyand lzare found in terms of the azimuth Aand elevation aas  lx ly lz  =  sinAcosa cosAcosa sina  (5.75) LX,LYand LZare obtained from lx,lyand lzby the coordinate transformation   LX LY LZ  =[Q]xX  lx ly lz  (5.76) where [ Q]xXis given by Equation 5.62b. Thus   LX LY LZ  = −sinθ−cosθsinφcosθcosφ cosθ−sinθsinφsinθcosφ 0c o s φ sinφ   sinAcosa cosAcosa sina  (5.77) Substituting Equation 5.71 we see that topoc entric right ascension/declination and azimuth/elevation are related by   cosαcosδ sinαcosδ sinδ  = −sinθ−cosθsinφcosθcosφ cosθ−sinθsinφsinθcosφ 0c o s φ sinφ   sinAcosa cosAcosa sina   Expanding the right-hand side and solving for sin δ, sinαand cos αwe get sinδ=cosφcosAcosa+sinφsina (5.78a) sinα=(cosφsina−cosAcosasinφ) sinθ+cosθsinAcosa cosδ(5.78b) cosα=(cosφsina−cosAcosasinφ)c o sθ−sinθsinAcosa cosδ(5.78c) We can simplify Equations 5.78b and 5.78c by introducing the hour angle h, h=θ−α (5.79) his the angular distance between the object and the local meridian. If his positive, the object is west of the meridian; if his negative, the object is east of the meridian. Using well-known trig identities we have sin(θ−α)=sinθcosα−cosθsinα (5.80a) cos(θ−α)=cosθcosα+sinθsinα (5.80b) Substituting Equations 5.78b and 5.78c on the right of 5.80a and simplifying yields sin(h)=−sinAcosa cosδ(5.81) 5.8 Orbit determination from angle and range measurements 231 Likewise, Equation 5.80b leads to cos(h)=cosφsina−sinφcosAcosa cosδ(5.82) We calculate hfrom this equation, resolving quadrant ambiguity by checking the sign of sin( h). That is, h=cos−1/parenleftbiggcosφsina−sinφcosAcosa cosδ/parenrightbigg if sin(h ) is positive. Otherwise, we must subtract hfrom 360◦. Since both the elevation angle aand the declension δlie between −90◦and+90◦, neither cos anor cos δcan be negative. It follows from Equation 5.81 that the sign of sin( h) depends only on that of sin A. T o summarize, given the topocentric azimuth Aand altitude aof the target together with the sidereal time θand latitude φof the tracking station, we compute the topocentric declension δand right ascension αas follows, δ=sin−1(cosφcosAcosa+sinφsina) (5.83a) h=  2π−cos−1/parenleftbiggcosφsina−sinφcosAcosa cosδ/parenrightbigg 0◦<A<180◦ cos−1/parenleftbiggcosφsina−sinφcosAcosa cosδ/parenrightbigg 180◦≤A≤360◦(5.83b) α=θ−h (5.83c) IfAand aare provided as functions of time, then αandδare found as functions of time by means of Equations 5.83. The rates ˙α,¨α,˙δand¨δare determined by differentiating α(t) andδ(t) and substituting the results into Equations 5.68 through 5.73 to calculate the direction cosine vector ˆ/rho1and its rates ˙ˆ/rho1and¨ˆ/rho1. It is a relatively simple matter to find ˙αand˙δin terms of ˙Aand˙a. Differentiating Equation 5.78a with respect to time yields ˙δ=1 cosδ[−˙AcosφsinAcosa+˙a(sinφcosa−cosφcosAsina)] (5.84) Differentiating Equation 5.81, we get ˙hcos(h)=−1 cos2δ[(˙AcosAcosa−˙asinAsina)c o sδ+˙δsinAcosasinδ] Substituting Equation 5.82 and simplifying leads to ˙h=−˙AcosAcosa−˙asinAsina+˙δsinAcosatanδ cosφsina−sinφcosAcosa But˙h=˙θ−˙α=ωE−˙α, so that, finally, ˙α=ωE+˙AcosAcosa−˙asinAsina+˙δsinAcosatanδ cosφsina−sinφcosAcosa(5.85) 232 Chapter 5 Preliminary orbit determination Algorithm 5.4.Given the range /rho1, azimuth A, angular elevation atogether with the rates ˙/rho1,˙Aand ˙arelative to an earth-based tracking station, calculate the state vectors rand vin the geocentric equatorial frame. A MATLAB script of this procedure appears in Appendix D.14. 1. Using the altitude H, latitude φand local sidereal time θof the site, calculate its geocentric position vector Rfrom Equation 5.56: R= Re/radicalBig 1−(2f−f2) sin2φ+H cosφ/parenleftBig cosθˆI+sinθˆJ/parenrightBig + Re(1−f)2 /radicalBig 1−(2f−f2) sin2φ+H sinφˆK where fis the earth’s flattening factor. 2. Calculate the topocentric declination δusing Equation 5.83a. 3. Calculate the topocentric right ascension αfrom Equations 5.83b and 5.83c. 4. Calculate the direction cosine unit vector ˆ/rho1from Equations 5.68 and 5.71, ˆ/rho1=cosδ(cosαˆI+sinαˆJ)+sinδˆK 5. Calculate the geocentric position vector rfrom Equation 5.63, r=R+/rho1ˆ/rho1 6. Calculate the inertial velocity ˙Rof the site from Equation 5.66. 7. Calculate the declination rate ˙δusing Equation 5.84. 8. Calculate the right ascension rate ˙αby means of Equation 5.85. 9. Calculate the direction cosine rate vector ˙ˆ/rho1from Equations 5.69 and 5.72: ˙ˆ/rho1=(−˙αsinαcosδ−˙δcosαsinδ)ˆI+(˙αcosαcosδ−˙δsinαsinδ)ˆJ+˙δcosδˆK 10. Calculate the geocentric velocity vector vfrom Equation 5.64: v=˙R+˙/rho1ˆ/rho1+/rho1˙ˆ/rho1 Example 5.10Atθ=300◦local sidereal time a sea-level ( H=0) tracking station at latitude φ=60◦ detects a space object and obtains the following data: Slant range : /rho1=2551 km Azimuth : A=90◦ Elevation : a=30◦ 5.8 Orbit determination from angle and range measurements 233 Range rate : ˙/rho1=0 Azimuth rate : ˙A=1.973×10−3rad/s( 0.1130◦/s) Elevation rate : ˙a=9.864×10−4rad/s( 0.05651◦/s) What are the orbital elements of the object? We must first employ Algorithm 5.4 to obtain the state vectors rand vin order to compute the orbital elements by means of Algorithm 4.1. Step 1: The equatorial radius of the earth is Re=6378 km and the flattening factor is f=0.003353. It follows from Equation 5.56 that the position vector of the observer is R=1598ˆI−2769ˆJ+5500ˆK(km) Step 2: δ=sin−1(cosφcosAcosa+sinφsina) =sin−1/parenleftbig cos 60◦cos 90◦cos 30◦+sin 60◦sin 30◦/parenrightbig =25.66◦ Step 3: Since the given azimuth lies between 0◦and 180◦, Equation 5.83b yields h=360◦−cos−1/parenleftbiggcosφsina−sinφcosAcosa cosδ/parenrightbigg =360◦−cos−1/parenleftbiggcos 60◦sin 30◦−sin 60◦cos 90◦cos 30◦ cos 25.66◦/parenrightbigg =360◦−73.90◦=286.1◦ Therefore, the right ascension is α=θ−h=300◦−286.1◦=13.90◦ Step 4: ˆ/rho1=cos 25.66(cos 13.90◦ˆI+sin 13.90◦ˆJ)+sinδˆK=0.8750 ˆI+0.2165 ˆJ+0.4330 ˆK Step 5: r=R+/rho1ˆ/rho1=(1598ˆI−2769ˆJ+5500ˆK)+2551(0.8750 ˆI+0.2165 ˆJ+0.4330 ˆK) r=3831ˆI−2216ˆJ+6605ˆK(km) Step 6: Recalling from Equation 2.57 that the angular velocity ωEof the earth is 72 .92× 10−6rad/s, ˙R=/Omega1×R=(72.92 ×10−6ˆK)×(1598ˆI−2769ˆJ+5500ˆK) =0.2019 ˆI+0.1166 ˆJ(km/s) 234 Chapter 5 Preliminary orbit determination (Example 5.10 continued)Step 7: ˙δ=1 cosδ/bracketleftbig −˙AcosφsinAcosa+˙a(sinφcosa−cosφcosAsina)/bracketrightbig =1 cos 25.66◦[−1.973×10−3·cos 60◦sin 90◦cos 30◦+9.864 ×10−4(sin 60◦cos 30◦−cos 60◦cos 90◦sin 30◦)] ˙δ=− 1.2696×10−4(rad/s) Step 8: ˙α−ωE=˙AcosAcosa−˙asinAsina+˙δsinAcosatanδ cosφsina−sinφcosAcosa =1.973×10−3cos 90◦cos 30◦−9.864×10−4sin 90◦sin 30◦ +(−1.2696×10−4) sin 90◦cos 30◦tan 25.66◦ cos 60◦sin 30◦−sin 60◦cos 90◦cos 30◦ =− 0.002184 ˙α=72.92×10−6−0.002184 =− 0.002111 (rad /s) Step 9: ˙ˆ/rho1=/parenleftbig −˙αsinαcosδ−˙δcosαsinδ/parenrightbigˆI+/parenleftbig ˙αcosαcosδ−˙δsinαsinδ/parenrightbigˆJ+˙δcosδˆK =/bracketleftbig −(−0.002111 )sin 13.90◦cos 25.66◦−(−0.1270)cos 13.90◦sin 25.66◦/bracketrightbigˆI +/bracketleftbig(−0.002111 )cos 13.90◦cos 25.66◦−(−0.1270)sin 13.90◦sin 25.66◦/bracketrightbigˆJ +/bracketleftbig −0.1270 cos 25 .66◦/bracketrightbigˆK ˙ˆ/rho1=(0.5104ˆI−1.834ˆJ−0.1144ˆK)(10−3)( r a d/s) Step 10: v=˙R+˙/rho1ˆ/rho1+/rho1˙ˆ/rho1 =(0.2019ˆI+0.1166ˆJ)+0·(0.8750ˆI+0.2165ˆJ+0.4330ˆK) +2551(0 .5104×10−3ˆI−1.834×10−3ˆJ−0.1144×10−3ˆK) v=1.504ˆI−4.562ˆJ−0.2920ˆK(km/s) Using the position and velocity vectors from steps 5 and 10, the reader can verify that Algorithm 4.1 yields the following orbital elements of the tracked object a=5170 km i=113.4◦ /Omega1=109.8◦ e=0.6195 ω=309.8◦ θ=165.3◦ This is a highly elliptical orbit with a semimajor axis less than the earth’s radius, so the object will impact the earth (at a true anomaly of 216◦). 5.9 Angles-only preliminary orbit determination 235 r R CB SunEarth /H9253/rho1 Figure 5.13 An object Borbiting the sun and tracked from earth. For objects orbiting the sun (planets, asteroids, comets and man-made inter- planetary probes), the fundamental vector triangle is as illustrated in Figure 5.13.The tracking station is on the earth but, of course, the sun rather than the earthis the center of attraction. The procedure for finding the heliocentric state vector r and vis similar to that outlined above. Because of the vast distances involved, the observer can usually be imagined to reside at the center of the earth. Dealing withRis different in this case. The daily position of the sun relative to the earth ( −Rin Figure 5.13) may be found in ephemerides, such as Astronomical Almanac (US Naval Observatory, 2004). A discussion of interplanetary trajectories appears in Chapter 8 ofthis text. 5.9 Angles-only preliminary orbit determination T o determine an orbit requires specify ing six independent quantities. These can be the six classical orbital elements or the total of six components the state vector, rand v,a t a given instant. T o determine an orbit solely from observations therefore requires sixindependent measurements. In the previous section we assumed the tracking station was able to measure simultaneously the six quantities: range and range rate; azimuthand azimuth rate; plus elevation and elevation rate. This data leads directly to thestate vector and, hence, to a complete determination of the orbit. In the absenceof range and range rate measuring capability, as with a telescope, we must rely onmeasurements of just the two angles, azimuth and elevation, to determine the orbit. A minimum of three observations of azimuth and elevation is therefore required to accumulate the six quantities we need to predict the orbit. We shall henceforth assumethat the angular measurements are converted to topocentric right ascension αand declination δ, as described in the previous section. We shall consider the classical method of angles-only orbit determination due to Carl Friedrich Gauss (1777–1855), a German mathematician who many consider 236 Chapter 5 Preliminary orbit determination was one of the greatest mathematicians ever. This method requires gathering angu- lar information over closely spaced intervals of time and yields a preliminary orbitdetermination based on those initial observations. We follow Boulet (1991). 5.10 Gauss’s method of preliminary orbit determination Suppose we have three observations of an orbiting body at times t1,t2andt3, as shown in Figure 5.14. At each time the geocentric position vector ris related to the observer’s position vector R, the slant range /rho1and the topocentric direction cosine vector ˆ/rho1by Equation 5.63, r1=R1+/rho11ˆ/rho11 (5.86a) r2=R2+/rho12ˆ/rho12 (5.86b) r3=R3+/rho13ˆ/rho13 (5.86c) The positions R1,R2and R3of the observer Oare known from the location of the tracking station and the time of the observations. ˆ/rho11,ˆ/rho12and ˆ/rho13are obtained by measuring the right ascension αand declination δof the body at each of the three times (recall Equation 5.57). Equations 5.86 are three vector equations, and therefore nine scalar equations, in 12 unknowns: the three components of each of the threevectors r 1,r2and r3, plus the three slant ranges /rho11,/rho12and/rho13. An additional three equations are ob tained by recalling from Chapter 2 that the conservation of angular momentum requires the vectors r1,r2and r3to lie in the r1 R1O Ct1t2t3 R2R3r2 r3B /H9253/rho13 /rho12 /rho11 Figure 5.14 Center of attraction C, observer Oand tracked body B. 5.10 Gauss’s method of preliminary orbit determination 237 same plane. As in our discussion of the Gibbs method in Section 5.2, that means r2is a linear combination r1and r3: r2=c1r1+c3r3 (5.87) Adding this equation to those in 5.86 introduces two new unknowns c1andc3. At this point we therefore have 12 scalar equations in 14 unknowns. Another consequence of the two-body eq uation of motion (Equation 2.15) is that the state vectors rand vof the orbiting body can be expressed in terms of the state vector at any given time by means of the Lagrange coefficients, Equations 2.125 and2.126. For the case at hand that means we can express the position vectors r 1and r3 in terms of the position r2and velocity v2at the intermediate time t2as follows, r1=f1r2+g1v2 (5.88a) r3=f3r2+g3v2 (5.88b) where f1and g1are the Lagrange coefficients evaluated at t1while f3and g3are those same functions evaluated at time t3. If the time intervals between the three observations are sufficiently small then Equations 2.163 reveal that fand gdepend approximately only on the distance from the center of attraction at the initial time.For the case at hand that means the coefficients in Equations 5.88 depend only on r 2. Hence, Equations 5.88 add six scalar equations to our previous list of 12 while addingto the list of 14 unknowns only four: the three components of v 2and the radius r2. We have arrived at 18 equations in 18 unknowns, so the problem is well posed and we can proceed with the solution. The ultimate objective is to determine the state vectorsr 2,v2at the intermediate time t2. Let us start out by solving for c1and c3in Equation 5.87. First take the cross product of each term in that equation with r3, r2×r3=c1(r1×r3)+c3(r3×r3) Since r3×r3=0, this reduces to r2×r3=c1(r1×r3) Taking the dot product of this result with r1×r3and solving for c1yields c1=(r2×r3)·(r1×r3) /bardblr1×r3/bardbl2(5.89) In a similar fashion, by forming the dot product of Equation 5.87 with r1,w ea r el e dt o c3=(r2×r1)·(r3×r1) /bardblr1×r3/bardbl2(5.90) Let us next use Equations 5.88 to eliminate r1and r3from the expressions for c1and c3. First of all, r1×r3=(f1r2+g1v2)×(f3r2+g3v2)=f1g3(r2×v2)+f3g1(v2×r2) But r2×v2=h,w h e r e his the constant angular momentum of the orbit (Equation 2.18). It follows that r1×r3=(f1g3−f3g1)h (5.91) 238 Chapter 5 Preliminary orbit determination and, of course, r3×r1=− (f1g3−f3g1)h (5.92) Therefore /bardblr1×r3/bardbl2=(f1g3−f3g1)2h2(5.93) Similarly r2×r3=r2×(f3r2+g3v2)=g3h (5.94) and r2×r1=r2×(f1r2+g1v2)=g1h (5.95) Substituting Equations 5.91, 5.93 and 5.94 into Equation 5.89 yields c1=g3h·(f1g3−f3g1)h (f1g3−f3g1)2h2=g3(f1g3−f3g1)h2 (f1g3−f3g1)2h2 or c1=g3 f1g3−f3g1(5.96) Likewise, substituting Equations 5.92, 5.93 and 5.95 into Equation 5.90 leads to c3=−g1 f1g3−f3g1(5.97) The coefficients in Equation 5.87 are now expressed solely in terms of the Lagrange functions, and so far no approximations have been made. However, we will have tomake some approximations in order to proceed. We must approximate c 1and c3under the assumption that the times between observations of the orbiting body are small. T o that end, let us introduce the notation τ1=t1−t2 τ3=t3−t2(5.98) τ1andτ3are the time intervals between the successive measurements of ˆ/rho11,ˆ/rho12andˆ/rho13. If the time intervals τ1andτ3are small enough, we can retain just the first two terms of the series expressions for the Lagrange coefficients fand gin Equations 2.163, thereby obtaining the approximations f1≈1−1 2µ r3 2τ2 1 (5.99a) f3≈1−1 2µ r3 2τ2 3 (5.99b) and g1≈τ1−1 6µ r3 2τ3 1 (5.100a) g3≈τ3−1 6µ r3 2τ3 3 (5.100b) 5.10 Gauss’s method of preliminary orbit determination 239 We want to exclude all terms in fandgbeyond the first two so that only the unknown r2appears in Equations 5.99 and 5.100. One can see from Equations 2.163 that the higher order terms include the unknown v2as well. Using Equations 5.99 and 5.100 we can calculate the denominator in Equations 5.96 and 5.97, f1g3−f3g1=/parenleftbigg 1−1 2µ r3 2τ2 1/parenrightbigg/parenleftbigg τ3−1 6µ r3 2τ3 3/parenrightbigg −/parenleftbigg 1−1 2µ r3 2τ2 3/parenrightbigg/parenleftbigg τ1−1 6µ r3 2τ3 1/parenrightbigg Expanding the right side and collecting terms yields f1g3−f3g1=(τ3−τ1)−1 6µ r3 2(τ3−τ1)3+1 12µ2 r6 2(τ2 1τ3 3−τ3 1τ2 3) Retaining terms of at most third order in the time intervals τ1andτ3, and setting τ=τ3−τ1 (5.101) reduces this expression to f1g3−f3g1≈τ−1 6µ r3 2τ3(5.102) From Equation 5.98 observe that τis just the time interval between the first and last observations. Substituting Equatio ns 5.100b and 5.102 into Equation 5.96, we get c1≈τ3−1 6µ r3 2τ3 3 τ−1 6µ r3 2τ3=τ3 τ/parenleftbigg 1−1 6µ r3 2τ2 3/parenrightbigg ·/parenleftbigg 1−1 6µ r3 2τ2/parenrightbigg−1 (5.103) We can use the binomial theorem to simplify (linearize) the last term on the right. Setting a=1,b=−1 6µ r3 2τ2and n=− 1 in Equation 5.44, and neglecting terms of higher order than 2 in τ, yields /parenleftbigg 1−1 6µ r3 2τ2/parenrightbigg−1 ≈1+1 6µ r3 2τ2 Hence Equation 5.103 becomes c1≈τ3 τ/bracketleftbigg 1+1 6µ r3 2(τ2−τ2 3)/bracketrightbigg (5.104) where only second order terms in the time have been retained. In precisely the same way it can be shown that c3≈−τ1 τ/bracketleftbigg 1+1 6µ r3 2(τ2−τ2 1)/bracketrightbigg (5.105) Finally, we have managed to obtain approximate formulas for the coefficients in Equation 5.87 in terms of just the time intervals between observations and the as yetunknown distance r 2from the center of attraction at the central time t2. The next stage of the solution is to seek formulas for the slant ranges /rho11,/rho12and/rho13 in terms of c1and c3. T o that end, substitute Equations 5.86 into Equation 5.87 to get R2+/rho12ˆ/rho12=c1(R1+/rho11ˆ/rho11)+c3(R3+/rho13ˆ/rho13) 240 Chapter 5 Preliminary orbit determination which we rearrange into the form c1/rho11ˆ/rho11−/rho12ˆ/rho12+c3/rho13ˆ/rho13=− c1R1+R2−c3R3 (5.106) Let us isolate the slant ranges /rho11,/rho12and/rho13in turn by taking the dot product of this equation with appropriate vectors. T o isolate /rho11take the dot product of each term in this equation with ˆ/rho12׈/rho13, which gives c1/rho11ˆ/rho11·(ˆ/rho12׈/rho13)−/rho12ˆ/rho12·(ˆ/rho12׈/rho13)+c3/rho13ˆ/rho13·(ˆ/rho12׈/rho13) =− c1R1·(ˆ/rho12׈/rho13)+R2·(ˆ/rho12׈/rho13)−c3R3·(ˆ/rho12׈/rho13) Since ˆ/rho12·(ˆ/rho12׈/rho13)=ˆ/rho13·(ˆ/rho12׈/rho13)=0, this reduces to c1/rho11ˆ/rho11·(ˆ/rho12׈/rho13)=(−c1R1+R2−c3R3)·(ˆ/rho12׈/rho13) (5.107) LetD0represent the scalar triple product of ˆ/rho11,ˆ/rho12and ˆ/rho13, D0=ˆ/rho11·(ˆ/rho12׈/rho13) (5.108) We will assume that D0is not zero, which means that ˆ/rho11,ˆ/rho12and ˆ/rho13do not lie in the same plane. Then we can solve Equation 5.107 for /rho11to get /rho11=1 D0/parenleftbigg −D11+1 c1D21−c3 c1D31/parenrightbigg (5.109a) where the Ds stand for the scalar triple products D11=R1·(ˆ/rho12׈/rho13)D21=R2·(ˆ/rho12׈/rho13)D31=R3·(ˆ/rho12׈/rho13) (5.109b) In a similar fashion, by taking the dot product of Equation 5.106 with ˆ/rho11׈/rho13and then ˆ/rho11׈/rho12we obtain /rho12and/rho13, /rho12=1 D0(−c1D12+D22−c3D32) (5.110a) where D12=R1·(ˆ/rho11׈/rho13)D22=R2·(ˆ/rho11׈/rho13)D32=R3·(ˆ/rho11׈/rho13) (5.110b) and /rho13=1 D0/parenleftbigg −c1 c3D13+1 c3D23−D33/parenrightbigg (5.111a) where D13=R1·(ˆ/rho11׈/rho12)D23=R2·(ˆ/rho11׈/rho12)D33=R3·(ˆ/rho11׈/rho12) (5.111b) T o obtain these results we used the fact that ˆ/rho12·(ˆ/rho11׈/rho13)=− D0and ˆ/rho13·(ˆ/rho11׈/rho12)= D0(Equation 2.32). Substituting Equations 5.104 and 5.105 into Equation 5.110a yields the approxi- mate slant range /rho12, /rho12=A+µB r3 2(5.112a) 5.10 Gauss’s method of preliminary orbit determination 241 where A=1 D0/parenleftBig −D 12τ3 τ+D22+D32τ1 τ/parenrightBig (5.112b) B=1 6D0/bracketleftBig D12(τ2 3−τ2)τ3 τ+D32(τ2−τ2 1)τ1 τ/bracketrightBig (5.112c) On the other hand, making the same substitutions into Equations 5.109 and 5.111 leads to the following approximate expressions for the slant ranges /rho11and/rho13, /rho11=1 D0 6/parenleftbigg D31τ1 τ3+D21τ τ3/parenrightbigg r3 2+µD 31(τ2−τ2 1)τ1 τ3 6r3 2+µ(τ2−τ2 3)−D11 (5.113) /rho13=1 D0 6/parenleftbigg D13τ3 τ1−D23τ τ1/parenrightbigg r3 2+µD 13(τ2−τ2 3)τ3 τ1 6r3 2+µ(τ2−τ2 3)−D33 (5.114) Equation 5.112a is a relation between the slant range /rho12and the geocentric radius r2. Another expression relating these two variables is obtained from Equation 5.86b, r2·r2=(R2+/rho12ˆ/rho12)·(R2+/rho12ˆ/rho12) or r2 2=/rho12 2+2E/rho12+R2 2 (5.115a) where E=R2·ˆ/rho12 (5.115b) Substituting Equation 5.112a into 5.115a gives r2 2=/parenleftbigg A+µB r3 2/parenrightbigg2 +2C/parenleftbigg A+µB r3 2/parenrightbigg +R2 2 Expanding and rearranging terms leads to an eighth order polynomial, x8+ax6+bx3+c=0 (5.116) where x=r2and the coefficients are a=− (A2+2AE+R2 2)b=− 2µB (A+E)c=−µ2B2(5.117) We solve Equation 5.116 for r2and substitute the result into Equations 5.112 through 5.114 to obtain the slant ranges /rho11,/rho12and/rho13. Then Equations 5.86 yield the position vectors r1,r2and r3. Recall that finding r2was one of our objectives. T o attain the other objective, the velocity v2, we first solve Equation 5.88a for r2 r2=1 f1r1−g1 f1v2 242 Chapter 5 Preliminary orbit determination Substitute this result into Equation 5.88b to get r3=f3 f1r1+/parenleftbiggf1g3−f3g1 f1/parenrightbigg v2 Solving this for v2yields v2=1 f1g3−f3g1(−f3r1+f1r3) (5.118) in which the approximate Lagrange functions appearing in Equations 5.99 and 5.100 are used. The approximate values we have found for r2and v2are used as the starting point for iteratively improving the accuracy of the computed r2and v2until convergence is achieved. The entire step-by-step procedure is summarized in Algorithms 5.5 and5.6 presented below. See also Appendix D.15. Algorithm 5.5Gauss’s method of preliminary orbit determination. Given the direction cosine vec- tors ˆ/rho11,ˆ/rho12and ˆ/rho13and the observer’s position vectors R1,R2and R3at the times t1,t2 and t3, proceed as follows. 1. Calculate the time intervals τ1,τ3andτusing Equations 5.98 and 5.101. 2. Calculate the cross products p1=ˆ/rho12׈/rho13,p2=ˆ/rho11׈/rho13and p3=ˆ/rho11׈/rho12. 3. Calculate D0=ˆ/rho11·p1(Equation 5.108). 4. From Equations 5.109b, 5.110b and 5.111b compute the six scalar quantities D11=R1·p1 D12=R1·p2 D13=R1·p3 D21=R2·p1 D22=R2·p2 D23=R2·p3 D31=R3·p1 D32=R3·p2 D33=R3·p3 5. Calculate Aand Busing Equations 5.112b and 5.112c. 6. Calculate E, using Equation 5.115b, and R2 2=R2·R2. 7. Calculate a,band cfrom Equation 5.117. 8. Find the roots of Equation 5.116 and select the most reasonable one as r2. Newton’s method can be used, in which case Equation 3.13 becomes xi+1=xi−x8 i+ax6 i+bx3 i+c 8x7 i+6ax5 i+3bx2 i(5.119) One must first print or graph the function F=x8+ax6+bx3+cforx>0 and choose as an initial estimate a value of xnear the point where Fchanges sign. If there is more than one physically reasonable root, then each one must be used and the resulting orbit checked against knowledge that may already be available about the general nature of the orbit. Alternatively, the analysis can be repeated using additional sets of observations. 9. Calculate /rho11,/rho12and/rho13using Equations 5.113, 5.112a and 5.114. 10. Use Equations 5.86 to calculate r1,r2and r3. 5.10 Gauss’s method of preliminary orbit determination 243 11. Calculate the Lagrange coefficients f1,g1,f3and g3from Equations 5.99 and 5.100. 12. Calculate v2using Equation 5.118. 13. (a) Use r2and v2from steps 10 and 12 to obtain the orbital elements from Algo- rithm 4.1. (b) Alternatively, proceed to Algorithm 5.6 to improve the preliminaryestimate of the orbit. Algorithm 5.6Iterative improvement of the orbit determined by Algorithm 5.5. Use the values of r2and v2obtained from Algorithm 5.5 to compute the ‘exact’ values of the fand gfunctions from their universal formulation, as follows: 1. Calculate the magnitude of r2(r2=√r2·r2) and v2(v2=√v2·v2). 2. Calculate α, the reciprocal of the semimajor axis: α=2/r 2−v2 2/µ. 3. Calculate the radial component of v2,vr2=v2·r2/r2. 4. Use Algorithm 3.3 to solve the universal Kepler’s equation (Equation 3.46) for the universal variables χ1andχ3at times t1and t3, respectively: √µτ1=r2vr2√µχ2 1C(αχ2 1)+(1−αr2)χ3 1S(αχ2 1)+r2χ1 √µτ3=r2vr2√µχ2 3C(αχ2 3)+(1−αr2)χ3 3S(αχ2 3)+r2χ3 5. Use χ1andχ3to calculate f1,g1,f3and g3from Equations 3.66: f1=1−χ2 1 r2C(αχ2 1) g1=τ1−1√µχ3 1S(αχ2 1) f3=1−χ2 3 r2C(αχ2 3) g3=τ3−1√µχ3 3S(αχ2 3) 6. Use these values of f1,g1,f3and g3to calculate c1and c3from Equations 5.96 and 5.97. 7. Use c1and c3to calculate updated values of /rho11,/rho12and/rho13from Equations 5.109 through 5.111. 8. Calculate updated r1,r2and r3from Equations 5.86. 9. Calculate updated v2using Equation 5.118 and the fand gvalues computed in step 5. 10. Go back to step 1 and repeat until, to the desired degree of precision, there is no further change in /rho11,/rho12and/rho13. 11. Use r2and v2to compute the orbital elements by means of Algorithm 4.1. Example 5.11A tracking station is located at φ=40◦north latitude at an altitude of H=1 km. Three observations of an earth satellite yield the values for the topocentric right ascension and declination listed in the following table, which also shows the local sidereal timeθof the observation site. 244 Chapter 5 Preliminary orbit determination (Example 5.11 continued)Use the Gauss Algorithm 5.5 to estimate the state vector at the second observation time. Recall that µ=398 600 km3/s2. T able 5.1 Data for Example 5.11 Observation Time Right ascension, αDeclination, δLocal sidereal time, θ (seconds) (degrees) (degrees) (degrees) 1 0 43.537 −8.7833 44.506 2 118.10 54.420 −12.074 45.000 3 237.58 64.318 −15.105 45.499 Recalling that the equatorial radius of the earth is Re=6378 km and the flattening factor is f=0.003353, we substitute φ=40◦,H=1 km and the given values of θinto Equation 5.56 to obtain the inertial position v ector of the tracking station at each of the three observation times: R1=3489.8ˆI+3430.2ˆJ+4078.5ˆK(km) R2=3460.1ˆI+3460.1ˆJ+4078.5ˆK(km) R3=3429.9ˆI+3490.1ˆJ+4078.5ˆK(km) Using Equation 5.57 we compute the direction cosine vectors at each of the three observation times from the right ascension and declination data: ˆ/rho11=cos(−8.7833◦) cos 43 .537◦ˆI+cos(−8.7833◦) sin 43 .537◦ˆJ+sin(−8.7833◦)ˆK =0.71643ˆI+0.68074ˆJ−0.15270 ˆK ˆ/rho12=cos(−12.074◦) cos 54 .420◦ˆI+cos(−12.074◦) sin 54 .420◦ˆJ+sin(−12.074◦)ˆK =0.56897ˆI+0.79531ˆJ−0.20917 ˆK ˆ/rho13=cos(−15.105◦) cos 64 .318◦ˆI+cos(−15.105◦) sin 64 .318◦ˆJ+sin(−15.105◦)ˆK =0.41841ˆI+0.87007ˆJ−0.26059 ˆK We can now proceed with Algorithm 5.5. Step 1: τ1=0−118.10=− 118.10 s τ3=237.58−118.10=119.47 s τ=119.47−(−118.1)=237.58 s Step 2: p1=ˆ/rho12׈/rho13=− 0.025258 ˆI+0.060753 ˆJ+0.16229 ˆK p2=ˆ/rho11׈/rho13=− 0.044538 ˆI+0.12281ˆJ+0.33853 ˆK p3=ˆ/rho11׈/rho12=− 0.020950 ˆI+0.062977 ˆJ+0.18246 ˆK 5.10 Gauss’s method of preliminary orbit determination 245 Step 3: D0=ˆ/rho11·p1=− 0.0015198 Step 4: D11=R1·p1=782.15 km D12=R1·p2=1646.5k m D13=R1·p3=887.10 km D21=R2·p1=784.72 km D22=R2·p2=1651.5k m D23=R2·p3=889.60 km D31=R3·p1=787.31 km D32=R3·p2=1656.6k m D33=R3·p3=892.13 km Step 5: A=1 −0.0015198/bracketleftbigg −1646.5119.47 237.58+1651.5 +1656.6(−118.10) 237.58/bracketrightbigg =− 6.6858 km B=1 6(−0.0015198)/braceleftbigg 1646.5(119.472−237.582)119.47 237.58 +1656.6[237.582−(−118.10)2](−118.10) 237.58/bracerightbigg =7.6667 ×109km·s2 Step 6: E=R2·ˆ/rho12=3875.8k m R2 2=R2·R2=4.058×107km2 Step 7: a=− [(−6.6858)2+2(−6.6858)(3875.8) +4.058×107]=− 4.0528 ×107km2 b=− 2(389 600)(7.6667 ×109)(−6.6858 +3875.8) =− 2.3597 ×1019km5 c=− (398 600)2(7.6667 ×109)2=− 9.3387 ×1030km8 Step 8: F(x)=x8−4.0528 ×107x6−2.3597 ×1019x3−9.3387 ×1030=0 The graph of F(x) in Figure 5.15 shows that it changes sign near x=9000 km. Let us use that as the starting value in Newton’s method for finding the roots of F(x). For the case at hand, Equation 5.119 is xi+1=xi−x8 i−4.0528 ×107x6 i−2.3622 ×1019x3 i−9.3186 ×1030 8x7 i−2.4317 ×108x5 i−7.0866 ×1019x2 i Stepping through Newton’s iterative procedure yields x0=9000 x1=9000−(−276.93) =9276.9 x2=9276.9 −34.526 =9242.4 246 Chapter 5 Preliminary orbit determination (Example 5.11 continued) 2000 4000 6000 8000/H110021/H11003 10/H110013101/H1100310/H11001312/H1100310/H1100131 10 000 0F x Figure 5.15 Graph of the polynomial in Equation (f). x3=9242.4−0.63428 =9241.8 x4=9241.8−0.00021048 =9241.8 Thus, after four steps we converge to r2=9241.8k m The other roots are either negative or complex and are therefore physically unacceptable. Step 9: /rho11=1 −0.0015198 ×  6/bracketleftbigg 787.31(−118.10) 119.47+784.72237.58 119.47/bracketrightbigg 9241.83 +398 600 ·787.31[237 .582−(−118.10)2]−118.10 119.47 6·9241.83+398 600(237 .582−119.472)−782.15   =3639.1k m /rho1 2=− 6.6858+398 600 ·7.6667×109 9241.83=3864.8k m /rho13=1 −0.0015198× 6/parenleftbigg 887.10119.47 −118.10−889.60237.58 −118.10/parenrightbigg 9241.83 +398 600 ·887.10(237 .582−119.472)119.47 −118.10 6·9241.83+398 600(237 .582−119.472)−892.13  =4156.9k m 5.10 Gauss’s method of preliminary orbit determination 247 Step 10: r1=(3489.8 ˆI+3430.2 ˆJ+4078.5 ˆK)+3639.1(0.71643 ˆI+0.68074 ˆJ−0.15270 ˆK) =6096.9 ˆI+5907.5 ˆJ+3522.9 ˆK(km) r2=(3460.1 ˆI+3460.1 ˆJ+4078.5 ˆK)+3864.8(0.56897 ˆI+0.79531 ˆJ−0.20917 ˆK) =5659.1 ˆI+6533.8 ˆJ+3270.1 ˆK(km) r3=(3429.9 ˆI+3490.1 ˆJ+4078.5 ˆK)+4156.9(0.41841 ˆI+0.87007 ˆJ−0.26059 ˆK) =5169.1 ˆI+7107.0 ˆJ+2995.3 ˆK(km) Step 11: f1≈1−1 2398 600 9241.83(−118.10)2=0.99648 f3≈1−1 2398 600 9241.83(119.47)2=0.99640 g1≈− 118.10 −1 6398 600 9241.83(−118.10)3=− 117.97 g3≈119.47 −1 6398 600 9241.83(119.47)3=119.33 Step 12: v2=−0.99640(6096.9 ˆI+5907.5 ˆJ+3522.9 ˆK)+0.99648(5169.1 ˆI+7107.0 ˆJ +2995.3 ˆK) 0.99648 ·119.33 −0.99640(−117.97) =− 3.9080 ˆI+5.0573 ˆJ−2.2222 ˆK(km/s) In summary, the state vector at time t2is, approximately, r2=5659.1 ˆI+6533.8 ˆJ+3270.1 ˆK(km) v2=− 3.9080 ˆI+5.0573 ˆJ−2.2222 ˆK(km/s) Example 5.12Starting with the state vector determined in Example 5.11, use Algorithm 5.6 to improve the vector to five significant figures. Step 1: r2=/bardbl r2/bardbl=/radicalbig 5659.12+6533.82+3270.12=9241.8k m v2=/bardbl v2/bardbl=/radicalbig (−3.9080)2+5.0573 +(−2.2222)2=6.7666 km /s Step 2: α=2 r2−v2 2 µ=2 9241.8−6.76662 398 600=1.0154 ×10−4km−1 248 Chapter 5 Preliminary orbit determination (Example 5.12 continued)Step 3: vr2=v2·r2 r2=(−3.9080) ·5659.1+5.0573·6533.8+(−2.2222) ·3270.1 9241.8 =0.39611 km /s Step 4: The universal Kepler’s equation at times t1and t3, respectively, becomes √ 398 600 τ1=9241.8·0.39611√ 398 600χ2 1C(1.0154×10−4χ2 1) +(1−1.0154×10−4·9241.8)χ3 1S(1.0154×10−4χ2 1)+9241.8χ1 √ 398 600 τ3=9241.8·0.39611√ 398 600χ2 3C(1.0154×10−4χ2 3) +(1−1.0154×10−4·9241.8)χ3 3S(1.0154×10−4χ2 3)+9241.8χ3 or 631.35τ1=5.7983χ2 1C(1.0154×10−4χ2 1)+0.061594 χ3 1S(1.0154×10−4χ2 1) +9241.8χ1 631.35τ3=5.7983χ2 3C(1.0154×10−4χ2 3)+0.061594 χ3 1S(1.0154×10−4χ2 3) +9241.8χ3 Applying Algorithm 3.3 to each of these equations yields χ1=− 8.0882√ km χ3=8.1404√ km Step 5: f1=1−χ2 1 r2C(αχ2 1)=1−(−8.0882)2 9241.8·0.49972/bracehtipdownleft /bracehtipupright/bracehtipupleft /bracehtipdownright C[1.0154×10−4(−8.0882)2]=0.99646 g1=τ1−1√µχ3 1S(αχ2 1)=− 118.1−1√ 398 600(−8.0882)3 ×0.16661/bracehtipdownleft /bracehtipupright/bracehtipupleft /bracehtipdownright S[1.0154×10−4(−8.0882)2]=− 117.96 s and f3=1−χ2 3 r2C(αχ2 3)=1−8.14042 9241.8·0.49972/bracehtipdownleft /bracehtipupright/bracehtipupleft /bracehtipdownright C[1.0154×10−4·8.14042]=0.99642 5.10 Gauss’s method of preliminary orbit determination 249 g3=τ3−1√µχ3 3S(αχ2 3)=− 118.1−1√ 398 6008.14043 ×0.16661/bracehtipdownleft /bracehtipupright/bracehtipupleft /bracehtipdownright S[1.0154 ×10−4(−8.0882)2]=119.33 It turns out that the procedure converges more rapidly if the Lagrange coefficients are set equal to the average of those computed for the current step and those computed for the previous step. Thus, we set f1=0.99648 +0.99646 2=0.99647 g1=−117.97 +(−117.96) 2=− 117.96 s f3=0.99642 +0.99641 2=0.99641 g3=119.3+119.3 2=119.3s Step 6: c1=119.3 (0.99647)(119.3) −(0.99641)(−117.96 s)=0.50467 c3=−−117.96 (0.99647)(119.3) −(0.99641)(−117.96)=0.49890 Step 7: /rho11=1 −0.0015198/parenleftbigg −782.15 +1 0.50467784.72 −0.49890 0.50467787.31/parenrightbigg =3650.7k m /rho12=1 −0.0015198(−0.50467 ·1646.5 +1651.5 −0.49890 ·1656.6 )=3877.2k m /rho13=1 −0.0015198/parenleftbigg −0.50467 0.49890887.10 +1 0.49890889.60 −892.13/parenrightbigg =4186.2k m Step 8: r1=(3489.8 ˆI+3430.2 ˆJ+4078.5 ˆK)+3650.7(0.71643 ˆI+0.68074 ˆJ−0.15270 ˆK) =6105.3 ˆI+5915.4 ˆJ+3521.1 ˆK(km) r2=(3460.1 ˆI+3460.1 ˆJ+4078.5 ˆK)+3877.2(0.56897 ˆI+0.79531 ˆJ−0.20917 ˆK) =5662.1 ˆI+6543.7 ˆJ+3267.5 ˆK(km) r3=(3429.9 ˆI+3490.1 ˆJ+4078.5 ˆK)+4186.2(0.41841 ˆI+0.87007 ˆJ−0.26059 ˆK) =5181.4 ˆI+7132.4 ˆJ+2987.6 ˆK(km) 250 Chapter 5 Preliminary orbit determination (Example 5.12 continued)Step 9: v2=1 0.99647 ·119.3−0.99641( −117.96)×[−0.99641(6105 .3ˆI+5915.4ˆJ +3521.1ˆK)+0.99647(5181 .4ˆI+7132.4ˆJ+2987.6ˆK)] =− 3.8918ˆI+5.1307ˆJ−2.2472ˆK(km/s) This completes the first iteration. The updated position r2and velocity v2are used to repeat the procedure begin- ning at step 1. The results of the first and subsequent iterations are shown in Table 5.2. Convergence to five significant figures in the slant ranges /rho11,/rho12and/rho13occurs in four steps, at which point the state vector is r2=5662.1ˆI+6538.0ˆJ+3269.0ˆK(km) v2=− 3.8856ˆI+5.1214ˆJ−2.2433ˆK(km/s) T able 5.2 Key results at each step of the iterative procedure Step χ1 χ3 f1 g1 f3 g3 /rho11 /rho12 /rho13 1−8.0882 8.1404 0.99647 −117.97 0.99641 119.33 3650.7 3877.2 4186.2 2−8.0818 8.1282 0.99647 −117.96 0.99642 119.33 3643.8 3869.9 4178.3 3−8.0871 8.1337 0.99647 −117.96 0.99642 119.33 3644.0 3870.1 4178.6 4−8.0869 8.1336 0.99647 −117.96 0.99642 119.33 3644.0 3870.1 4178.6 Using Algorithm 4.1 we find that the orbital elements are a=10 000 km ( h=62 818 km2/s) e=0.1000 i=30◦ /Omega1=270◦ ω=90◦ θ=45.01◦ Problems 5.1 The geocentric equatorial position vectors of a satellite at three separate times are r1=5887ˆI−3520ˆJ−1204ˆK(km) r2=5572ˆI−3457ˆJ−2376ˆK(km) r3=5088ˆI−3289ˆJ−3480ˆK(km) Use Gibbs’ method to find v2. {Partial ans.: v2=7.59 km/s} Problems 251 5.2 Calculate the orbital elements and perigee altitude of the space object in the previous problem.{Partial ans.: z p=567 km} 5.3 At a given instant the altitude of an earth satellite is 600 km. Fifteen minutes later the altitude is 300 km and the true anomaly has increased by 60◦. Find the perigee altitude. {Ans.: zp=298 km} 5.4 At a given instant, the geocentric equatorial position vector of an earth satellite is r1=− 3600ˆI+3600ˆJ+5100ˆK(km) Thirty minutes later the position is r2=− 5500ˆI−6240ˆJ−520ˆK(km) Calculate v1and v2. {Partial ans.: v1=7.711 km/s, v2=6.670 km/s} 5.5 Compute the orbital elements and perigee altitude for the previous problem. {Partial ans.: zp=648 km} 5.6 At a given instant, the geocentric equatorial position vector of an earth satellite is r1=5644ˆI−2830ˆJ−4170ˆK(km) Twenty minutes later the position is r2=− 2240ˆI+7320ˆJ−4980ˆK(km) Calculate v1and v2. {Partial ans.: v1=10.84 km/s, v2=9.970 km/s} 5.7 Compute the orbital elements and perigee altitude for the previous problem. {Partial ans.: zp=224 km} 5.8 Calculate the Julian day number ( JD) for the following epochs: (a) 5:30 UT on August 14, 1914. (b) 14:00 UT on April 18, 1946. (c) 0:00 UT on September 1, 2010. (d) 12:00 UT on October 16, 2007. (e) Noon today, your local time. {Ans.: (a) 2 420 358.729; (b) 2 431 929.083; (c) 2 455 440.500; (d) 2 454 390.000} 5.9 Calculate the number of days from 12:00 UT on your date of birth to 12:00 UT on today’s date. 5.10 Calculate the local sidereal time (in degrees) at: (a) Stockholm, Sweden (east longitude 18◦03/prime) at 12:00 UT on 1 January 2008. (b) Melbourne, Australia (east longitude 144◦58/prime) at 10:00 UT on 21 December 2007. (c) Los Angeles, California (west longitude 118◦15/prime) at 20:00 UT on 4 July 2005. (d) Rio de Janeiro, Brazil (west longitude 43◦06/prime) at 3:00 UT on 15 February 2006. (e) Vladivostok, Russia (east longitude 131◦56/prime) at 8:00 UT on 21 March 2006. (f) At noon today, your local time and place. {Ans.: (a) 298.6◦, (b) 24.6◦, (c) 104.7◦, (d) 146.9◦, (e) 70.6◦} 5.11 Relative to a tracking station whose local sidereal time is 117◦and latitude is +51◦, the azimuth and elevation angle of a satellite are 27.5156◦and 67.5556◦, respectively. Calculate the topocentric right ascension and declination of the satellite.{Ans.: α=145.3 ◦,δ=68.24◦} 252 Chapter 5 Preliminary orbit determination 5.12 A sea-level tracking station at whose local sidereal time is 40◦and latitude is 35◦makes the following observations of a space object: Azimuth: 36.0◦ Azimuth rate: 0.590◦/s Elevation: 36.6◦ Elevation rate: −0.263◦/s Range: 988 kmRange rate: 4.86 km/s What is the state vector of the object? {Partial ans.: r=7003.3 km,v=10.922 km /s} 5.13 Calculate the orbital elements of the satellite in the previous problem. {Partial ans.: e=1.1,i=40 ◦} 5.14 A tracking station at latitude −20◦and elevation 500 m makes the following observations of a satellite at the given times. Time Local sidereal time Azimuth Elevation angle Range(min) (degrees) (degrees) (degrees) (km) 0 60.0 165.932 8.81952 1212.482 60.5014 145.970 44.2734 410.5964 61.0027 2.40973 20.7594 726.464 Use the Gibbs method to calculate the state vector of the satellite at the centralobservation time.{Partial ans.: r 2=6684 km, v2=7.7239 km/s} 5.15 Calculate the orbital elements of the satellite in the previous problem. {Partial ans.: e=0.001, i=95◦} 5.16 A sea-level tracking station at latitude +29◦makes the following observations of a satellite at the given times. Time Local sidereal time T opocentric T opocentric(min) (degrees) right ascension declination (degrees) (degrees) 0.0 0 0 51.51101.0 0.250684 65.9279 27.99112.0 0.501369 79.8500 14.6609 Use the Gauss method without iterative improvement to estimate the state vector of thesatellite at the middle observation time.{Partial ans.: r=6700.9 km,v=8.0757 km/s} 5.17 Refine the estimate in the previous problem using iterative improvement. {Partial ans.: r=6701.5 km,v=8.0881 km/s} 5.18 Calculate the orbital elements from the state vector obtained in the previous problem. {Partial ans.: e=0.10,i=30 ◦} Problems 253 5.19 A sea-level tracking station at latitude +29◦makes the following observations of a satellite at the given times. Time Local sidereal time T opocentric T opocentric(min) (degrees) right ascension declination (degrees) (degrees) 0.0 90 15.0394 20.7487 1.0 90.2507 25.7539 30.1410 2.0 90.5014 48.6055 43.8910 Use the Gauss method without iterative improvement to estimate the state vector of the satellite.{Partial ans.: r=6999.1 km, v=7.5541 km/s} 5.20 Refine the estimate in the previous problem using iterative improvement. {Partial ans.: r=7000.0 km, v=7.5638 km/s} 5.21 Calculate the orbital elements from the state vector obtained in the previous problem. {Partial ans.: e=0.0048, i=31 ◦} 5.22 The position vector Rof a tracking station and the direction cosine vector ˆ/rho1of a satellite relative to the tracking station at three times are as follows: t1=0 min R1=− 1825.96 ˆI+3583.66 ˆJ+4933.54 ˆK(km) ˆ/rho11=− 0.301687 ˆI+0.200673 ˆJ+0.932049 ˆK t2=1 min R2=− 1816.30 ˆI+3575.63 ˆJ+4933.54 ˆK(km) ˆ/rho12=− 0.793090 ˆI−0.210324 ˆJ+0.571640 ˆK t3=2 min R3=− 1857.25 ˆI+3567.54 ˆJ+4933.54 ˆK(km) ˆ/rho13=− 0.873085 ˆI−0.362969 ˆJ+0.325539 ˆK Use the Gauss method without iterative improvement to estimate the state vector of the satellite at the central observation time.{Partial ans.: r=6742.3 km, v=7.6799 km/s} 5.23 Refine the estimate in the previous problem using iterative improvement. {Partial ans.: r=6743.0 km, v=7.6922 km/s} 5.24 Calculate the orbital elements from the state vector obtained in the previous problem. {Partial ans.: e=0.001, i=52 ◦} 5.25 A tracking station at latitude 60°N and 500 m elevation obtains the following data: Time Local sidereal time T opocentric T opocentric (min) (degrees) right ascension declination (degrees) (degrees) 0.0 150 157.783 24.2403 5.0 151.253 159.221 27.2993 10.0 152.507 160.526 29.8982 254 Chapter 5 Preliminary orbit determination Use the Gauss method without iterative improvement to estimate the state vector of the satellite.{Partial ans.: r=25 132 km, v=6.0588 km/s} 5.26 Refine the estimate in the previous problem using iterative improvement. {Partial ans.: r=25 169 km, v=6.0671 km/s} 5.27 Calculate the orbital elements from the state vector obtained in the previous problem. {Partial ans.: e=1.09,i=63 ◦} 5.28 The position vector Rof a tracking station and the direction cosine vector ˆ/rho1of a satellite relative to the tracking station at three times are as follows: t1=0 min R1=5582.84ˆI+3073.90ˆK(km) ˆ/rho11=0.846428 ˆI+0.532504 ˆK t2=5 min R2=5581.50ˆI+122.122ˆJ+3073.90ˆK(km) ˆ/rho12=0.749290 ˆI+0.463023 ˆJ+0.473470 ˆK t3=10 min R3=5577.50ˆI+244.186ˆJ+3073.90ˆK(km) ˆ/rho13=0.529447 ˆI+0.777163 ˆJ+0.340152 ˆK Use the Gauss method without iterative improvement to estimate the state vector of the satellite.{Partial ans.: r=9729.6 km,v=6.0234 km/s} 5.29 Refine the estimate in the previous problem using iterative improvement. {Partial ans.: r=9759.8 km,v=6.0713 km/s} 5.30 Calculate the orbital elements from the state vector obtained in the previous problem. {Partial ans.: e=0.1,i=30 ◦} 6Chapter Orbital maneuvers Chapter outline 6.1 Introduction 255 6.2 Impulsive maneuvers 256 6.3 Hohmann transfer 257 6.4 Bi-elliptic Hohmann transfer 264 6.5 Phasing maneuvers 268 6.6 Non-Hohmann transfers with a common apse line 2736.7 Apse line rotation 279 6.8 Chase maneuvers 285 6.9 Plane change maneuvers 290 Problems 304 6.1 Introduction Orbital maneuvers transfer a spacecraft from one orbit to another. Orbital changes can be dramatic, such as the transfer from a low-earth parking orbit to an interplanetary trajectory. They can also be quite small, as in the final stages ofthe rendezvous of one spacecraft with another. Changing orbits requires the firingof onboard rocket engines. We will be concerned solely with impulsive maneuversin which the rockets fire in relatively short bursts to produce the required velocitychange (delta-v). We start with the classical, energy-efficient Hohmann transfer maneuver, and generalize it to the bi-elliptic Hohmann transfer to see if even more efficiency can beobtained. The phasing maneuver, a form of Hohmann transfer, is considered next. 255 256 Chapter 6 Orbital maneuvers This is followed by a study of non-Hohmann transfer maneuvers with and without rotation of the apse line. We then analyze chase maneuvers, which involves solvingLambert’s problem as explained in Chapter 5. The energy-demanding chase maneu-vers may be impractical for low-earth orbits, but they are necessary for interplanetarymissions, as we shall see in Chapter 8. Up to this point, all of the maneuvers are transfers between coplanar orbits. The chapter ends with an introduction to plane change maneuvers and an explanation ofwhy they require such large delta-vs compared to coplanar maneuvers. 6.2 Impulsive maneuvers Orbital maneuvers transfer a spacecraft from one orbit to another. Orbital changes can be dramatic, such as the transfer from a low-earth parking orbit to an inter-planetary trajectory. They can also be quite small, as in the final stages of therendezvous of one spacecraft with another. Impulsive maneuvers are those in whichbrief firings of onboard rocket motors change the magnitude and direction of thevelocity vector instantaneously. During an impulsive maneuver, the position ofthe spacecraft is considered to be fixed; only the velocity changes. The impulsivemaneuver is an idealization by means of which we can avoid having to solve theequations of motion (Equation 2.6) with the rocket thrust included. The idealiza-tion is satisfactory for those cases in which the position of the spacecraft changesonly slightly during the time that the maneuvering rockets fire. This is true forhigh-thrust rockets with burn times short compared with the coasting time of thevehicle. Each impulsive maneuver results in a change /Delta1vin the velocity of the spacecraft. /Delta1vcan represent a change in the magnitude (‘pumping maneuver’) or the direction (‘cranking maneuver’) of the velocity vector, or both. The magnitude /Delta1vof the velocity increment is related to /Delta1m, the mass of propellant consumed, by the formula (see Equation 11.30) /Delta1m m=1−e−/Delta1v Ispgo (6.1) where mis the mass of the spacecraft before the burn, gois the sea-level standard acceleration of gravity, and Ispis the specific impulse of the propellants. Specific impulse is defined as follows: Isp=thrust sea-level weight rate of fuel consumption Specific impulse has units of seconds, and it is a measure of the performance of a rocket propulsion system. Ispfor some common propellant combinations are shown in Table 6.1. Figure 6.1 is a graph of Equation 6.1 for a range of specific impulses.Note that for /Delta1vs on the order of 1 km/s or higher, the required propellant exceeds 25 percent of the spacecraft mass prior to the burn. There are no refueling stations in space, so a mission’s delta-v schedule must be carefully planned to minimize the propellant mass carried aloft in favor ofpayload. 6.3 Hohmann transfer 257 T able 6.1 Some typical specific impulses Propellant Isp(seconds) Cold gas 50 Monopropellant hydrazine 230 Solid propellant 290 Nitric acid/monomethylhydrazine 310 Liquid oxygen/liquid hydrogen 455 0.0010.010.1 10 100 1000 1 10 000 5 50 500 5000 2 20 200 2000m/H9004m /H9004y, m/s Isp /H11005 250 Isp /H11005 350 Isp /H11005 450 Figure 6.1 Propellant mass fraction versus /Delta1vfor typical specific impulses. 6.3 Hohmann transfer The Hohmann transfer (Hohmann, 1925) is the most energy efficient two-impulse maneuver for transferring between two coplanar circular orbits sharing a commonfocus. The Hohmann transfer is an elliptical orbit tangent to both circles at its apseline, as illustrated in Figure 6.2. The periapse and apoapse of the transfer ellipse arethe radii of the inner and outer circles, respectively. Obviously, only one-half of theellipse is flown during the maneuver, which can occur in either direction, from theinner to the outer circle, or vice versa. It may be helpful in sorting out orbit transfer strategies to use the fact that the energy of an orbit depends only on its semimajor axis a. Recall that for an ellipse (Equation 2.70), the specific energy is negative, ε=−µ 2a Increasing the energy requires reducing its magnitude, in order to make εless negative. Therefore, the larger the semimajor axis is, the more the energy the orbit has. InFigure 6.2, the energies increase as we move from the inner to the outer circle. Starting at Aon the inner circle, a velocity increment /Delta1v Ain the direction of flight is required to boost the vehicle onto the higher-energy elliptical trajectory. Aftercoasting from AtoB, another forward velocity increment /Delta1v Bplaces the vehicle on the still higher-energy, outer circular orbit. Without the latter delta-v burn, the spacecraftwould, of course, remain on the Hohmann transfer ellipse and return to A. The total energy expenditure is reflected in the total delta-v requirement, /Delta1v total=/Delta1vA+/Delta1vB. 258 Chapter 6 Orbital maneuvers 12 r1 r2Periapse ApoapseHohmann transfer ellipse A B Figure 6.2 Hohmann transfer. The same total delta-v is required if the transfer begins at Bon the outer circular orbit. Since moving to the lower-energy inner circle requires lowering the energy ofthe spacecraft, the /Delta1vs must be accomplished by retrofires. That is, the thrust of the maneuvering rocket is directed opposite to the flight direction in order to act as abrake on the motion. Since /Delta1vrepresents the same propellant expenditure regardless of the direction the thruster is aimed, when summing up /Delta1vs, we are concerned only with their magnitudes. Example 6.1A spacecraft is in a 480 km by 800 km earth orbit (orbit 1 in Figure 6.3). Find (a) the /Delta1vrequired at perigee Ato place the spacecraft in a 480 km by 16 000 km transfer orbit (orbit 2); and (b) the /Delta1v(apogee kick) required at Bof the transfer orbit to establish a circular orbit of 16 000 km altitude (orbit 3). Earth 1 3/H9004yA /H9004yBAB CHohmann transfer ellipse Circular orbit of radius 22 378 kmPerigee of orbit 1 (z /H11005 480 km)Apogee of orbit 1 (z /H11005 800 km)2 Figure 6.3 Hohmann transfer between two earth orbits. 6.3 Hohmann transfer 259 (a) First, let us establish the primary orbital parameters of the original orbit 1. The perigee and apogee radii are rA=RE+zA=6378+480=6858 km rC=RE+zC=6378+800=7178 km Therefore, the eccentricity of orbit 1 is e1=rC−rA rC+rA=0.022799 Applying the orbit equation at perigee of orbit 1, we calculate the angular momentum, rA=h2 1 µ1 1+e1cos(0)⇒ h1=52 876 km2/slashbig s With the angular momentum, we can calculate the speed at Aon orbit 1, vA)1=h1 rA=7.7102 km /s( a) Moving to the transfer orbit 2, we proceed in a similar fashion to get rB=RE+zB=6378+16 000 =22 378 km e2=rB−rA rB+rA=0.53085 rA=h2 2 µ1 1+e2cos(0)⇒ h2=64 690 km Thus, the speed at Aon orbit 2 is vA)2=h2 rA=64 690 6858=9.4327 km /s (b) The required forward velocity increment at Ais now obtained from (a) and (b) as /Delta1vA=vA)2−vA)1=1.7225 km /s (b) We use the angular momentum formula to find the speed at Bon orbit 2, vB)2=h2 rB=64 690 22 378=2.8908 km /s( c) Orbit 3 is circular, so its constant orbital speed is obtained from Equation 2.53, vB)3=/radicalbigg 398 600 22 378=4.2204 km /s( d) 260 Chapter 6 Orbital maneuvers (Example 6.1 continued)Thus, the delta-v requirement at Bto climb from orbit 2 to orbit 3 is /Delta1vB=vB)3−vB)2=4.2204−2.8908=1.3297 km /s Observe that the total delta-v requirement for this Hohmann transfer is /Delta1vtotal=|/Delta1vA|+|/Delta1vB|=1.7225+1.3297=3.0522 km /s In the previous example the initial orbit of the Hohmann transfer sequence was an ellipse, rather than a circle. Since no real orbit is perfectly circular, we must generalizethe notion of a Hohmann transfer to include two impulsive transfers between ellipticalorbits that are coaxial, i.e., share the same apse line, as shown in Figure 6.4. Thetransfer ellipse must be tangent to both the initial and target ellipses 1 and 2. As can be seen, there are two such transfer orbits, 3 and 3 /prime. It is not immediately obvious which of the two requires the lowest energy expenditure. T o find out which is the best transfer orbit in general, we must calculate the individual total delta-v requirement for orbits 3 and 3/prime. This requires finding the velocities at A,A/prime,BandB/primefor each pair of orbits having those points in common. T o do so, recall from Equation 2.74 that for an ellipse, e=ra−rp ra+rp where rpand raare the radii to periapse and apoapse, respectively. Evaluating the orbit equation at periapse rp=h2 µ1 1+e=h2 µ1 1+ra−rp ra+rp A BA /H110322 3/H110323 1rB rArB/H11032 B/H11032 rA/H11032 Figure 6.4 Hohmann transfers between coaxial elliptical orbits. In this illustration, rA/prime/r0=3,rB/r0=8 and rB/prime/r0=4. 6.3 Hohmann transfer 261 yields the angular momentum in terms of the periapse and apoapse radii, h=/radicalbig 2µ/radicalBigg rarp ra+rp(6.2) Equation 6.2 is used to evaluate the angular momentum of each of the four orbits in Figure 6.4: h1=/radicalbig 2µ/radicalbiggrArA/prime rA+rA/primeh3=/radicalbig 2µ/radicalbiggrArB rA+rB h2=/radicalbig 2µ/radicalbiggrBrB/prime rB+rB/primeh3/prime=/radicalbig 2µ/radicalbiggrA/primerB/prime rA/prime+rB/prime From these we obtain the velocities, vA)1=h1 rAvA)3=h3 rA vB)2=h2 rBvB)3=h3 rB vA/prime)1=h1 rA/primevA/prime)3/prime=h3/prime rA/prime vB/prime)2=h2 rB/primevB/prime)3/prime=h3/prime rB/prime These lead to the delta-vs /Delta1vA=|vA)3−vA)1| /Delta1vB=|vB)2−vB)3| /Delta1vA/prime=|vA/prime)3/prime−vA/prime)1|/Delta1vB/prime=|vB/prime)2−vB/prime)3/prime| and, finally, to the total delta-v requirement for the two possible transfer trajectories, /Delta1vtotal)3=/Delta1vA+/Delta1vB/Delta1vtotal)3/prime=/Delta1vA/prime+/Delta1vB/prime If/Delta1vtotal)3/prime//Delta1v total)3>1, then orbit 3 is the most efficient. On the other hand, if /Delta1vtotal)3/prime//Delta1v total)3<1, then orbit 3/primeis more efficient than orbit 3. Three contour plots of /Delta1vtotal)3/prime//Delta1v total)3are shown in Figure 6.5, for three different shapes of the inner orbit 1 of Figure 6.4. Figure 6.5(a) is for rA/prime/rA=3, which is the situation represented in Figure 6.4, in which point Ais the periapse of the initial ellipse. In Figure 6.5(b) rA/prime/rA=1, which means the starting ellipse is a circle. Finally, in Figure 6.5(c) rA/prime/rA=1/3, which corresponds to an initial orbit of the same shape as orbit 1 in Figure 6.4, but with point Abeing the apoapse instead of periapse. Figure 6.5(a), for which rA/prime>rA, implies that if point Ais the periapse of orbit 1, then transfer orbit 3 is the most efficient. Figure 6.5(c), for which rA/prime<rA, shows that if point A/primeis the periapse of orbit 1, then transfer orbit 3/primeis the most efficient. T ogether, these results lead us to conclude that it is most efficient for the transferorbit to begin at the periapse on the inner orbit 1, where its kinetic energy is greatest,regardless of shape of the outer target orbit. If the starting orbit is a circle, thenFigure 6.5(b) shows that transfer orbit 3 /primeis most efficient if rB/prime>rB. That is, from an inner circular orbit, the transfer ellipse should terminate at apoapse of the outer target ellipse, where the speed is slowest. 262 Chapter 6 Orbital maneuvers 468 1 0 2246810 0.6 0.7 0.8 0.9 1.0 468 1 0 2246810 0.90.85 1.0 1.1 1.2246810 468 1 01.1 1.2 1.3 1.4 1.5 (a) (b) (c)rB/H11032/rArA/H11032/rA/H110053 rB/rA rB/rA rB/rArA/H11032/rA/H110051/3 rA/H11032/rA/H110051 rB/H11032/rArB/H11032/rA Figure 6.5 Contour plots of /Delta1vtotal)3/prime//Delta1v total)3for different relative sizes of the ellipses in Figure 6.4. Note that rB>rA/primeand rB/prime>rA. If the Hohmann transfer is in the reverse direction, i.e., to a lower-energy inner orbit, the above analysis still applies, since the same total delta-v is required whetherthe Hohmann transfer runs forwards or backwards. Thus, from an outer circle orellipse to an inner ellipse, the most energy-efficient transfer ellipse terminates atperiapse of the inner target orbit. If the inner orbit is a circle, the transfer ellipseshould start at apoapse of the outer ellipse. We close this section with an illustration of the careful planning required for one spacecraft to rendezvous with another at the end of a Hohmann transfer. Example 6.2A spacecraft returning from a lunar mission approaches earth on a hyperbolic trajec- tory. At its closest approach Ait is at an altitude of 5000 km, traveling at 10 km/s. At Aretrorockets are fired to lower the spacecraft into a 500 km altitude circular orbit, where it is to rendezvous with a space station. Find the location of the space station at retrofire so that rendezvous will occur at B. The time of flight from AtoBis one-half the period T2of the elliptical transfer orbit 2. While the spacecraft coasts from AtoB, the space station coasts through the angle φCBfrom CtoB. Hence, this mission has to be carefully planned and executed, going all the way back to lunar departure, so that the two vehicles meet at B. T o calculate the period T2, we must first obtain the primary orbital parameters, eccentricity and angular momentum. The apogee and perigee of orbit 2, the transfer ellipse, are rA=5000+6378=11 378 km rB=500+6378=6878 km Therefore, the eccentricity is e2=11 378 −6878 11 378 +6878=0.24649 Evaluating the orbit equation at perigee yields the angular momentum, rB=h2 2 µ1 1+e2⇒ 6878=h2 2 398 6001 1+0.24649⇒ h2=58 458 km2/s 6.3 Hohmann transfer 263 A B 5000 kmEarth500 km circular orbit 1 2/H9004yB3 Position of space station when spacecraft is at AC /H9004/H9271A fCB Figure 6.6 Relative position of spacecraft and space station at beginning of the transfer ellipse. Now we can use Equation 2.72 to find the period of the transfer ellipse, T2=2π µ2 h2/radicalBig 1−e2 2 3 =2π 398 6002/parenleftbigg58 458√ 1−0.246492/parenrightbigg3 =8679.1 s (a) The period of circular orbit 3 is, according to Equation 2.54, T3=2π√µr3 2 B=2π√ 398 60068783 2=5676.8 s (b) The time of flight from CtoBon orbit 3 must equal the time of flight from AtoB on orbit 2. /Delta1tCB=1 2T2=1 2·8679.1 =4339.5s Since orbit 3 is a circle, its angular velocity, unlike an ellipse, is constant. Therefore, we can write φCB /Delta1tCB=360◦ T3⇒φCB=4339.5 5676.8·360=275.2◦ (The student should verify that the total delta-v required to lower the spacecraft from the hyperbola into the parking orbit is 6.415 km/s. A glance at Figure 6.1 reveals thetremendous amount of propellant this would require.) 264 Chapter 6 Orbital maneuvers A BC12 34rArCrB FD Figure 6.7 Bi-elliptic transfer from inner orbit 1 to outer orbit 4. 6.4 Bi-elliptic Hohmann transfer A Hohmann transfer from circular orbit 1 to circular orbit 4 in Figure 6.7 is the dotted ellipse lying inside the outer circle, outsid e the inner circle, and tangent to both. The bi-elliptical Hohmann transfer uses two co axial semi-ellipses, 2 and 3, which extend beyond the outer target orbit. Each of the two ellipses is tangent to one of the circularorbits, and they are tangent to each other at B, which is the apoapse of both. The idea is to place Bsufficiently far from the focus that the /Delta1v Bwill be very small. In fact, as rBapproaches infinity, /Delta1vBapproaches zero. For the bi-elliptical scheme to be more energy efficient than the Hohmann transfer, it must be true that /Delta1vtotal)bi-elliptical </Delta1 v total)Hohmann (6.3) Delta-v analyses of the Hohmann and bi-elliptical transfers lead to the following results, /Delta1v)Hohmann =/bracketleftBigg 1√α−√ 2(1−α)√α(1+α)−1/bracketrightBigg/radicalbiggµ rA /Delta1v)bi-elliptical =/bracketleftBigg/radicalBigg 2(α+β) αβ−1+√α√α−/radicalBigg 2 β(1+β)(1−β)/bracketrightBigg/radicalbiggµ rA(6.4a) where α=rC rAβ=rB rA(6.4b) 6.4 Bi-elliptic Hohmann transfer 265 5 10 15 20 2520406080100 11.94rB rA rC rArB /H11005 r C∆ybi-elliptical /H11005 ∆y Hohmann ∆ybi-elliptical /H11021 ∆y Hohmann ∆ybi-elliptical /H11022 ∆y Hohmann Figure 6.8 Orbits for which the bi-elliptical transfer is either less efficient or more efficient than the Hohmann transfer. Plotting the difference between Hohmann and bi-elliptical /Delta1vtotalas a function of α andβreveals the regions in which the difference is positive, negative and zero. These are shown in Figure 6.8. From the figure we see that if the radius rCof the outer circular target orbit is less than about 11.9 times that of the inner one (r A), the standard Hohmann maneuver is the more energy efficient. If the ratio exceeds about 15, then the bi-elliptical strategyis better in that regard. Between those two ratios, large values of the apoapse radiusr Bfavor the bi-elliptical transfer, while smaller values favor the Hohmann transfer. Small gains in energy efficiency may be more than offset by the much longer flight times around the bi-elliptical trajectories as compared with the time of flight on thesingle semi-ellipse of the Hohmann transfer. Example 6.3Find the total delta-v requirement for a bi-elliptical Hohmann transfer from a geo- centric circular orbit of 7000 km radius to one of 105 000 km radius. Let the apogee of the first ellipse be 210 000 km. Compare the delta-v schedule and total flight time with that for an ordinary single Hohmann transfer ellipse. Since rA=7000 km rB=210 000 km rC=rD=105 000 km we have rB/rA=30 and rC/rA=15, so that from Figure 6.8 it is apparent right away that the bi-elliptic transfer will be the more energy efficient. T o do the delta-v analysis requires analyzing each of the five orbits. Orbit 1: Since this is a circular orbit, we have, simply, vA)1=/radicalbiggµ rA=/radicalbigg 398 600 7000=7.546 km /s( a) 266 Chapter 6 Orbital maneuvers (Example 6.3 continued) A D B4 105 000 km 210 000 km3Circular target orbit 57000 km radius initial orbit2 Hohmann transfer ellipse1Bi-elliptic trajectoriesC Figure 6.9 Bi-elliptic transfer. Orbit 2: For this transfer ellipse, Equation 6.2 yields h2=/radicalbig 2µ/radicalbiggrArB rA+rB=√ 2·398 600/radicalbigg 7000·210 000 7000+210 000=73 487 km2/s Therefore, vA)2=h2 rA=73 487 7000=10.498 km /s (b) vB)2=h2 rB=73 487 210 000=0.34994 km /s( c ) Orbit 3: For the second transfer ellipse, we have h3=√ 2·398 600/radicalbigg 105 000 ·210 000 105 000 +210 000=236 230 km2/s From this we obtain vB)3=h3 rB=236 230 210 000=1.1249 km /s( d ) vC)3=h3 rC=236 230 105 000=2.2498 km /s( e ) 6.4 Bi-elliptic Hohmann transfer 267 Orbit 4: The target orbit, like orbit 1, is a circle, which means vC)4=vD)4=/radicalbigg 398 600 105 000=1.9484 km /s( f) For the bi-elliptical maneuver, the total delta-v is, therefore, /Delta1vtotal)bi-elliptical =/Delta1vA+/Delta1vB+/Delta1vC =|vA)2−vA)1|+|vB)3−vB)2|+|vC)4−vC)3| =|10.498 −7.546|+| 1.1249 −0.34994|+| 1.9484 −2.2498| =2.9521 +0.77496 +0.30142 or, /Delta1vtotal)bi-elliptical =4.0285 km /s (g) The semimajor axes of transfer orbits 2 and 3 are a2=1 2(7000+210 000 )=108 500 km a3=1 2(105 000 +210 000 )=157 500 km With this information and the period formula, Equation 2.73, the time of flight for the two semi-ellipses of the bi-elliptical transfer is found to be tbi-elliptical =1 2/parenleftbigg2π√µa3 2 2+2π√µa3 2 3/parenrightbigg =488 870 s =5.66 days (h) For the Hohmann transfer ellipse 5, h5=√ 2·398 600/radicalbigg 7000·105 000 7000+105 000=72 330 km2/s Hence, vA)5=h5 rA=72 330 7000=10.333 km /s( i) vD)5=h5 rD=72 330 105 000=0.68886 km /s (j) It follows that /Delta1vtotal)Hohmann =|vA)5−vA)1|+|vD)5−vD)1| =(10.333 −7.546) +(1.9484 −0.68886) =2.7868 +1.2595 268 Chapter 6 Orbital maneuvers (Example 6.3 continued)or /Delta1vtotal)Hohmann =4.0463 km /s (k) This is only slightly (0.44 percent) larger than that of the bi-elliptical transfer. Since the semimajor axis of the Hohmann semi-ellipse is a5=1 2(7000+105 000 )=56 000 km the time of flight from AtoDis tHohmann =1 2/parenleftbigg2π√µa3 2 5/parenrightbigg =65 942 s =0.763 days (l) The time of flight of the bi-elliptical maneuver is over seven times longer than that of the Hohmann transfer. 6.5 Phasing maneuvers A phasing maneuver is a two-impulse Hohmann transfer from and back to the same orbit, as illustrated in Figure 6.10. The Hohmann transfer ellipse is the phasing orbitwith a period selected to return the spacecraft to the main orbit within a specifiedtime. Phasing maneuvers are used to change the position of a spacecraft in its orbit.If two spacecraft, destined to rendezvous, are at different locations in the same orbit,then one of them may perform a phasing maneuver in order to catch the otherone. Communications and weather satellites in geostationary earth orbit use phasingmaneuvers to move to new locations above the equator. In that case, the rendezvous 012 T00.8606 T0 1.146 T0PA Figure 6.10 Main orbit (0) and two phasing orbits, faster (1) and slower (2). T0is the period of the main orbit. 6.5 Phasing maneuvers 269 is with an empty point in space rather than with a physical target. In Figure 6.10, phasing orbit 1 might be used to return to Pin less than one period of the main orbit. This would be appropriate if the target is ahead of the chasing vehicle. Note that aretrofire is required to enter orbit 1 at P. That is, it is necessary to slow the spacecraft down in order to speed it up, relative to the main orbit. If the chaser is ahead of thetarget, then phasing orbit 2 with its longer period might be appropriate. A forwardfire of the thruster boosts the spacecraft’s speed in order to slow it down. Once the period Tof the phasing orbit is established, then Equation 2.73 should be used to determine the semimajor axis of the phasing ellipse, a=/parenleftbiggT√ µ 2π/parenrightbigg2 3 (6.5) With the semimajor axis established, the radius of point Aopposite to Pis obtained from the fact that 2 a=rP+rA. It is then apparent whether Pis periapse or apoapse, so that Equation 2.74 can be used to calculate the eccentricity of the phasing orbit.The orbit equation, Equation 2.35, may then be applied at either PorAto obtain the angular momentum, whereupon the phasing orbit is characterized completely. Example 6.4Spacecraft at Aand Bare in the same orbit (1). At the instant shown, the chaser vehicle at Aexecutes a phasing maneuver so as to catch the target spacecraft back at A after just one revolution of the chaser’s phasing orbit (2). What is the required total delta-v? 13 600 km 6800 kmAB 21 (Phasing orbit) C EarthD Figure 6.11 Phasing maneuver. From the figure, rA=6800 km rC=13 600 km 270 Chapter 6 Orbital maneuvers (Example 6.4 continued)Orbit 1: The eccentricity of orbit 1 is e1=rC−rA rC+rA=0.33333 Evaluating the orbit equation at A, we find rA=h2 1 µ1 1+e1cos(0)⇒ 6800=h2 1 398 6001 1+0.3333⇒ h1=60 116 km2/s The period is found using Equation 2.72, T1=2π µ2 h1/radicalBig 1−e2 1 3 =2π 398 6002/parenleftbigg60 116√ 1−0.333332/parenrightbigg3 =10 252 s Since Ais perigee, there is no radial velocity component there. The speed, directed entirely in the transverse direction, is found from the angular momentum formula, vA1=h1 rA=60 116 6800=8.8406 km /s The phasing orbit must have a period T2equal to the time it takes the target vehi- cle at Bto coast around to point Aon orbit 1. We can determine the flight time by calculating the time /Delta1tABfrom AtoBand subtracting that result from the period T1of orbit 1. At Bthe true anomaly is θA=90◦. Therefore, according to Equation 3.10a, tanEB 2=/radicalBigg 1−e1 1+e1tanθB 2=/radicalbigg 1−0.33333 1+0.33333tan90◦ 2 =0.70711 ⇒EB=1.2310 rad Then, from Kepler’s equation (Equations 3.5 and 3.11), we get /Delta1tAB=T1 2π(EB−e1sinEB)=10 252 2π(1.231−0.33333 ·sin 1.231)=1495.7s Thus, the time of flight of the target spacecraft from BtoAis /Delta1tBA=T1−/Delta1tAB=10 252 −1495.7=8756.3s Orbit 2: The period of orbit 2 must equal /Delta1tBAso that the chaser will arrive at Awhen the target does. That is, T2=8756.3s 6.5 Phasing maneuvers 271 This, together with the period formula, Equation 2.73, yields the semimajor axis of orbit 2, T2=2π√µa3 2 2⇒ 8756.2 =2π√ 398 600a3 2 2⇒ a2=9182.1k m ( a ) Since 2a 2=rA+rD, we find rD=2a2−rA=2·9182.1 −6800=11 564 km Therefore, point Ais indeed the perigee of orbit 2, the eccentricity of which can now be determined: e2=rD−rA rD+rA=0.25943 Evaluating the orbit equation at point Aof orbit 2 yields its angular momentum, rA=h2 2 µ1 1+e2cos(0)⇒ 6800=h2 2 398 6001 1+0.25943⇒ h2=58 426 km2/s Finally, we can calculate the speed at perigee of orbit 2, vA2=h2 rA=58 426 6800=8.5921 km /s At the beginning of the phasing maneuver, /Delta1vA=vA2−vA1=8.5921 −8.8406 =− 0.24851 km /s At the end of the phasing maneuver, /Delta1vA=vA1−vA2=8.8406 −8.5921 =0.24851 km /s The total delta-v, therefore, is /Delta1vtotal=| − 0.24851|+| 0.24851|=0.4970 km /s Example 6.5It is desired to shift the longitude of a GEO satellite 12◦westward in three revolutions of its phasing orbit. Calculate the delta-v requirement. This problem is illustrated in Figure 6.12. It may be recalled from Equations 2.57, 2.58 and 2.59 that the angular velocity of the earth, the radius to GEO and the speed in GEO are, respectively, ωE=ωGEO=72.922 ×10−6rad/s rGEO=42 164 km vGEO=3.0747 km /s(a) 272 Chapter 6 Orbital maneuvers (Example 6.5 continued)Let/Delta1/Lambda1 be the change in longitude in radians. Then the period T2of the phasing orbit can be obtained from the following formula, ωE(3T2)=3·2π+/Delta1/Lambda1 (b) which states that after three circuits of the phasing orbit, the original position of the satellite will be /Delta1/Lambda1 radians east of P. In other words, the satellite will end up /Delta1/Lambda1 radians west of its original position in GEO, as desired. From (b) we obtain, T2=1 3/Delta1/Lambda1+6π ωE=1 3·12◦·π 180◦+6π 72.922×10−6=87 121 s North Pole12°Original position Target positionP WestEarthEast A BC12 Phasing orbitGEO Figure 6.12 GEO repositioning. Note that the period of GEO is TGEO=2π ωGEO=86 163 s The satellite in its slower phasing orbit appears to drift westward at the rate ˙/Lambda1=/Delta1/Lambda1 3T2=8.0133×10−7rad/s=3.9669◦/day Having the period, we can use Equation 6.5 to obtain the semimajor axis of orbit 2, a=/parenleftbiggT√µ 2π/parenrightbigg2 3 =/parenleftBigg 87 121√ 398 600 2π/parenrightBigg2 3 =42 476 km From this we find the radial coordinate of C, 2a2=rP+rC⇒ rC=2·42 476 −42 164 =42 787 km 6.6 Non-Hohmann transfers with a common apse line 273 Now we can find the eccentricity of orbit 2, e2=rC−rA rC+rA=42 787 −42 164 42 787 +42 164=0.0073395 and the angular momentum follows from applying the orbit equation at P(orC)o f orbit 2: rP=h2 2 µ1 1+e2cos (0)⇒42 164 =h2 2 398 6001 1+0.0073395⇒h2=130 120 km2/s AtPthe speed in orbit 2 is vP2=130 120 42 164=3.0859 km /s Therefore, at the beginning of the phasing orbit, /Delta1v=vP2−vGEO=3.0859 −3.0747 =0.01126 km /s at the end of the phasing maneuver, /Delta1v=vGEO−vP2=3.0747 −3.08597 =− 0.01126 km /s Therefore, /Delta1vtotal=|0.01126|+| − 0.01126|=0.022525 km /s 6.6 Non-Hohmann transfers with a common apse line Figure 6.13 illustrates a transfer between tw o coaxial elliptical orbits in which the transfer trajectory shares the apse line but is not necessarily tangent to either the initialor target orbit. The problem is to determine whether there exists such a trajectoryjoining points Aand B, and, if so, to find the total delta-v requirement. r Aand rBare given, as are the true anomalies θAandθB. Because of the com- mon apse line assumption, θAandθBare the true anomalies of points Aand B on the transfer orbit as well. Applying the orbit equation to Aand Bon orbit 3 yields rA=h2 3 µ1 1+e3cosθA rB=h2 3 µ1 1+e3cosθB 274 Chapter 6 Orbital maneuvers AB rB F Common apse line2 13 uAuBrA ˆpˆ q Figure 6.13 Non-Hohmann transfer (3) between two coaxial elliptical orbits. Solving these two equations for e3and h3,w eg e t e3=rB−rA rAcosθA−rBcosθB h3=√µrArB/radicalBigg cosθA−cosθB rAcosθA−rBcosθB(6.6) With these, the transfer orbit is determined and velocity may be found at any true anomaly. For a Hohmann transfer, in which θA=0 andθB=π, Equations 6.6 become e3=rB−rA rB+rAh3=/radicalbig 2µ/radicalbiggrArB rA+rB(Hohmann transfer) (6.7) When a delta-v calculation is done at a point which is not on the apse line, care must be taken to include the change in direction as well as the magnitude of the velocity vector. Figure 6.14 shows a point where an impulsive maneuver changes the velocityvector from v 1on orbit 1 to v2on orbit 2. The difference in length of the two vectors shows the change in the speed, and the difference in the flight path angles indicatesthe change in the direction. It is important to observe that the /Delta1vwe seek is the magnitude of the change in the velocity vector, not the change in its magnitude(speed). That is, /Delta1v=| | v 2−v1|| (6.8) Only if v1and v2are parallel, as in Hohmann transfers, is it true that /Delta1v=| | v2|| − ||v1||. From Figure 6.14 and the law of cosines, we find that /Delta1v=/radicalBig v2 1+v2 2−2v1v2cos/Delta1γ (6.9) where v1=/bardblv1/bardbl,v2=/bardblv2/bardbland/Delta1γ=γ2−γ1. 6.6 Non-Hohmann transfers with a common apse line 275 v1g1g2 v2 F12B rBφ Local horizon∆v ∆γ Figure 6.14 Vector diagram of the change in velocity and flight path angle at the intersection of two orbits. The direction of /Delta1vshows the required alignment of the thruster that produces the impulse. The orientation of /Delta1vrelative to the local horizon is found by replacing vrandv⊥in Equation 2.41 by /Delta1vrand/Delta1v⊥, so that tanφ=/Delta1vr /Delta1v⊥(6.10) where φis the angle from the local horizon to the /Delta1vvector. Finally, recall the formula for spe cific mechanical energy of an orbit, Equation 2.47, ε=v·v 2−µ r(v2=v·v) An impulsive maneuver results in a change of orbit and, therefore, a change in the specific energy ε. If the expenditure of propellant /Delta1m is negligible compared to the initial mass m1of the vehicle, then /Delta1ε=ε2−ε1. For the situation illustrated in Figure 6.14, ε1=v2 1 2−µ rB and ε2=(v1+/Delta1v)·(v 1+/Delta1v) 2−µ rB=v2 1+2v1·/Delta1v+/Delta1v2 2−µ rB 276 Chapter 6 Orbital maneuvers Hence /Delta1ε=v1·/Delta1v+/Delta1v2 2 From Figure 6.14 it is apparent that v1·/Delta1v=v1/Delta1vcos/Delta1γ, so that /Delta1ε=v1/Delta1vcos/Delta1γ+/Delta1v2 2=v1/Delta1v/parenleftbigg cos/Delta1γ+1 2/Delta1v v1/parenrightbigg For consistency with our assumption that /Delta1m<<m1, it must be true (recall Figure 6.1) that /Delta1v<<v 1. It follows that /Delta1ε≈v1/Delta1vcos/Delta1γ (6.11) This shows that, for a given /Delta1v, the change in specific energy is larger the faster the spacecraft is moving (unless, of course, the change in flight path angle is 90◦). The larger the /Delta1εassociated with a given /Delta1v, the more efficient the maneuver. As we know, a spacecraft has its greatest speed at periapsis. Example 6.6A geocentric satellite in orbit 1 of Figure 6.15 executes a delta-v maneuver at Awhich places it on orbit 2, for re-entry at D. Calculate /Delta1vatAand its direction relative to the local horizon. A1 2Earth150° 10 000 km 20 000 kmB CvA2 vA1 DLocal horizon g2g1∆vA ∆γ Figure 6.15 Non-Hohmann transfer with a common apse line. From the figure we see that rB=20 000 km rC=10 000 km rD=6378 km 6.6 Non-Hohmann transfers with a common apse line 277 Orbit 1: The eccentricity is e1=rB−rC rB+rC=0.33333 The angular momentum is obtained from the orbit equation, noting that point Cis perigee: rC=h2 1 µ1 1+e1cos(0)⇒ 10 000 =h2 1 398 6001 1+0.33333⇒ h1=72 902 km2/s With the angular momentum and the eccentricity, we can use the orbit equation to find the radial coordinate of point A, rA=72 9022 398 6001 1+0.33333 ·cos 150◦=18 744 km Equations 2.21 and 2.38 yield the transverse and radial components of velocity at A on orbit 1, v⊥A)1=h1 rA=3.8893 km /s( a) vrA)1=µ h1e1sin 150◦=0.91127 km /s From these we find the speed at A vA)1=/radicalBig v⊥A)2 1+vrA)2 1=3.9946 km /s and the flight path angle, γ1=tan−1vrA)1 v⊥A)1=tan−10.91127 3.8893=13.187◦ Orbit 2: The radius and true anomaly of points Aand Don orbit 2 are known. Applying the orbit equation at A,w eg e t 18 744 =h2 2 398 6001 1+e2cos 150◦⇒ h2 2=7.4715 ×109−6.4705 ×109e2(b) Likewise, at point D, which is perigee of orbit 2, 6378=h2 2 398 6001 1+e2⇒ h2 2=2.5423 ×109+2.5423 ×109e2 (c) Equating the expressions for h2 2in (b) and (c), and solving for e2, yields e2=0.54692 278 Chapter 6 Orbital maneuvers (Example 6.6 continued)whereupon either (b) or (c) may be used to find h2=62 711 km2/s Now we can calculate the radial and perpendicular components of velocity on orbit 2 at point A: v⊥A)2=h2 rA=3.3456 km /s vrA)2=µ h2e2sin 150◦=1.7381 km /s( d ) Hence, the speed and flight path angle at Aon orbit 2 are vA)2=/radicalBig v⊥A)2 2+vrA)2 2=3.7702 km /s γ2=tan−1vrA)2 v⊥A)2=tan−11.7381 3.3456=27.453◦ The change in the flight path angle as a result of the impulsive maneuver is /Delta1γ=γ2−γ1=27.453◦−13.187◦=14.266◦ With this we can use Equation 6.9 to finally obtain /Delta1vA, /Delta1vA=/radicalBig vA)2 1+vA)2 2−2vA)1vA)2cos/Delta1γ =/radicalbig 3.99462+3.77022−2·3.9946·3.7702·cos 14.266 /Delta1vA=0.9896 km /s (e) Note that /Delta1vAis the magnitude of the change in velocity vector /Delta1vAatA. That is not the same as the change in the magnitude of the velocity (i.e., the change in speed),which is v A)2−vA)1=3.9946−3.7702=0.2244 km /s. T o find the orientation of /Delta1vA, we use Equation 6.10, tanφ=/Delta1vr)A /Delta1v⊥)A=vrA)2−vrA)1 v⊥A)2−v⊥A)1=1.7381−0.9113 3.3456−3.8893=− 1.5207 ˆ u⊥123.3 °A/H9004vA 150°rA Local horizon Figure 6.16 Orientation of /Delta1vAto the local horizon. 6.7 Apse line rotation 279 so that φ=123.3◦ This angle is illustrated in Figure 6.16. Prior to firing, the spacecraft would have to be rotated so that the centerline of the rocket motor coincides with the line of action of/Delta1v A, with the nozzle aimed in the opposite direction. 6.7 Apse line rotation Figure 6.17 shows two intersecting orbits which have a common focus, but their apse lines are not collinear. A Hohmann transfer between them is clearly impossible. Theopportunity for transfer from one orbit to the other by a single impulsive maneuveroccurs where they intersect, at points Iand Jin this case. As can be seen from the figure, the rotation ηof the apse line is the difference between the true anomalies of the point of intersection, measured from periapse of each orbit. That is, η=θ 1−θ2 (6.12) We will consider two cases of apse line rotation. The first case is that in which the apse line rotation ηis given as well as the orbital parameters eand hof both orbits. The problem is then to find the true anomaly of Iand Jrelative to both orbits. The radius of the point of intersection Iis given by either of the following rI)1=h2 1 µ1 1+e1cosθ1rI)2=h2 2 µ1 1+e2cosθ2 FP AP' A'JI 12 ηu1 r Apse line of orbit 1 Apse line of orbit 2u2 Figure 6.17 Two intersecting orbits whose apse lines do not coincide. 280 Chapter 6 Orbital maneuvers Since rI)1=rI)2, we can equate these two expressions and rearrange terms to get e1h2 2cosθ1−e2h2 1cosθ2=h2 1−h2 2 Setting θ2=θ1−ηand using the trig identity cos ( θ1−η)=cosθ1cosη+sinθ1sinη leads to an equation for θ1 acosθ1+bsinθ1=c (6.13a) where a=e1h2 2−e2h2 1cosη b=− e2h2 1sinη c=h2 1−h2 2 (6.13b) Equation 6.13a has two roots (see Problem 3.9), corresponding to the two points of intersection Iand Jof the two orbits: θ1=φ±cos−1/parenleftBigc acosφ/parenrightBig (6.14a) where φ=tan−1b a(6.14b) Having found θ1we obtain θ2from Equation 6.12. /Delta1vfor the impulsive maneuver may then be computed as illustrated in the following example. Example 6.7An earth satellite is in an 8000 km by 16 000 km radius orbit (orbit 1 of Figure 6.18). Calculate the delta-v and the true anomaly θ1required to obtain a 7000 km by 21 000 km radius orbit (orbit 2) whose apse line is rotated 25◦counterclockwise. Indicate the orientation φof/Delta1vto the local horizon. 1 25° P1 A1 A2P2 Earth2 φI J15 175 km u1∆v Figure 6.18 /Delta1vproduces a rotation of the apse line. 6.7 Apse line rotation 281 The eccentricities of the two orbits are e1=rA1−rP1 rA1+rP1=16 000 −8000 16 000 +8000=0.33333 e2=rA2−rP2 rA2+rP2=21 000 −7000 21 000 +7000=0.5( a) The orbit equation yields the angular momenta rP1=h2 1 µ1 1+e1cos(0)⇒ 8000=h2 1 398 6001 1+0.33333⇒ h1=65 205 km2/s rP2=h2 2 µ1 1+e2cos(0)⇒ 7000=h2 2 398 6001 1+0.5⇒ h2=64 694 (b) Using these orbital parameters and the fact that η=25◦, we calculate the terms in Equations 6.13b, a=e1h2 2−e2h2 1cosη=0.3333 ·64 6942−0.5·65 2052·cos 25◦ =− 5.3159 ×108km4/s2 b=− e2h2 1sinη=− 0.5·65 2052sin 25◦=− 8.9843 ×108km4/s2 c=h2 1−h2 2=65 2052−64 6942=6.6433 ×107km4/s2 Then Equations 6.14 yield φ=tan−1−8.9843 ×108 −5.3159 ×108=59.39◦ θ1=59.39◦±cos−1/parenleftbigg6.6433 ×107 −5.3159 ×108cos 59.39◦/parenrightbigg =59.39◦±93.65◦ Thus, the true anomaly of point I, the point of interest, is θ1=153.04◦(c) (For point J,θ1=325.74◦.) With the true anomaly available, we can evaluate the radial coordinate of the maneuver point, r=h2 1 µ1 1+e1cos 153.04◦=15 175 km The velocity components and flight path angle for orbit 1 at point Iare v⊥1=h1 r=65 205 15 175=4.2968 km /s vr1=µ h1e1sin 153.04◦=398 600 65 205·0.33333 ·sin 153.04◦=0.92393 km /s γ1=tan−1vr1 v⊥1=12.135◦ 282 Chapter 6 Orbital maneuvers (Example 6.7 continued)The speed of the satellite in orbit 1 is, therefore, v1=/radicalBig v2r1+v2 ⊥1=4.3950 km /s Likewise, for orbit 2, v⊥2=h2 r=64 694 15 175=4.2631 km /s vr2=µ h2e2sin(153 .04◦−25◦)=398 600 64 694·0.5·sin 128 .04◦=2.4264 km /s γ2=tan−1vr2 v⊥2=29.647◦ v2=/radicalBig v2r2+v2 ⊥2=4.9053 km /s Equation 6.9 is used to find /Delta1v, /Delta1v=/radicalBig v2 1+v2 2−2v1v2cos(γ2−γ1) =/radicalbig 4.39502+4.90532−2·4.3950·4.9053 cos(29 .647◦−12.135◦) /Delta1v=1.503 km /s The angle φwhich the vector /Delta1vmakes with the local horizon is given by Equation 6.10, φ=tan−1/Delta1vr /Delta1v⊥=tan−1vr2−vr1 v⊥2−v⊥1=tan−12.4264−0.92393 4.2631−4.2968=91.28◦ The second case of apse line rotation is that in which the impulsive maneuver takes place at a given true anomaly θ1on orbit 1. The problem is to determine the angle of rotation ηand the eccentricity e2of the new orbit. The impulsive maneuver creates a change in the radial and transverse velocity components at point Iof orbit 1. From the angular momentum formula, h=rv⊥,w e obtain the angular momentum of orbit 2, h2=r(v⊥+/Delta1v⊥)=h1+r/Delta1v⊥ (6.15) The formula for radial velocity, vr=(µ/h)esinθ, applied to orbit 2 at point I,w h e r e vr2=vr1+/Delta1vrand θ2=θ1−η, yields vr1+/Delta1vr=µ h2e2sinθ2 6.7 Apse line rotation 283 Substituting Equation 6.15 into this expression and solving for sin θ2leads to sinθ2=1 e2(h1+r/Delta1v⊥)(µe 1sinθ1+h1/Delta1vr) µh1(6.16) From the orbit equation, we have at point I r=h2 1 µ1 1+e1cosθ1(orbit 1) r=h2 2 µ1 1+e2cosθ2(orbit 2) Equating these two expressions for r, substituting Equation 6.15, and solving for cosθ2, yields cosθ2=1 e2(h1+r/Delta1v⊥)2e1cosθ1+(2h 1+r/Delta1v⊥)r/Delta1v⊥ h2 1(6.17) Finally, substituting Equations 6.16 and 6.17 into the trig identity tan θ2=sinθ2/ cosθ2, we obtain tanθ2=h1 µ(h1+r/Delta1v⊥)(µe 1sinθ1+h1/Delta1vr) (h1+r/Delta1v⊥)2e1cosθ1+(2h 1+r/Delta1v⊥)r/Delta1v⊥(6.18a) Equation 6.18a can be simplified a bit by replacing µe1sinθ1with h1vr1and h1with rv⊥1, so that tanθ2=(v⊥1+/Delta1v⊥)(vr1+/Delta1vr) (v⊥1+/Delta1v⊥)2e1cosθ1+(2v⊥1+/Delta1v⊥)/Delta1v⊥v2 ⊥1 (µ/r )(6.18b) Equations 6.18 show how the apse line rotation, η=θ1−θ2, is completely determined by the components of /Delta1vimparted at the true anomaly θ1. After solving Equation 6.18 (a or b), we substitute θ2into Equation 6.16 or 6.17 to calculate the eccentricity e2of orbit 2. Therefore, with h2from Equation 6.15, the rotated orbit 2 is completely specified. If the impulsive maneuver takes place at the periapse of orbit 1, so that θ1=vr=0, and if it is also true that /Delta1v⊥=0, then Equation 6.18b yields tanη=−rv⊥1 µe1/Delta1vr(with radial impulse at periapse) Thus, if the velocity vector is given an outward radial component at periapse, then η<0, which means the apse line of the resulting orbit is rotated clockwise relative tothe original one. That makes sense, since having acquired v r>0 means the spacecraft is now flying away from its new periapse. Likewise, applying an inward radial velocity component at periapse rotates the apse line counterclockwise. 284 Chapter 6 Orbital maneuvers Example 6.8An earth satellite in orbit 1 of Figure 6.19 undergoes the indicated delta-v maneuver at its perigee. Determine the rotation ηof its apse line. 60° 7000 km17 000 km2 Earth P1 P2A1A2 η1 ∆υ /H11005 2 km/s Figure 6.19 Apse line rotation maneuver. From the figure rA1=17 000 km rP1=7000 km The eccentricity of orbit 1 is e1=rA1−rP1 rA1+rP1=0.41667 (a) As usual, we use the orbit equation to find the angular momentum, rP1=h2 1 µ1 1+e1cos(0)⇒ 7000=h2 1 398 6001 1+0.41667⇒ h1=62 871 km2/s At the maneuver point P1, the angular momentum formula and the fact that P1is perigee of orbit 1( θ1=0) imply that v⊥1=h1 rP1=62 871 7000=8.9816 km /s vr1=0 (b) From Figure 6.18 it is clear that /Delta1v⊥=/Delta1vcos 60◦=1k m/s /Delta1vr=/Delta1vsin 60◦=1.7321 km /s( c ) 6.8 Chase maneuvers 285 T o compute θ2, we use Equation 6.18b together with (a), (b) and (c): tanθ2=(v⊥1+/Delta1v⊥)(vr1+/Delta1vr) (v⊥1+/Delta1v⊥)2e1cosθ1+(2v⊥1+/Delta1v⊥)/Delta1v⊥v2 ⊥1 (µ/r P1) =(8.9816 +1)(0+1.7321) (8.9816 +1)2·0.41667· cos(0) +(2·8.9816 +1)·1·8.98162 (398 600/7000) =0.4050 It follows that θ2=22.047◦, so that Equation 6.12 yields η=− 22.05◦ which means the rotation of the apse line is clockwise, as indicated in Figure 6.19. From Equation 6.17 we obtain the eccentricity of orbit 2, e2=(h1+rP1/Delta1v⊥)2e1cosθ1+(2h 1+rP1/Delta1v⊥)rP1/Delta1v⊥ h2 1cosθ2 =(62 871 +7000·1)2·0.41667 ·cos(0) +(2·62 871 +7000·1)·7000·1 62 8712·cos 22.047◦ =0.808830 With this and the angular momentum we find using the orbit equation that the perigee and apogee radii of orbit 2 are rP2=h2 2 µ1 1+e2=69 8712 398 6001 1+0.808830=6771.1k m rA2=69 8712 398 6001 1−0.808830=64 069 km 6.8 Chase maneuvers Whereas Hohmann transfers and phasing maneuvers are leisurely, energy-efficient procedures that require some preconditions (e.g., coaxial elliptical, orbits) in orderto work, a chase or intercept trajectory is one which answers the question, ‘How doI get from point Ato point Bin space in a given amount of time?’ The nature of the orbit lies in the answer to the question rather than being prescribed at the outset.Intercept trajectories near a planet are likely to require delta-vs beyond the capabilitiesof today’s technology, so they are largely of theoretical rather than practical interest. We might refer to them as ‘star wars maneuvers.’ Chase trajectories can be found as solutions to Lambert’s problem (Section 5.3). 286 Chapter 6 Orbital maneuvers Example 6.9Spacecraft BandCare both in the geocentric elliptical orbit (1) shown in Figure 6.20, from which it can be seen that the true anomalies are θB=45◦andθC=150◦. At the instant shown, spacecraft Bexecutes a delta-v maneuver, embarking upon a trajectory (2) which will intercept vehicle Cin precisely one hour. Find the orbital parameters (eand h) of the intercept trajectory and the total delta-v required for the chase maneuver. 45° 30° 8100 km 18 900 kmBC P AEarth1 2 C'ˆ p ˆ q Figure 6.20 Intercept trajectory (2) required for Bto catch Cin one hour. First, we must determine the parameters of orbit 1 in the usual way. The eccentricity is found using the orbit’s perigee and apogee, shown in Figure 6.20, e1=18 900 −8100 18 900 +8100=0.4000 From the orbit equation, rP=h2 1 µ1 1+e1cos(0)⇒ 8100=h2 1 398 6001 1+0.4⇒ h1=67 232 km2/s Using Equation 2.72 yields the period, T1=2π µ2 h1/radicalBig 1−e2 1 3 =2π 398 6002/parenleftbigg67 232√ 1−0.42/parenrightbigg3 =15 610 s 6.8 Chase maneuvers 287 In perifocal coordinates (Equat ion 2.109) the position vector of Bis rB=h2 1 µ1 1+e1cosθB(cosθBˆp+sinθBˆq) =67 2322 398 6001 1+0.4 cos 45◦(cos 45◦ˆp+sin 45◦ˆq) or rB=6250.6 ˆp+6250.6 ˆq(km) (a) Likewise, according to Equation 2.115, the velocity at Bon orbit 1 is vB1=µ h[−sinθBˆp+(e+cosθB)ˆq]=398 600 67 232[−sin 45◦ˆp+(0.4+cos 45◦)ˆq] so that vB1=− 4.1922 ˆp+6.5637 ˆq(km/s) (b) Now we need to move spacecraft Calong orbit 1 to the position C/primethat it will occupy one hour later ( /Delta1t), when it will presumably be met by spacecraft B. T o do that, we must first calculate the time since perigee passage at C. Since we know the true anomaly, the eccentric anomaly follows from Equation 3.10a, tanEC 2=/radicalBigg 1−e1 1+e1tanθC 2=/radicalbigg 1−0.4 1+0.4tan150◦ 2=2.4432 ⇒EC=2.3646 rad Substituting this value into Kepler’s equation (Equations 3.5 and 3.11) yields the time since perigee passage, tC=T1 2π(EC−e1sinEC)=15 610 2π(2.3646 −0.4·sin 2.3646) =5178 s One hour later ( /Delta1t=3600 s), the spacecraft will be in intercept position at C/prime, tC/prime=tC+/Delta1t=5178+3600=8778 s The corresponding mean anomaly is Me)C/prime=2πtC/prime T1=2π8778 15 610=3.5331 rad With this value of the mean anomaly, Kepler’s equation becomes EC/prime−e1sinEC/prime=3.5331 Applying Algorithm 4.1 to the solution of this equation we get EC/prime=3.4223 rad Substituting this result into Equation 3.10a yields the true anomaly at C/prime, tanθC/prime 2=/radicalbigg 1+0.4 1−0.4tan3.4223 2=− 10.811 ⇒θC/prime=190.57◦ 288 Chapter 6 Orbital maneuvers (Example 6.9 continued)We are now able to calculate the perifocal position and velocity vectors at C/primeon orbit 1: rC/prime=67 2322 398 6001 1+0.4 cos 190 .57◦(cos 190 .57◦ˆp+sin 4190 .57◦ˆq) =− 18 372 ˆp−3428.1ˆq(km) vC/prime 1=398 600 67 232[−sin 190 .57◦ˆp+(0.4+cos 190 .57◦)ˆq] =1.0875ˆp−3.4566ˆq(km/s) (c) The intercept trajectory connecting points Band C/primeare found by solving Lambert’s problem. Substituting rBand rC/primealong with /Delta1t=3600 s into Algorithm 5.2 yields vB2=− 8.1349ˆp+4.0506ˆq(km/s) (d) vC/prime 2=− 3.4745ˆp−4.7943ˆq(km/s) (e) These velocities are most easily obtained by running the following MATLAB script, which executes Algorithm 5.2 by means of the function M-file lambert.m(Appendix D.11). clearglobal mu deg = pi/180; mu = 398600; e = 0.4;h = 67232; theta1 = 45*deg;theta2 = 190.57*deg; delta_t = 3600;rB = hˆ2/mu/(1 + e*cos(theta1)) ... *[cos(theta1),sin(theta1),0]; rC_prime = hˆ2/mu/(1 + e*cos(theta2)) ... *[cos(theta2),sin(theta2),0]; string = 'pro';[vB2 vC_prime_2] = lambert(rB, rC_prime, ... delta_t, string) From (b) and (d) we find /Delta1vB=vB2−vB1=− 3.9426ˆp−2.5132ˆq(km/s) whereas (c) and (e) yield /Delta1vC/prime=vC/prime 1−vC/prime 2=4.5620ˆp+1.3376ˆq(km/s) 6.8 Chase maneuvers 289 The anticipated, extremely large, delta-v requirement for this chase maneuver is the sum of the magnitudes of these two vectors, /Delta1v=| |/Delta1vB| |+| |/Delta1vC/prime|| = 4.6755 +4.7540 =9.430 km /s We know that orbit 2 is an ellipse. T o pin it down a bit more, we can use rBand vB2 to obtain the orbital elements from Algorithm 4.1, which yields h2=76 167 km2/s e2=0.8500 a2=52 449 km θB2=319.52◦ These may be found quickly by running the following MATLAB script, in which the M-function coe_from_sv.m is Algorithm 4.1 (see Appendix D.8): clearglobal mu mu = 398600;rB = [6250.6 6250.6 0]; vB2 = [-8.1349 4.0506 0];orbital_elements = coe_from_sv(rB, vB2) The details of the intercept trajectory and the delta-v maneuvers are shown in Fig-ure 6.21. A far less dramatic though more leisurely (and realistic) way for Bto catch up with Cwould be to use a phasing maneuver. B 45˚ EarthC B', C'Intercept trajectory (ellipse)Perigee 94.51˚1 4.754 km/s3.624 km/s5.92 km/s4.675 km/s 7.79 km/s 9.09 km/s 2 10.57˚ 18 689 km8840 km Apse lineApse line Figure 6.21 Details of the large elliptical orbit, a portion of which serves as the intercept trajectory. 290 Chapter 6 Orbital maneuvers 6.9 Plane change maneuvers Orbits having a common focus Fneed not, and generally do not, lie in a common plane. Figure 6.22 shows two such orbits and their line of intersection BD.Aand Pdenote the apoapses and periapses. Since the common focus lies in every orbital plane, it must lie on the line of intersection of any two orbits. For a spacecraft inorbit 1 to change its plane to that of orbit 2 by means of a single delta-v maneuver(cranking maneuver), it must do so when it is on the line of intersection of the orbitalplanes. Those two opportunities occur only at points Band Din Figure 6.22(a). A view down the line of intersection, from Btowards D, is shown in Figure 6.22(b). Here we can see in true view the dihedral angle δbetween the two planes. The trans- verse component of velocity v ⊥atBis evident in this perspective, whereas the radial component vr, lying as it does on the line of intersection, is normal to the view plane (thus appearing as a dot). It is apparent that changing the plane of orbit 1 requiressimply rotating v ⊥around the intersection line, through the dihedral angle. If the magnitudes of v⊥and vrremain unchanged in the process, then we have a rigid body rotation of the orbit. That is, except for its new orientation in space, the orbit remainsunchanged. If the magnitudes of v rand v⊥change in the process, then the rotated orbit acquires a new size and shape. T o find the delta-v associated with a plane change, let v1be the velocity before and v2the velocity after the impulsive maneuver. Then v1=vr1ˆur+v⊥1ˆu⊥1 v2=vr2ˆur+v⊥2ˆu⊥2 where ˆuris the radial unit vector directed along the line of intersection of the two orbital planes. ˆurdoes not change during the maneuver. As we know, the transverse B δ⊥12 PA P'A' 21 FB C Dv (a) (b)vr v Figure 6.22 (a) Two non-coplanar orbits about F. (b) A view down the line of intersection of the two orbital planes. 6.9 Plane change maneuvers 291 unit vector ˆu⊥is perpendicular to ˆurand lies in the orbital plane. Therefore it rotates through the dihedral angle δfrom its initial orientation ˆu⊥1to its final orientation ˆu⊥2. The change /Delta1vin the velocity vector is /Delta1v=v2−v1=(vr2−vr1)ˆur+v⊥2ˆu⊥2−v⊥1ˆu⊥1 /Delta1vis found by taking the dot product of /Delta1vwith itself, /Delta1v2=/Delta1v·/Delta1v =/bracketleftbig (vr2−vr1)ˆur+v⊥2ˆu⊥2−v⊥1ˆu⊥1/bracketrightbig ·/bracketleftbig (vr2−vr1)ˆur+v⊥2ˆu⊥2−v⊥1ˆu⊥1/bracketrightbig Carrying out the dot products while noting that ˆur·ˆur=ˆu⊥1·ˆu⊥1=ˆu⊥2·ˆu⊥2=1 andˆur·ˆu⊥1=ˆur·ˆu⊥2=0, yields /Delta1v2=(vr2−vr1)2+v2 ⊥1+v2 ⊥2−2v⊥1v⊥2(ˆu⊥1·ˆu⊥2) Butˆu⊥1·ˆu⊥2=cosδ, so that we finally obtain a general formula for /Delta1vwith plane change, /Delta1v=/radicalBig (vr2−vr1)2+v2 ⊥1+v2 ⊥2−2v⊥1v⊥2cosδ (6.19) From the definition of the flight path angle (cf. Figure 2.11), vr1=v1sinγ1v⊥1=v1cosγ1 vr2=v2sinγ2v⊥2=v2cosγ2 Substituting these relations into Equation 6.19, expanding and collecting terms, and using the trig identities sin2γ1+cos2γ1=sin2γ2+cos2γ2=1 cos(γ2−γ1)=cosγ2cosγ1+sinγ2sinγ1 leads to another version of the same equation, /Delta1v=/radicalBig v2 1+v2 2−2v1v2[cos/Delta1γ−cosγ2cosγ1(1−cosδ)] (6.20) where /Delta1γ=γ2−γ1. If there is no plane change ( δ=0), then cos δ=1 and Equation 6.20 reduces to /Delta1v=/radicalBig v2 1+v2 2−2v1v2cos/Delta1γ which is the cosine law we have been using to compute /Delta1vin coplanar maneuvers. Therefore, Equation 6.19 contains Equation 6.9 as a special case. To ke e p /Delta1vat a minimum, it is clear from Equation 6.19 that the radial velocity should remain unchanged during a plane change maneuver. For the same reason, it isapparent that the maneuver should occur where v ⊥is smallest, which is at apoapse. Figure 6.23 illustrates a plane change maneuver at apoapse. In this case vr1=vr2=0, so that v⊥1=v1andv⊥2=v2, thereby reducing Equation 6.19 to /Delta1v=/radicalBig v2 1+v2 2−2v1v2cosδPlane change at apoapse (or periapse) .(6.21) 292 Chapter 6 Orbital maneuvers v1v2∆v 12 F Apoapsis Figure 6.23 Plane change at apoapse. (a)υ1 υ1∆υυ2 (b)υ1υ2 2υ1 sin2 (c)υ2 υ1 υ22υ2 sin2 δ δ δδδ Figure 6.24 Plane changes at apoapse or periapse. (a) Speed change accompanied by plane change. (b) Plane change followed by speed change. (c) Speed change followed by plane change. Equation 6.21 is for a speed change acco mpanied by a plane change, as illustrated in Figure 6.24(a). Using the trig identity cosδ=1−2 sin2δ 2 we can rewrite Equation 6.21 as follows for a plane change together with a speed change at apoapse or periapse, /Delta1vI=/radicalbigg (v2−v1)2+4v1v2sin2δ 2(6.22) If there is no change in the speed, so that v2=v1, Equation 6.22 yields /Delta1vδ=2vsinδ 2(6.23) The subscript δreminds us that this is the delta-v for a pure rotation of the velocity vector through the angle δ. Another plane-change strategy, illustrated in Figure 6.24(b), is to rotate the velocity vector and then change its magnitude. In that case, the delta-v is /Delta1vII=2v1sinδ 2+|v2−v1| 6.9 Plane change maneuvers 293 20 40 60 8050100150 δ, degrees∆υ (% of υ) Figure 6.25 /Delta1vrequired to rotate the velocity vector through an angle δ. Y et another possibility is to change the speed first, and then rotate the velocity vector (Figure 6.24(c)). Then /Delta1vIII=|v2−v1|+2v 2sinδ 2 It is easy to show that /Delta1vII=/radicalBigg /Delta1v2 1+4v1|v2−v1|sinδ 2/parenleftbigg 1−sinδ 2/parenrightbigg >/Delta1 v I /Delta1vIII=/radicalBigg /Delta1v2 1+4v2|v2−v1|sinδ 2/parenleftbigg 1−sinδ 2/parenrightbigg >/Delta1 v I It follows that plane change accompanied by speed change is the most efficient of the above three maneuvers. Equation 6.23, the delta-v formula for pure rotation of the velocity vector, is plot- ted in Figure 6.25, which shows why significant plane changes are so costly in termsof propellant expenditure. For example, a plane change of just 24 ◦requires a delta- v equal to that needed for an escape trajectory (41.4 percent). A 60◦plane change requires a delta-v equal to the speed of the spacecraft itself, which in earth orbit oper-ations is about 7.5 km/s. This would require placing in orbit, a spacecraft about thesize of that which launched the satellite into orbit. Of course, this launch-vehicle sizedsatellite would itself have to be launched atop a vehicle of monstrous proportions.The space shuttle is capable of a plane change in orbit of only about 3 ◦, a maneuver which would exhaust its entire fuel capacity. Orbit plane adjustments are thereforemade during the powered ascent phase when the energy is available to do so. For some missions, however, plane changes must occur in orbit. A common example is the maneuvering of GEO satellites into position. These must orbit theearth in the equatorial plane, but it is impossible to throw a satellite directly into anequatorial orbit from a launch site which is not on the equator. That is not difficultto understand when we realize that the plane of the orbit must contain the center ofthe earth (the focus) as well as the point at which the satellite is inserted into orbit, asillustrated in Figure 6.26. So if the insertion point is anywhere but on the equator, the 294 Chapter 6 Orbital maneuvers N SFEquator28.6°N EastiN SFEquator28.6°N East (a) (b)Insertion Insertion 90° Orbital plane Figure 6.26 Two views of the orbit of a satellite launched directly east at 28.6◦north latitude. (a) Edge-on view of the orbital plane. (b) View towards insertion point meridian. N SFEquator28.6°N EastAN SFEquator28.6˚N EastA (a) (b)Insertion Insertion i i Figure 6.27 (a) Northeasterly launch (0◦<A<270) from a latitude of 28.6◦N. (b) Southeasterly launch (90◦<A<270). plane of the orbit will be tilted away from the earth’s axis. As we know from Chapter 4, the angle between the equatorial plane and the plane of the orbiting satellite is calledthe inclination i. Launching a satellite due east takes full advantage of the earth’s rotational velocity, which is about 0.5 km/s at the equator and diminishes towards the poles. Figure 6.26shows a spacecraft launched due east into low earth orbit at a latitude φof 28.6 ◦ north, which is the latitude of Kennedy Space Flight Center (KSC). As can be seen from the figure, the inclination of the orbit will be 28.6◦. One-fourth of the way around the earth the satellite will cross the equator. Halfway around the earth itreaches its southernmost latitude, φ=28.6 ◦south. It then heads north, crossing over the equator at the three-quarters point and returning after one complete revolutiontoφ=28.6 ◦north. Launch azimuth Ais the flight direction at insertion, measured clockwise from north on the local meridian. Thus A=90◦is due east. If the launch direction is not directly eastward, then the orbit will have an inclination greater than the launch latitude, as illustrated in Figure 6.27 for φ=28.6◦N. Northeasterly (0 <A<90◦)o r southeasterly (90◦<A<180◦) launches take only partial advantage of the earth’s 6.9 Plane change maneuvers 295 90 180 2703090120 36060150180i, degrees A, degrees φ /H11005 0°φ /H11005 60°φ /H11005 50° φ /H11005 40° φ /H11005 30° φ /H11005 20° φ /H11005 10° Figure 6.28 Orbit inclination iversus launch azimuth Afor several latitudes φ. 90˚ A /H11005 0˚64˚ A /H11005 30˚40˚ A /H11005 60˚28˚ A /H11005 90˚ 40˚ A /H11005 120˚64˚ A /H11005 150˚90˚ A /H11005 180˚116˚ A /H11005 210˚ 140˚ A /H11005 240˚152˚ A /H11005 270˚140˚ A /H11005 300˚116˚ A /H11005 330˚ Figure 6.29 Variation of orbit inclinations with launch azimuth at φ=28◦. Note the retrograde orbits for A>180◦. 296 Chapter 6 Orbital maneuvers rotational speed and both produce an inclination igreater than the launch latitude but less than 90◦. Since these orbits have an eastward velocity component, they are called prograde orbits. Launches to the west produce retrograde orbits with inclinationsbetween 90 ◦and 180◦. Launches directly north or directly south result in polar orbits. Spherical trigonometry is required to obtain the relationship between orbital inclination i, launch platform latitude φ, and launch azimuth A. It turns out that cosi=cosφsinA (6.24) From this we verify, for example, that i=φwhen A=90◦, as pointed out above. A plot of this relation is presented in Figure 6.28, while Figure 6.29 illustrates the orientation of orbits for a range of launch azimuths at φ=28◦. Example 6.10Determine the required launch azimuth for the sun-synchronous satellite of Example 4.7 if it is launched from Vandenburgh AFB on the California coast(latitude =34.5 ◦N). In Example 4.7 the inclination of the sun-synchronous orbit was determined to be 98.43◦. Equation 6.24 is used to calculate the launch azimuth, sinA=cosi cosl=cos 98.43◦ cos 34.5◦=− 0.1779 From this, A=190.2◦, a launch to the south or A=349.8◦, a launch to the north. Figure 6.30 shows the effect that the choice of launch azimuth has on the orbit. It does not change the fact that the orbit is retrograde; it simply determines whether theascending node will be in the same hemisphere as the launch site or on the oppositeside of the earth. Actually, a launch to the north from Vandenburgh is not an optionbecause of the safety hazard to the populated land lying below the ascent trajectory.Launches to the south, over open water, are not a hazard. Working this problem forKennedy Space Center (latitude 28 .6 ◦N) yields nearly the same values of A. Since safety considerations on the Florida east coast limit launch azimuths to between 35◦ and 120◦, polar and sun-synchronous satellites cannot be launched from the US eastern test range. The projection of a satellite’s orbit onto the earth’s surface is called its ground track. Because the satellite reaches a maximum and minimum latitude (‘amplitude’)during each orbit while passing over the equator twice, on a Mercator projection theground track of a satellite in low-earth orbit resembles a sine curve. If the earth did notrotate, there would be just one sinusoid-like track, traced over and over again as thesatellite orbits the earth. However, the earth rotates eastward beneath the satellite orbitat 15.04 ◦per hour, so the ground track advances westward at that rate. Figure 6.31 shows about two and a half orbits of a satellite, with the beginning and end of thisportion of the ground track labeled. The distance between two successive crossings ofthe equator is measured to be 23.2 ◦, which is the amount of earth rotation in one orbit of the spacecraft. Therefore, the ground track reveals that the period of the satellite is T=23.2◦ 15.04◦/hr=1.54 hr=92.6 min This is a typical low earth orbital period. 6.9 Plane change maneuvers 297 98.43° 98.43°34.5°N 34.5°N SNN SLaunch site Launch site Launch azimuth A /H11005 349.8° Launch azimuth A /H11005 190.2°Ascending nodeDescending nodeEquator Equator Figure 6.30 Effect of launch azimuth on the position of the orbit. StartFinish51.6N 51.6S 180W 150W 120W 90W 60W 30W 0 60E 90E 120E 150E 180E 30E180W 150W 120W 90W 60W 30W 0 60E 90E 120E 150E 180E 30E 90N 60N 30N 0 30S 60S 90S90N 60N 30N 0 30S 60S 90S 23.2° Figure 6.31 Ground track of a satellite. Example 6.11Find the delta-v required to transfer a satellite from a circular, 300 km altitude low- earth orbit of 28◦inclination to a geostationary equatorial orbit. Circularize and change the inclination at altitude. Compare that delta-v requirement with the one inwhich the plane change is done in the low-earth orbit. Figure 6.32 shows the 28◦inclined low-earth parking orbit (1), the coplanar Hohmann transfer ellipse (2), and the coplanar GEO orbit (3). From the figure we see that rB=6678 km rc=42 164 km 298 Chapter 6 Orbital maneuvers (Example 6.11 continued)Orbit 1: For this circular orbit the speed at Bis vB1=/radicalbiggµ rB=/radicalbigg 398 600 6678=7.7258 km /s 6678 km 42 164 kmEarth1 C BLEO 2Orbits 1, 2 and 3 all have 28° inclination 3 Figure 6.32 Transfer from LEO to GEO in an orbit of 28◦inclination. Orbit 2: The eccentricity of the transfer orbit is e2=rC−rB rC+rB=0.72655 Let us evaluate the orbit equation at Bto find the angular momentum of the Hohmann transfer orbit 2, rB=h2 2 µ1 1+e2cos(0)⇒ 6678=h2 2 398 6001 1+0.72655⇒ h2=67 792 km /s 6.9 Plane change maneuvers 299 The velocities at perigee and apogee of orbit 2 are, from the angular momentum formula, vB2=h2 rB=10.152 km /s vC2=h2 rC=1.6078 km /s At this point we can calculate /Delta1vB, /Delta1vB=vB2−vB1=10.152 −7.7258 =2.4258 km /s( a ) Orbit 3: For this orbit, which is circular, the speed at Cis vC3=/radicalbiggµ rC=3.0747 km /s so that /Delta1vC=vC3−vC2=3.0747 −1.6078 =1.4668 km /s (b) We can now calculate the total delta-v for the Hohmann transfer: /Delta1vHohmann =/Delta1vB+/Delta1vC=2.4258 +1.4668 =3.8926 km /s This places the satellite in a circular orbit of the correct radius, but the wrong incli- nation. The velocity vector at Cmust be rotated into the plane of the equator, as illustrated in Figure 6.33. According to Equation 6.30, the delta-v required to rotate that velocity through the change in inclination of 28◦is /Delta1viC=2vC3sin/Delta1i 2=2·3.0747 ·sin28◦ 2=1.4877 km /s Therefore, the total delta-v requirement is /Delta1vtotal=/Delta1vHohmann +/Delta1viC=5.3803 km /s 28° 4 GEO3.0747 km/s CN SEarthEquatorial plane 3 2Plane ofand∆υ Figure 6.33 Plane change maneuver required after the Hohmann transfer. 300 Chapter 6 Orbital maneuvers (Example 6.11 continued)Suppose we make the plane change at LEO instead of at GEO. T o rotate the velocity vector vB1through 28◦requires /Delta1vBi=2vB1sin/Delta1i 2=2·7.7258·sin28◦ 2=3.7381 km /s This, together with (a) and (b), yields the total delta-v schedule for insertion into GEO: /Delta1vtotal=/Delta1vBi+/Delta1vB+/Delta1vC=3.7381+2.4258+1.4668=7.6307 km /s This is a 42 percent increase over the total delta-v with plane change at GEO. Clearly, it is best to do plane change maneuvers at the largest possible distance (apoapse) from the primary attractor, where the velocities are smallest. Example 6.12Suppose in the previous example that part of the plane change, /Delta1i, takes place at B, the perigee of the Hohmann transfer ellipse, and the remainder, 28◦−/Delta1i, occurs at the apogee C. What is the value of /Delta1iwhich results in the minimum /Delta1vtotal? We found in Example 6.11 that if /Delta1i=0, then /Delta1vtotal=5.3803 km /s, whereas /Delta1i=28◦made /Delta1vtotal=7.6307 km /s. Here we are to determine if there is a value of/Delta1ibetween 0◦and 28◦that yields a /Delta1vtotal which is smaller than either of those two. In this case a plane change occurs at both Band C. Recall that the most efficient strategy is to combine the plane change with the speed change, so that the delta-vs at those points are (Equation 6.21) /Delta1vB=/radicalBig v2 B1+v2 B2−2vB1vB2cos/Delta1i =/radicalbig 7.72582+10.1522−2·7.7258·10.152·cos/Delta1i =√ 162.74−156.86 cos /Delta1i and /Delta1vC=/radicalBig v2 C2+v2 C3−2vC2vC3cos(28◦−/Delta1i) =/radicalbig 1.60782+3.07472−2·1.6078·3.0747·cos(28◦−/Delta1i) =/radicalbig 12.039−9.8871 cos(28◦−/Delta1i) Thus, /Delta1vtotal=/Delta1vB+/Delta1vC =√ 162.74−156.86 cos /Delta1i+/radicalbig 12.039−9.8871 cos(28◦−/Delta1i)( a ) 6.9 Plane change maneuvers 301 T o determine if there is a /Delta1iwhich minimizes /Delta1vtotal, we take its derivative with respect to /Delta1iand set it equal to zero: d/Delta1vtotal d/Delta1i=78.43 sin /Delta1i√ 162.74 −156.86 cos /Delta1i−4.9435 sin(28◦−/Delta1i)√12.039 −9.8871 cos(28◦−/Delta1i)=0 This is a transcendental equation which must be solved iteratively. The solution, as the reader may verify, is /Delta1i=2.1751◦(b) That is, an inclination change of 2.1751◦should occur in low-earth orbit, while the rest of the plane change, 25.825◦, is done at GEO. Substituting (b) into (a) yields /Delta1vtotal=4.2207 km /s This is 21 percent less than the smallest /Delta1vtotalcomputed in Example 6.11. Example 6.13A spacecraft is in a 500 km by 10 000 km altitude geocentric orbit which intersects the equatorial plane at a true anomaly of 120◦(see Figure 6.34). If the inclination to the equatorial plane is 15◦, what is the minimum velocity increment required to make this an equatorial orbit? The orbital parameters are e=rA−rP rA+rP=(6378 +10 000) −(6378 +500) (6378 +10 000) +(6378 +500)=0.4085 F120° 16 378 km 6878 kmB CA P2.2692 km/s 2.2692 km/sAscending node12 174 km 5.1043 km/s 7.7246 km/s8044.6 km Figure 6.34 An orbit which intersects the equatorial plane along line BC. The equatorial plane makes an angle of 15◦with the plane of the page. 302 Chapter 6 Orbital maneuvers (Example 6.13 continued)rP=h2 µ1 1+ecos(0)⇒ 6878=h2 398 6001 1+0.4085⇒ h=62 141 km /s The radial coordinate and velocity components at points Band C, on the line of intersection with the equatorial plane, are rB=h2 µ1 1+ecosθB=62 1412 398 6001 1+0.4085·cos 120◦=12 174 km v⊥B=h rB=62 141 12 174=5.1043 km /s vrB=µ hesinθB=398 600 62 141·0.4085·sin 120◦=2.2692 km /s and rC=h2 µ1 1+ecosθC=62 1412 398 6001 1+0.4085·cos 300◦=8044.6k m v⊥C=h rC=62 141 8044.6=7.7246 km /s vrC=µ hesinθC=398 600 62 141·0.4085·sin 300◦=− 2.2692 km /s All we wish to do here is rotate the plane of the orbit rigidly around the node line BC. The impulsive maneuver must occur at either BorC. Equation 6.19 applies, and since the radial and perpendicular velocity components remain fixed, it reduces to /Delta1v=v⊥/radicalbig 2(1−cosδ) where δ=15◦. For the minimum /Delta1v, the maneuver must be done where v⊥is smallest, which is at B, the point farthest from the center of attraction F. Thus, /Delta1v=5.1043/radicalbig 2(1−cos 15◦)=1.3325 km /s Example 6.14Orbit 1 has angular momentum hand eccentricity e. The direction of motion is shown. Calculate the /Delta1vrequired to rotate the orbit 90◦about its latus rectum BC without changing hand e. The required direction of motion in orbit 2 is shown in Figure 6.35. By symmetry, the required maneuver may occur at either BorC, and it involves a rigid body rotation of the ellipse, so that vrandv⊥remain unaltered. Because of the directions of motion shown, the true anomalies of Bon the two orbits are θB1=− 90◦θB2=+ 90◦ The radial coordinate of Bis rB=h2 µ1 1+ecos(±90)=h2 µ 6.9 Plane change maneuvers 303 BC1 2F P1P2 Figure 6.35 Identical ellipses intersecting at 90◦along their common latus rectum, BC. For the velocity components at B,w eh a v e v⊥B)1=v⊥B)2=h rB=µ h vrB)1=µ hesin(θ B1)=−µe hvrB)2=µ hesin(θ B2)=µe h Substituting these into Equation 6.19, yields /Delta1vB=/radicalBig/bracketleftbig vrB)2−vrB)1/bracketrightbig2+v⊥B)2 1+v⊥B)2 2−2v⊥B)1v⊥B)2cos 90◦ =/radicalbigg/bracketleftBigµe h−/parenleftBig −µe h/parenrightBig/bracketrightBig2 +/parenleftBigµ h/parenrightBig2 +/parenleftBigµ h/parenrightBig2 −2/parenleftBigµ h/parenrightBig/parenleftBigµ h/parenrightBig ·0 =/radicalbigg 4µ2 h2e2+2µ2 h2 so that /Delta1vB=√ 2µ h/radicalbig 1+2e2 (a) If the motion on ellipse 2 were opposite to that shown in Figure 6.35, then the radial velocity components at B(and C) would be in the same rather than in the opposite direction on both ellipses, so that instead of (a) we would find a smaller velocity increment, /Delta1vB=√ 2µ h 304 Chapter 6 Orbital maneuvers Problems 6.1 The shuttle orbiter has a mass of 125 000 kg. The two orbital maneuvering engines produce a combined (non-throttleable) thrust of 53.4 kN. The orbiter is in a 300 kmcircular orbit. A delta-v maneuver transfers the spacecraft to a coplanar 250 km by300 km elliptical orbit. Neglecting pr opellant loss and using elementary physics (linear impulse equals change in linear momentum, distance equals speed times time), estimate (a) the time required for the /Delta1vburn, and (b) the distance traveled by the orbiter during the burn. (c) Calculate the ratio of your answer for (b) to the circumference of the initial circular orbit. {Ans.: (a) /Delta1t=34 s; (b) 263 km; (c) 0.0063} 6.2 A satellite traveling at 8.2 km/s at perigee fires a retrorocket at perigee altitude of 480 km. What delta-v is necessary to reach a minimum altitude of 100 miles during the next orbit? {Ans.: −66.8 m/s} 6.3 A spacecraft is in a 300 km circular earth orbit. Calculate (a) the total delta-v required for a Hohmann transfer to a 3000 km coplanar circular earth orbit, and (b) the transfer orbit time. {Ans.: (a) 1.198 km/s; (b) 59 min 39 s} 312 ∆υ2 ∆υ1 300 km3000 km Figure P .6.3 6.4 A spacecraft Sis in a geocentric hyperbolic trajectory with a perigee radius of 7000 km and a perigee speed of 1 .3vesc. At perigee, the spacecraft releases a projectile Bwith a speed of 7.1 km/s parallel to the spacecraft’s velocity. How far dfrom the earth’s surface isSat the instant Bimpacts the earth? Neglect the atmosphere. {Ans.: d=8978 km} 6.5 Assuming the orbits of earth and Mars are circular and coplanar, calculate (a) the time required for a Hohmann transfer from earth to Mars, and (b) the initial position of Mars ( α) in its orbit relative to earth for interception to occur. Radius of earth orbit =1.496×108km. Radius of Mars orbit =2.279×108km. µsun=1.327×1011km3/s2. {Ans.: (a) 259 days; (b) α=44·3◦} Problems 305 S B Earth 7000 kmPerigee of impact ellipseSeparationd Impact Figure P .6.4 Mars at launch Mars at encounterSun Earth at launchαHohmann transfer orbit Figure P .6.5 6.6 Two geocentric elliptical orbits have common apse lines and their perigees are on the same side of the earth. The first orbit has a perigee radius of rp=7000 km and e=0.3, whereas for the second orbit rp=32 000 km and e=0.5 (a) Find the minimum total delta-v and the time of flight for a transfer from the perigee of the inner orbit to the apogee of the outer orbit. (b) Do part (a) for a transfer from the apogee of the inner orbit to the perigee of the outer orbit. {Ans.: (a) /Delta1vtotal=2.388 km /s, TOF =16.2 hr; (b) /Delta1vtotal=3.611km /s, TOF =4.66 hr} 6.7 A spacecraft is in a 500 km altitude circular earth orbit. Neglecting the atmosphere, find the delta-v required at Ain order to impact the earth at (a) point B (b) point C. {Ans.: (a) 192 m/s; (b) 7.61 km/s} 306 Chapter 6 Orbital maneuvers 7000 km 32 000 km12 e /H11005 0.3e /H11005 0.5 Earth A BC D3 4 Figure P .6.6 60˚B CAEarth 500 km Figure P .6.7 6.8 A spacecraft is in a 200 km circular earth orbit. At t=0, it fires a projectile in the direction opposite to the spacecraft’s motion. Thirty minutes after leaving the spacecraft,the projectile impacts the earth. What delta-v was imparted to the projectile? Neglect the atmosphere.{Ans.: /Delta1v=77.2m/s} 6.9 The space shuttle was launched on a 15-day mission. There were four orbits after injection, all of them at 39 ◦inclination. Orbit 1: 302 by 296 kmOrbit 2 (day 11): 291 by 259 kmOrbit 3 (day 12): 259 km circularOrbit 4 (day 13): 255 by 194 km Problems 307 Calculate the total delta-v, which should be as small as possible, assuming Hohmann transfers.{Ans.: /Delta1v total=43.5m /s} 6.10 A space vehicle in a circular orbit at an altitude of 500 km above the earth executes a Hohmann transfer to a 1000 km circular orbit. Calculate the total delta-v requirement.{Ans.: 0.2624 km/s} 500 km 1000 kmA B3 12 Figure P .6.10 6.11 Calculate the total delta-v required for a Hohmann transfer from a circular orbit of radius rto a circular orbit of radius 12 r. {Ans.: 0.5342√µ/r} A B3 12 12rr Figure P .6.11 308 Chapter 6 Orbital maneuvers 6.12 A spacecraft in circular orbit 1 of radius rleaves for infinity on parabolic trajectory 2 and returns from infinity on a parabolic trajectory 3 to a circular orbit 4 of radius 12 r. Find the total delta-v required for this non-Hohmann orbit change maneuver.{Ans.: 0 .5338√ µ/r} B A1 12rr 342 Figure P .6.12 6.13 Calculate the total delta-v required for a Hohmann transfer from the smaller circular orbit to the larger one.{Ans.: 0 .394v 1,w h e r e v1is the speed in orbit 1} A3r r123 B Figure P .6.13 Problems 309 6.14 A spacecraft is in a 300 km circular earth orbit. Calculate (a) the total delta-v required for the bi-elliptical transfer to a 3000 km altitude coplanar circular orbit shown, and (b) the total transfer time. {Ans.: (a) 2.039 km/s; (b) 2.86 hr} 300 kme /H11005 0.3 14 3000 km 2 3A BC ∆υB ∆υC∆υA Figure P .6.14 6.15 (a) With a single delta-v maneuver, the earth orbit of a satellite is to be changed from a circle of radius 15 000 km to a coplanar ellipse with perigee altitude of 500 kmand apogee radius of 22 000 km. Calculate the magnitude of the required delta-vand the change in the flight path angle /Delta1γ. (b) What is the minimum total delta-v if the orbit change is accomplished instead by a Hohmann transfer? {Ans.: (a) ||/Delta1v|| =2.77 km /s,/Delta1γ=31.51 ◦; (b)/Delta1vHohmann =1.362 km /s} 6.16 An earth satellite has a perigee altitude of 1270 km and a perigee speed of 9 km/s. It is required to change its orbital eccentricity to 0.4, without rotating the apse line, bya delta-v maneuver at θ=100 ◦. Calculate the magnitude of the required /Delta1vand the change in flight path angle /Delta1γ. {Ans.: ||/Delta1v|| =0.915 km /s;/Delta1γ=− 8.18◦} 6.17 At point Aon its earth orbit, the radius, speed and flight path angle of a satellite are rA=12 756 km, vA=6.5992 km /s and γA=20◦. At point B, at which the true anomaly is 150◦, an impulsive maneuver causes /Delta1v⊥=+ 0.75820 km /s and /Delta1vr=0. (a) What is the time of flight from AtoB? (b) What is the rotation of the apse line as a result of this maneuver? {Ans.: (a) 2.045 hr; (b) 43.39◦counterclockwise} 6.18 A satellite is in elliptical orbit 1. Calculate the true anomaly θ(relative to the apse line of orbit 1) of an impulsive maneuver which rotates the apse line at an angle η counterclockwise but leaves the eccentricity and the angular momentum unchanged.{Ans.: θ=η/2} 310 Chapter 6 Orbital maneuvers 15 000 km 22 000 km 6878 km2A BC D Eγ2∆v Common apse lineEarth3 1 4vA1vA2 Figure P .6.15 Original apse line 21η θ Figure P .6.18 6.19 A satellite in orbit 1 undergoes a delta-v maneuver at perigee P1such that the new orbit 2 has the same eccentricity e, but its apse line is rotated 90◦clockwise from the original one. Calculate the specific angular momentum of orbit 2 in terms of that of orbit 1 andthe eccentricity e. {Ans.: h 2=h1/√1+e} Problems 311 12 P1 P2F Figure P .6.19 1 2 FA Figure P .6.20 6.20 Calculate the delta-v required at Ain orbit 1 for a single impulsive maneuver to rotate the apse line 180◦counterclockwise (to become orbit 2), but keep the eccentricity eand the angular momentum hthe same. {Ans.: /Delta1v=2µe/h} 6.21 The space station and spacecraft Aand Bare all in the same circular earth orbit of 350 km altitude. Spacecraft Ais 600 km behind the space station and spacecraft Bis 600 km ahead of the space station. At the same instant, both spacecraft apply a /Delta1v⊥so as to arrive at the space station in one revolution of their phasing orbits. (a) Calculate the times required for each spacecraft to reach the space station. (b) Calculate the total delta-v requirement for each spacecraft. {Ans.: (a) spacecraft A: 90.2 min; spacecraft B: 92.8 min; (b) /Delta1vA=73.9m /s; /Delta1vB=71.5m /s} 6.22 Satellites AandBare in the same circular orbit of radius r.Bis 180◦ahead of A. Calculate the semimajor axis of a phasing orbit in which Awill rendezvous with Bafter just one revolution in the phasing orbit.{Ans.: a=0.63r } 6.23 Two spacecraft are in the same elliptical earth orbit with perigee radius 8000 km and apogee radius 13 000 km. Spacecraft 1 is at perigee and spacecraft 2 is 30 ◦ahead. Calcu- late the total delta-v required for spacecraft 1 to intercept and rendezvous with spacecraft2 when spacecraft 2 has traveled 60 ◦. {Ans.: /Delta1vtotal=6.24 km /s} 312 Chapter 6 Orbital maneuvers 350 km600 km 600 km EarthSpace stationSpacecraft A Spacecraft B Circular orbit Figure P .6.21 A Br F Figure P .6.22 6.24 An earth satellite has the following orbital elements: a=15 000 km, e=0.5,W=45◦, w=30◦,i=10◦. What minimum delta-v is required to reduce the inclination to zero? {Ans.: 0.588 km/s} 6.25 With a single impulsive maneuver, an earth satellite changes from a 400 km circular orbit inclined at 60◦to an elliptical orbit of eccentricity e=0.5 with an inclination of 40◦. Calculate the minimum required delta-v.{Ans.: 3.41 km/s} 6.26 An earth satellite is in an elliptical orbit of eccentricity 0.3 and angular momentum 60 000 km 2/s. Find the delta-v required for a 90◦change in inclination at apogee (no change in speed).{Ans.: 6.58 km/s} 6.27 A spacecraft is in a circular, equatorial orbit (1) of radius r oabout a planet. At point B it impulsively transfers to polar orbit (2), whose eccentricity is 0.25 and whose perigeeis directly over the North Pole. Calculate the minimum delta-v required at Bfor this maneuver.{Ans.: 1 .436√ µ/ro} Problems 313 PCD 30˚60˚ Spacecraft 1Spacecraft 2intercept 8000 km 13 000 kmA 12 Figure P .6.23 ro BN S1 2Orbit 1 shown edge-on Figure P .6.27 6.28 A spacecraft is in a 300 km circular parking orbit. It is desired to increase the altitude to 600 km and change the inclination by 20◦. Find the total delta-v required if (a) the plane change is made after insertion into the 600 km orbit (so that there are a total of three delta-v burns); (b) the plane change and insertion into the 600 km orbit are accomplished simultane- ously (so that the total number of delta-v burns is two); (c) the plane change is made upon departing the lower orbit (so that the total number of delta-v burns is two). {Ans.: (a) 2.793 km/s; (b) 2.696 km/s; (c) 2.783 km/s} 314 Chapter 6 Orbital maneuvers 6.29 At time t=0, manned spacecraft aand unmanned spacecraft bare at the positions shown in circular earth orbits 1 and 2, respectively. For assigned values of θ(a) 0andθ(b) 0, design a series of impulsive maneuvers by means of which spacecraft atransfers from orbit 1 to orbit 2 so as to rendezvous with spacecraft b(i.e., occupy the same position in space). The total time and total delta-v required for the transfer should be as small aspossible. Consider earth’s gravity only. 20 000 km 210 000 kmab 2 1(b)θ0 (a)θ0 Figure P .6.29 6.30 What must the launch azimuth be if the satellite in Example 4.8 is launched from (a) Kennedy Space Center (latitude =28.5◦N); (b) Vandenburgh AFB (latitude =34.5◦N); (c) Kourou, French Guiana (latitude 5.5◦N). {Ans.: (a) 329.4◦; (b) 327.1◦; (c) 333.3◦} 7Chapter Relative motion and rendezvous Chapter outline 7.1 Introduction 315 7.2 Relative motion in orbit 316 7.3 Linearization of the equations of relative motion in orbit 322 7.4 Clohessy–Wiltshire equations 324 7.5 Two-impulse rendezvous maneuvers 330 7.6 Relative motion in close-proximity circular orbits 338Problems 340 7.1 Introduction Up to now we have mostly referenced the motion of orbiting objects to a non- rotating coordinate system fixed to the center of attraction (e.g., the center of the earth). This platform served as an inertial frame of reference, in which Newton’ssecond law can be written F net=maabsolute An exception to this rule was the discussion of the restricted three-body problem at the end of Chapter 2, in which we made use of the relative motion equations developedin Chapter 1. In a rendezvous maneuver, two orbiting vehicles observe one anotherfrom each of their own free-falling, rotating, clearly non-inertial frames of reference.T o base impulsive maneuvers on observations made from a moving platform requirestransforming relative velocity and acceleration measurements into an inertial frame. 315 316 Chapter 7 Relative motion and rendezvous Otherwise, the true thrusting forces cannot be sorted out from the fictitious ‘inertial forces’ that appear in Newton’s law when it is written incorrectly as Fnet=marel The purpose of this chapter is to use relative motion analysis to gain some familiarity with the problem of maneuvering one spacecraft relative to another, especially whenthey are in close proximity. 7.2 Relative motion in orbit A rendezvous maneuver usually involves a target vehicle, which is passive and non- maneuvering, and a chase vehicle which is act ive and performs the maneuvers required to bring itself alongside the target. An obvious example is the space shuttle, the chaser,rendezvousing with the international space station, the target. The position vector ofthe target in the geocentric equatorial frame is r 0. This outward radial is sometimes called ‘ r-bar’ . The moving frame of reference has its origin at the target, as illustrated in Figure 7.1. The xaxis is directed along r0, the outward radial to the target. The yaxis is perpendicular to r0and points in the direction of the target satellite’s local horizon. The xand yaxes therefore lie in the target’s orbital plane, and the zaxis is normal to that plane. The angular velocity of the xyzaxes attached to the target is just the angular velocity of the position vector r0, and it is obtained from the fact that h=r0×v0=(r0v0⊥)ˆk=/parenleftbig r2 0/Omega1/parenrightbigˆk=r2 0/Omega1 XYZ /H9253xy zˆiˆj ˆkB rrel r0A Target orbit Figure 7.1 Co-moving reference frame attached to A, from which body Bis observed. 7.2 Relative motion in orbit 317 which means that /Omega1=r0×v0 r2 0(7.1) T o find the angular acceleration ˙/Omega1, we take the derivative of /Omega1in Equation 7.1 to obtain ˙/Omega1=1 r2 0(˙r0×v0+r0×˙v0)−2 r3 0˙r0(r0×v0) (7.2) But ˙r0×v0=v0×v0=0 (7.3) According to Equation 2.15, the acceleration ˙v0of the target satellite is given by ˙v0=−µ r3 0r0 Hence, r0×˙v0=r0×/parenleftbigg −µ r3 0r0/parenrightbigg =−µ r3 0(r0×r0)=0 (7.4) Substituting Equations 7.1, 7.3 and 7.4 into Equation 7.2 yields ˙/Omega1=−2 r0˙r0/Omega1 Finally, recalling from Equation 2.25a that ˙r0=v0·r0/r0, we obtain ˙/Omega1=−2(r0·v0) r2 0/Omega1 (7.5) Equations 7.1 and 7.5 are the means of determining the angular velocity and accelera- tion of the co-moving frame for use in the relative velocity and acceleration formulas,Equations 1.38 and 1.42. Example 7.1Spacecraft Ais in an elliptical earth orbit having the following parameters: h=52 059 km2/s,e=0.025724, i=60◦,/Omega1=40◦,ω=30◦,θ=40◦(a) Spacecraft Bis likewise in an orbit with these parameters: h=52 362km2/s,e=0.0072696, i=50◦,/Omega1=40◦,ω=120◦,θ=40◦(b) Calculate the velocity vrel)xyzand acceleration arel)xyzof spacecraft Brelative to spacecraft A, measured along the xyzaxes of the co-moving coordinate system of spacecraft A, as defined in Figure 7.1. 318 Chapter 7 Relative motion and rendezvous (Example 7.1 continued) XYZ rA rB vBvA /H9253rrelAB Figure 7.2 Spacecraft Aand Bin slightly different orbits. From the orbital elements in (a) and (b) we can use Algorithm 4.2 to find the position and velocity of the spacecraft relative to the geocentric equatorial reference frame. Omitting those calculations, we find, for spacecraft A, rA=− 266.74ˆI+3865.4ˆJ+5425.7ˆK(km) (a) vA=− 6.4842ˆI−3.6201ˆJ+2.4159ˆK(km/s) (b) and for spacecraft B, rB=− 5890.0ˆI−2979.4ˆJ+1792.0ˆK(km) (c) vB=0.93594ˆI−5.2409ˆJ−5.5016ˆK(km/s) (d) According to Equation 2.15, the accelerations of the two spacecraft are aA=−µrA /bardblrA/bardbl3=0.00035876 ˆI−0.0051989 ˆJ−0.0072975 ˆK(km/s2)( e ) aB=−µrB /bardblrB/bardbl3=0.0073377 ˆI+0.0037117 ˆJ−0.0022325 ˆK(km/s2)( f ) The unit vector ˆialong the xaxis of spacecraft A’s rigid, co-moving frame of reference is ˆi=rA rA=− 0.040008 ˆI+0.57977ˆJ+0.81380 ˆK (g) 7.2 Relative motion in orbit 319 Since the zaxis is in the direction of hA, and hA=rA×vA=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆI ˆJ ˆK −266.74 3865.4 5425.7−6.4842 −3.6201 2.4159/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =28 980 ˆI−34 537 ˆJ+26 030 ˆK(km/s 2) we obtain ˆk=hA hA=0.55667 ˆI−0.66341 ˆJ+0.5000 ˆK (h) Finally, ˆj=ˆk׈i, so that ˆj=− 0.82977 ˆI−0.47302 ˆJ+0.29620 ˆK (i) The angular velocity /Omega1of the xyz frame attached to spacecraft Ais given by Equation 7.1, /Omega1=28 980 ˆI−34 537 ˆJ+26 030 ˆK 6667.12 (j) =0.00065196 ˆI−0.00077698 ˆJ+0.00058559 ˆK(rad/s) We find the angular acceleration ˙/Omega1using Equation 7.5, ˙/Omega1=−2(844.41) 6667.12(0.00065196 ˆI−0.00077698 ˆJ+0.00058559 ˆK) (k) =− 2.4763(10−8)ˆI+2.9512(10−8)ˆJ−2.2242(10−8)ˆK(rad/s2) According to Equation 1.38, the relative velocity relation is vB=vA+/Omega1×rrel+vrel (l) where rreland vrelare the position and velocity of Bas measured relative to the moving xyzframe attached to A. From (a) and (b), we have rrel=rB−rA=− 5623.3 ˆI−6844.8 ˆJ−3633.7 ˆK(km) (m) Substituting this, together with (b), (d) and (j) into (l), we get 0.93594 ˆI−5.2409 ˆJ−5.5016K =(−6.4842 ˆI−3.6201 ˆJ+2.4159 ˆK) +/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆI ˆJ ˆK 0.00065196 −0.00077698 0.00058559 −5623.3 −6844.8 −3633.7/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle+v rel Solving for vrelyields vrel=0.58865 ˆI−0.69692 ˆJ+0.91414 ˆK(km/s) (n) The relative acceleration formula, Equation 1.42, is aB=aA+˙/Omega1×rrel+/Omega1×(/Omega1×rrel)+2/Omega1×vrel+arel (o) 320 Chapter 7 Relative motion and rendezvous (Example 7.1 continued)Substituting (e), (f), (j), (k), (m), and (n) into (o), we get 0.0073377 ˆI+0.0037117 ˆJ−0.0022325 ˆK =0.00035876 ˆI−0.0051989 ˆJ−0.0072975 K +/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆI ˆJ ˆK −2.4770(10 −8)2.9520(10−8)−2.2248(10−8) −5623.3 −6844.8 −3633.7/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle +(0.00065196 ˆI−0.00077698 ˆJ+0.00058559 ˆK) ×/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆI ˆJ ˆK 0.00065196 −0.00077698 0 .00058559 −5623.3 −6844.8 −3633.7/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle +2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆI ˆJ ˆK 0.00065196 −0.00077698 0 .00058559 0.58865 −0.69692 0 .91414/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle+a rel Carrying out the cross products, combining terms and solving for arelyields arel=0.00043984 ˆI−0.00038019 ˆJ+0.000017988 ˆK(km/s2) (p) From (g), (h), and (i), we see that the orthogonal transformation matrix [ Q]Xxfrom the inertial XYZ frame into the co-moving xyzframe is [Q]Xx= −0.040008 0 .57977 0 .81380 −0.82977 −0.47302 0 .29620 0.55667 −0.66341 0 .5000  T o get the components of rrel,vrel, and arelalong the axes of the co-moving xyzframe of spacecraft A, we multiply each of their expressions as components in the XYZ frame [(m), (n) and (p), respectively] by [ Q]Xxas follows: rrel)xyz= −0.040008 0 .57977 0 .81380 −0.82977 −0.47302 0 .29620 0.55667 −0.66341 0 .5000   −5623.3 −6844.8 −3633.7  =  −6700.5 6827.4 −406.22  (km) vrel)xyz= −0.040008 0 .57977 0 .81380 −0.82977 −0.47302 0 .29620 0.55667 −0.66341 0 .5000   0.58865 −0.69692 0.91414  =  0.31632 0.11199 1.2471  (km/s) arel)xyz= −0.040008 0 .57977 0 .81380 −0.82977 −0.47302 0 .29620 0.55667 −0.66341 0 .5000   0.00043984 −0.00038019 0.000017988   =  −0.00022338 −0.00017980 0.00050607  (km/s2) 7.2 Relative motion in orbit 321 xy IIIIIIIV V VIVIIVIIIxy 19000 km 7000 km IIIIII IV V VI VIIVIII e /H11005 0 e /H11005 0.125 As viewed in the inertial frame. As viewed from the co-moving frame in circular orbit 1.yxyPeriod of both orbits /H11005 1.97797 hr 6378 km8000 km2 ABStartxyxxyx yxyx y Figure 7.3 The spacecraft Bin elliptical orbit 2 appears to orbit the observer Ain circular orbit 1. The motion of one spacecraft relative to another in orbit may be hard to visualize at first. Figure 7.3 is offered as an assist. Orbit 1 is circular and orbit 2 is an ellipsewith eccentricity 0.125. Both orbits were chosen to have the same semimajor axislength, so they both have the same period. A co-moving frame is shown attached tothe observers Ain circular orbit 1. At epoch Ithe spacecraft Bin elliptical orbit 2 is directly below the observers. In other words, Amust draw an arrow in the negative local xdirection to determine the position vector of Bin the lower orbit. The figure shows eight different epochs ( I,II,III,...), equally spaced around the circular orbit, at which observers Aconstruct the position vector pointing from them to Bin the elliptical orbit. Of course, A’s frame is rotating, because its xaxis must always be directed away from the earth. Observers Acannot sense this rotation and record the set of observations in their (to them) fixed xycoordinate system, as shown at the bottom of the figure. Coasting at a uniform speed along his circular orbit, A sees the other vehicle orbiting them clockwise in a sort of bean-shaped path. The 322 Chapter 7 Relative motion and rendezvous distance between the two spacecraft in this case never becomes so great that the earth intervenes. IfAdeclared theirs to be an inertial frame of reference, they would be faced with the task of explaining the physical origin of the force holding Bin its bean-shaped orbit. Of course, there is no such force. The apparent path is due to the actual,combined motion of both spacecraft in their free fall towards the earth. When B is below A(having a negative xcoordinate), conservation of angular momentum demands that Bmove faster than A, thereby speeding up in A’s positive ydirection until the orbits cross ( x=0). When B’sxcoordinate becomes positive, i.e., Bis above A, the laws of momentum dictate that Bslow down, which it does, progressing in A’s negative ydirection until the next crossing of the orbits. Bthen falls below and begins to pick up speed. The process repeats over and over. From inertial space, the process is the motion of two satellites on inter secting orbits, appearing not at all like the orbiting motion seen by the moving observers A. 7.3 Linearization of the equations of relative motion in orbit Figure 7.4 shows two spacecraft in earth orbit. The inertial position vector of the target vehicle Aisr0, and that of the chase vehicle Bisr. The position vector of the chase vehicle relative to the target is δr, so that r=r0+δr (7.6) The symbol δis used to represent the fact that the relative position vector has a magnitude which is very small compared to the magnitude of r0(and r); i.e., δr r0/lessmuch1 (7.7) XYZ r0rAB Inertial framegdr Figure 7.4 Position of chaser Brelative to the target A. 7.3 Linearization of the equations of relative motion in orbit 323 where δr=/bardblδr/bardbland r0=/bardblr0/bardbl. This is true if the two vehicles are in close proximity to each other, as is the case in a rendezvous maneuver. Our purpose in this section isto seek the equations of motion of the chase vehicle relative to the target. The equation of motion of the chase vehicle Bis ¨r=−µr r3(7.8) where r=/bardblr/bardbl. Substituting Equation 7.6 into Eq uation 7.8 yields the equation of motion of the chaser relative to the target, δ¨r=− ¨ r0−µr0+δr r3(7.9) We will simplify this equation by making use of the fact that /bardblδr/bardblis very small, as expressed in Equation 7.7. First, note that r2=r·r=(r0+δr)·(r0+δr)=r0·r0+2r0·δr+δr·δr Since r0·r0=r2 0andδr·δr=δr2, we can factor out r2 0on the right to obtain r2=r2 0/bracketleftBigg 1+2r0·δr r2 0+/parenleftbiggδr r0/parenrightbigg2/bracketrightBigg By virtue of Equation 7.7, we can neglect the last term in the brackets, so that r2=r2 0/parenleftbigg 1+2r0·δr r2 0/parenrightbigg (7.10) In fact, we will neglect all powers of δr/r0greater than unity, wherever they appear. Since r−3=(r2)−3/2, it follows from Equation 7.10 that r−3=r−3 0/parenleftbigg 1+2r0·δr r2 0/parenrightbigg−3 2 (7.11) Using the binomial theorem (Equation 5.52) and neglecting terms of higher order than 1 in δr/r0, we obtain /parenleftbigg 1+2r0·δr r2 0/parenrightbigg−3 2 =1+/parenleftbigg −3 2/parenrightbigg/parenleftbigg2r0·δr r2 0/parenrightbigg Therefore, Equation 7.11 becomes r−3=r−3 0/parenleftbigg 1−3 r2 0r0·δr/parenrightbigg which can be written 1 r3=1 r3 0−3 r5 0r0·δr (7.12) 324 Chapter 7 Relative motion and rendezvous Substituting Equation 7.12 into Equatio n 7.9 (the equation of motion), we get δ¨r=− ¨ r0−µ/parenleftbigg1 r3 0−3 r5 0r0·δr/parenrightbigg (r0+δr) =− ¨ r0−µ/bracketleftbiggr0+δr r3 0−3 r5 0(r0·δr)(r0+δr)/bracketrightbigg =− ¨ r0−µ r0 r3 0+δr r3 0−3 r5 0(r0·δr)r0+neglect/bracehtipdownleft /bracehtipupright/bracehtipupleft /bracehtipdownright terms of higher order than 1 in δr  That is, δ¨r=− ¨ r0−µr0 r3 0−µ r3 0/bracketleftbigg δr−3 r2 0(r0·δr)r0/bracketrightbigg (7.13) But the equation of motion of the target vehicle is ¨r0=−µr0 r3 0 Substituting this into Equation 7.13 finally yields δ¨r=−µ r3 0/bracketleftbigg δr−3 r2 0(r0·δr)r0/bracketrightbigg (7.14) This is the linearized version of Equation 7.8, the equation which governs the motion of the chaser with respect to the target. The expression is linear because δrappears only in the numerator and only to the first power throughout. We achieved this bydropping a lot of terms that are insignificant when Equation 7.7 is valid. 7.4 Clohessy–Wiltshire equations Let us attach a moving frame of reference xyzto the target vehicle A, as shown in Figure 7.5. This is similar to Figure 7.1, the difference being that δris restricted by Equation 7.7. The origin of the moving system is at A.T h e xaxis lies along r0, so that ˆi=r0 r0(7.15) The yaxis is in the direction of the local horizon, and the zaxis is normal to the orbital plane of A, such that ˆk=ˆi׈j. The inertial angular velocity of the moving frame of reference is /Omega1, and the inertial angular acceleration is ˙/Omega1. According to the relative acceleration formula (Equation 1.42), we have ¨r=¨r0+˙/Omega1×δr+/Omega1×(/Omega1×δr)+2/Omega1×δvrel+δarel (7.16) where, in terms of their components in the moving frame, the relative position, velocity and acceleration are given by δr=δxˆi+δyˆj+δzˆk (7.17a) 7.4 Clohessy–Wiltshire equations 325 XZ r0xy z A Inertial frameCo-moving frameˆiˆ j ˆ k ˆ I ˆ J ˆ K Ydr Figure 7.5 Co-moving Clohessy–Wiltshire frame. δvrel=δ˙xˆi+δ˙yˆj+δ˙zˆk (7.17b) δarel=δ¨xˆi+δ¨yˆj+δ¨zˆk (7.17c) For simplicity, we assume at this point that the orbit of the target vehicle Ais circular. (Note that for a space station in low-earth orbit, this is a very good assumption.)Then ˙/Omega1=0. Substituting this together with Equation 7.6 into Equation 7.16 yields δ¨r=/Omega1×(/Omega1×δr)+2/Omega1×δv rel+δarel Applying the bac−cabrule to the first term on the right-hand side, we get δ¨r=/Omega1(/Omega1·δr)−/Omega12δr+2/Omega1×δvrel+δarel (7.18) Since the orbit of Ais circular, we may write the angular velocity as /Omega1=nˆk (7.19) where n, the mean motion, is constant. Thus, /Omega1·δr=nˆk·(δxˆi+δyˆj+δzˆk)=nδz (7.20) and /Omega1×δvrel=nˆk×(δ˙xˆi+δ˙yˆj+δ˙zˆk)=− nδ˙yˆi+nδ˙xˆj (7.21) Substituting Equations 7.19, 7.20 and 7.21, along with Equations 7.17, into Equation 7.18 yields δ¨r=nˆk(nδz)−n2(δxˆi+δyˆj+δzˆk)+2(−nδ ˙yˆi+nδ˙xˆj)+δ¨xˆi+δ¨yˆj+δ¨zˆk Finally, collecting terms leads to δ¨r=(−n2δx−2nδ˙y+δ¨x)ˆi+(−n2δy+2nδ˙x+δ¨y)ˆj+δ¨zˆk (7.22) 326 Chapter 7 Relative motion and rendezvous This expression gives the components of the chaser’s absolute relative acceleration vector in terms of quantities measured in the moving reference. Since the orbit of Ais circular, the mean motion is found as n=v r0=1 r0/radicalbiggµ r0=/radicalbiggµ r3 0 Therefore, µ r3 0=n2(7.23) Recalling Equations 7.15 and 7.17a, we also note that r0·δr=(r0ˆi)·(δxˆi+δyˆj+δzˆk)=r0δx (7.24) Substituting Equations 7.17a, 7.23 and 7.24 into Equation 7.14 (the equation of motion) yields δ¨r=− n2/bracketleftbigg δxˆi+δyˆj+δzˆk−3 r2 0(r0δx)r0ˆi/bracketrightbigg =2n2δxˆi−n2δyˆj−n2δzˆk (7.25) Combining Equation 7.22 (a kinematic relat ionship) and Equation 7.25 (the equation of motion), we obtain (−n2δx−2nδ˙y+δ¨x)ˆi+(−n2δy+2nδ˙x+δ¨y)ˆj+δ¨zˆk=2n2δxˆi−n2δyˆj−n2δzˆk Upon collecting terms to the left-hand side, we get (δ¨x−3n2δx−2nδ˙y)ˆi+(δ¨y+2nδ˙x)ˆj+(δ¨z+n2δz)ˆk=0 That is, δ¨x−3n2δx−2nδ˙y=0 (7.26a) δ¨y+2nδ˙x=0 (7.26b) δ¨z+n2δz=0 (7.26c) These are the Clohessy–Wiltshire (CW) equations. When using these equations we will refer to the moving frame of reference in which they were derived as theClohessy–Wiltshire frame (or CW frame). Equations 7.26 are a set of coupled, secondorder differential equations with constant coefficients. The initial conditions are Att=0δx=δx 0δy=δy0δz=δz0 (7.27) δ˙x=δ˙x0δ˙y=δ˙y0δ˙z=δ˙z0 From Equation 7.26b, d dt(δ˙y+2nδx)=0 which means δ˙y+2nδx=const 7.4 Clohessy–Wiltshire equations 327 We find the constant by evaluating the left-hand side at t=0. Therefore, δ˙y+2nδx=δ˙y0+2nδx0 so that δ˙y=δ˙y0+2n(δ x0−δx) (7.28) Substituting this result into Equation 7.26a yields δ¨x−3n2δx−2n[δ˙y0+2n(δ x0−δx)]=0 which, upon rearrangement, becomes, δ¨x+n2δx=2nδ˙y0+4n2δx0 (7.29) The solution of this differential equation is δx=complementary solution/bracehtipdownleft/bracehtipupright/bracehtipupleft /bracehtipdownright Asinnt+Bcosnt+particular solution/bracehtipdownleft /bracehtipupright/bracehtipupleft /bracehtipdownright 1 n2(2nδ˙y0+4n2δx0) or δx=Asinnt+Bcosnt+2 nδ˙y0+4δx0 (7.30) Differentiating this equation o nce with respect to time, we obtain δ˙x=nAcosnt−nBsinnt (7.31) Evaluating Equation 7.30 at t=0w efi n d δx0=B+2 nδ˙y0+4δx0⇒ B=− 3δx0−2δ˙y0 n Evaluating Equation 7.31 at t=0 yields δ˙x0=nA⇒ A=δ˙x0 n Substituting these values of Aand Bback into Equation 7.30 leads to δx=δ˙x0 nsinnt+/parenleftbigg −3δ x0−2δ˙y0 n/parenrightbigg cosnt+2 nδ˙y0+4δx0 which, upon combining terms, becomes δx=(4−3c o s nt)δx0+sinnt nδ˙x0+2 n(1−cosnt)δ˙y0 (7.32) Therefore, δ˙x=3nsinntδx0+cosntδ˙x0+2 sin ntδ˙y0 (7.33) 328 Chapter 7 Relative motion and rendezvous Substituting Equation 7.32 into Equation 7.28 yields δ˙y=δ˙y0+2n/bracketleftbigg δx0−(4−3c o s nt)δx0−sinnt nδ˙x0−2 n(1−cosnt)δ˙y0/bracketrightbigg which simplifies to become δ˙y=6n(cos nt−1)δx0−2 sin ntδ˙x0+(4 cos nt−3)δ˙y0 (7.34) Integrating this expression with respect to time, we find δy=6n/parenleftbigg1 nsinnt−t/parenrightbigg δx0+2 ncosntδ˙x0+/parenleftbigg4 nsinnt−3t/parenrightbigg δ˙y0+C (7.35) Evaluating δyatt=0 yields δy0=2 nδ˙x0+C⇒ C=δy0−2 nδ˙x0 Substituting this value for Cinto Equation 7.35, we get δy=6(sin nt−nt)δx0+δy0+2 n(cos nt−1)δ˙x0+/parenleftbigg4 nsinnt−3t/parenrightbigg δ˙y0(7.36) Finally, the solution of Equation 7.26c is δz=Dcosnt+Esinnt (7.37) so that δ˙z=− nDsinnt+nEcosnt (7.38) We evaluate these two expressions at t=0 to obtain the constants of integration: δz0=D δ˙z0=nE Putting these values of Dand Eback into Equations 7.36 and 7.38 yields δz=cosntδz0+1 nsinntδ˙z0 (7.39) δ˙z=− nsinntδz0+cosntδ˙z0 (7.40) Now that we have finished solving the Clohessy–Wiltshire equations, let us change our notation a bit and denote the x,yand zcomponents of relative velocity in the moving frame as δu,δvandδw, respectively. That is, δu=δ˙xδv=δ˙yδw=δ˙z The initial conditions on the relative velocity components are then written δu0=δ˙x0δv0=δ˙y0δw0=δ˙z0 7.4 Clohessy–Wiltshire equations 329 Using this notation we write Equations 7.32, 7.33, 7.34, 7.36, 7.39 and 7.40 as δx=(4−3c o s nt)δx0+sinnt nδu0+2 n(1−cosnt)δv 0 δy=6(sin nt−nt)δx0+δy0+2 n(c o s nt−1)δu0+1 n(4 sin nt−3nt)δv 0 δz=cosntδz0+1 nsinntδw0 (7.41) δu=3nsinntδx0+cosntδu0+2 sin ntδv0 δv=6n(cos nt−1)δx0−2 sin ntδu0+(4 cos nt−3)δv 0 δw=− nsinntδz0+cosntδw0 Let us introduce matrix notation to define the relative position and velocity vectors {δr(t)}=  δx(t) δy(t) δz(t)  {δv(t)}=  δu(t) δv(t ) δw(t)   and their initial values (at t=0) {δr0}=  δx0 δy0 δz0  {δv0}=  δu0 δv0 δw0   Observe that we have dropped the subscript rel introduced in Equations 7.17 because it is superfluous in rendezvous analysis, where all kinematic quantities are relativeto the Clohessy–Wiltshire frame. In matrix notation Equations 7.41 appear morecompactly as {δr(t)}=[/Phi1 rr(t)]{δ r0}+[/Phi1 rv(t)]{δ v0} (7.42a) {δv(t)}=[/Phi1 vr(t)]{δ r0}+[/Phi1 vv(t)]{δ v0} (7.42b) where the Clohessy–Wiltshire matrices are [/Phi1rr(t)]= 4−3c o s nt 00 6(sin nt−nt)1 0 00 c o s nt  [/Phi1rv(t)]= 1 nsinnt2 n(1−cosnt)0 2 n(cos nt−1)1 n(4 sin nt−3nt)0 001 nsinnt  (7.43) [/Phi1vr(t)]= 3nsinnt 00 6n(cos nt−1) 0 0 00 −nsinnt  [/Phi1vv(t)]= cosnt 2 sin nt 0 −2 sin nt 4c o s nt−30 00 c o s nt  330 Chapter 7 Relative motion and rendezvous 7.5 Two-impulse rendezvous maneuvers Figure 7.6 illustrates the rendezvous problem. At time t=0−(the instant preceding t=0), the position δr0and velocity δv− 0of the chase vehicle Brelative to the target Aare known. At t=0 an impulsive maneuver instantaneously changes the relative velocity to δv+ 0att=0+(the instant after t=0). The components of δv+ 0are shown in Figure 7.6. We must determine the values of δu+ 0,δv+ 0,δw+ 0, at the beginning of the rendezvous trajectory, so that Bwill arrive at the target in a specified time tf.T h e delta-v required to place Bon the rendezvous trajectory is {/Delta1v0}={δv+ 0}−{δv− 0}=  δu+ 0 δv+ 0 δw+ 0  −  δu− 0 δv− 0 δw− 0  (7.44) At time tf,Barrives at A, at the origin of the co-moving frame, which means {δrf}={δr(tf)}={ 0}. Evaluating Equation 7.42a at tf, we find {0}=[/Phi1rr(tf)]{δr0}+[/Phi1rv(tf)]{δv+ 0} (7.45) Solving this for {δv+ 0}yields {δv+ 0}=− [/Phi1rv(tf)]−1[/Phi1rr(tf)]{δr0} (7.46) Earthr0xz y AB Rendezvous trajectory Orbit of A /H9254v/H11001 /H9254u/H11001 /H9254v/H11001 /H9254u/H1100200 0 /H9254y/H11002 f/H9254v/H11002 f fdr0 Figure 7.6 Rendezvous with a target Ain the neighborhood of the chase vehicle B. 7.5 Two-impulse rendezvous maneuvers 331 where [ /Phi1rv(tf)]−1is the matrix inverse of [ /Phi1rv(tf)]. We know the velocity δv+ 0at the beginning of the rendezvous path substitut ing Equation 7.46 into Equation 7.42b we obtain the velocity δv− fat which Barrives at the target A, when t=t− f: {δv− f}=[/Phi1 vr(tf)]{δ r0}+[/Phi1 vv(tf)]{δ v+ 0} =[/Phi1vr(tf)]{δ r0}+[/Phi1 vv(tf)](−[/Phi1 rv(tf)]−1[/Phi1rr(tf)]{δ r0}) Simplifying, we get {δv− f}=([/Phi1 vr(tf)]−[/Phi1vv(tf)][/Phi1 rv(tf)]−1[/Phi1rr(tf)]){δ r0} (7.47) Obviously, an impulsive delta-v maneuver is required at t=tfto bring vehicle Bto rest relative to A(δv+ f=0): {/Delta1v f}={δv+ f}−{δv− f}={ 0}−{δv− f}=− { δv− f} (7.48) Note that in Equations 7.44 and 7.48 we are using the difference between relative velocities to calculate delta-v, which is the difference in absolute velocities. T o showthat this is valid, use Equation 1.38, to write v −=v− 0+/Omega1−×r− rel+v− rel (7.49) v+=v+ 0+/Omega1+×r+ rel+v+ rel Since the target is passive, the impulsive maneuver has no effect on its state of motion, which means v+ 0=v− 0and/Omega1+=/Omega1−. Furthermore, by definition of an impulsive maneuver, there is no change in the position, i.e., r+ rel=r− rel. It follows from Equation 7.49 that v+−v−=v+ rel−v− relor/Delta1v=/Delta1vrel example 7.2A space station and spacecraft are in orbits with the following parameters: Space station Spacecraft Perigee ×apogee (altitude) 300 km circular 318.50 ×515.51 km Period (computed using above data) 1.508 hr 1.548 hr True anomaly, θ 60◦349.65◦ Inclination, i 40◦40.130◦ RA,/Omega1 20◦19.819◦ Argument of perigee, ω 0◦(arbitrary) 70.662◦ Compute the total delta-v required for an eight-hour, two-impulse rendezvous trajectory. We use the given data in Algorithm 4.1 to obtain the state vectors of the two spacecraft in the geocentric equatorial frame. Space station: r0=1622.39 ˆI+5305.10 ˆJ+3717.44 ˆK(km) v0=− 7.29977 ˆI+0.492357 ˆJ+2.48318 ˆK(km/s) 332 Chapter 7 Relative motion and rendezvous (Example 7.2 continued)Spacecraft: r=1612.75ˆI+5310.19ˆJ+3750.33ˆK(km) v=− 7.35321ˆI+0.463856 ˆJ+2.46920 ˆK(km/s) The space station reference frame unit vectors (at this instant) are, by definition: ˆi=r0 /bardblr0/bardbl=0.242945 ˆI+0.794415 ˆJ+0.556670 ˆK ˆj=v0 /bardblv0/bardbl=− 0.944799 ˆI+0.063725 ˆJ+0.321394 ˆK ˆk=ˆi׈j=0.219846 ˆI−0.604023 ˆJ+0.766044 ˆK Therefore, the transformation matrix from the geocentric equatorial frame into space station frame is (at this instant) [Q]Xx= 0.242945 0 .794415 0 .556670 −0.944799 0 .063725 0 .321394 0.219846 −0.604023 0 .766044  The position vector of the spacecraft relative to the space station (in the geocentric equatorial frame) is δr=r−r0=− 9.63980ˆI+5.08240ˆJ+32.8821ˆK(km) The relative velocity is given by the formula (Equation 1.38) δv=v−v0−/Omega1space station ×δr where /Omega1space station =nˆkand n, the mean motion of the space station, is n=v0 r0=7.72627 6678=0.00115697 rad /s( a ) Thus δv=− 7.35321ˆI+0.463856 ˆJ+2.46920 ˆK−(−7.29977ˆI+0.492357 ˆJ+2.48318 ˆK) −(0.00115697)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆI ˆJ ˆK 0.219846 −0.604023 0 .766044 −9.63980 5 .08240 32 .8821/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle so that δv=− 0.024854 ˆI−0.01159370 ˆJ−0.00853577 ˆK(km/s) In space station coordinates, the relative position vector δr0at the beginning of the rendezvous maneuver is {δr0}=[Q]Xx{δr}= 0.242945 0 .794415 0 .556670 −0.944799 0 .063725 0 .321394 0.219846 −0.604023 0 .766044   −9.63980 5.08240 32.8821   =  20 20 20  (km) (b) 7.5 Two-impulse rendezvous maneuvers 333 Likewise, the relative velocity δv− 0justbefore launch into the rendezvous trajectory is {δv− 0}=[Q ]Xx{δv}= 0.242945 0.794415 0.556670 −0.944799 0.063725 0.321394 0.219846 −0.604023 0.766044   −0.024854 −0.0115937 −0.00853578   =  −0.02000 0.02000 −0.005000  (km/s) The Clohessy–Wiltshire matrices, for t=tf=8h r= 28 800 s and n=0.00115697 rad/s [from (a)], are [/Phi1rr]= 4−3c o s nt 00 6(sin nt−nt)1 0 00 c o s nt = 4.98383 0 0 −194.257 1.000 0 00 −0.327942  [/Phi1rv]= 1 nsinnt2 n(1−cosnt)0 2 n(cos nt−1)1 n(4 sin nt−3nt)0 001 nsinnt  = 816.525 2295.54 0 −2295.54 −83 133.90 0 0 816.525  [/Phi1vr]= 3nsinnt 00 6n(cos nt−1) 0 0 00 −nsinnt = 0.00327897 0 0 −0.00921837 0 0 00 −0.00109299  [/Phi1vv]= cosnt 2 sin nt 0 −2 sin nt 4c o s nt−30 00 c o s nt  = −0.327942 1.88940 0 −1.88940 −4.31177 0 00 −0.327942  From Equation 7.46 and (b) we find δv+ 0:   δu+ 0 δv+ 0 δw+ 0  =− 816.525 2295.54 0 −2295.54 −83 133.90 0 0 816.525 −1 × 4.98383 0 0 −194.257 1.000 0 00 −0.327942   20 20 20   334 Chapter 7 Relative motion and rendezvous (Example 7.2 continued) =− 816.525 2295 .54 0 −2295.54−83 133 .90 0 0 816 .525 −1  99.6765 −3865.14 −6.55884   =  0.00936084 −0.0467514 0.00803263  (km/s) (c) From Equation 7.42b, evaluated at t=tf,w eh a v e {δvf}=[/Phi1vr(tf)]{δr0}+[/Phi1vv(tf)]{δv+ 0} Substituting (b) and (c),   δu− f δv− f δw− f  = 0.00327897 0 0 −0.00921837 0 0 00 −0.00109299   20 2020   + −0.327942 1 .88940 0 −1.88940 −4.31177 0 00 −0.327942   0.00936084 −0.0467514 0.00803263     δu − f δv− f δw− f  =  −0.0258223 −0.000472444 −0.0222449  (km/s) (d) Delta-v at the beginning of the rendezvous maneuver is found as {/Delta1v0}={δv+ 0}−{δv− 0}=  0.00936084 −0.0467514 0.00803263  −  −0.02 0.02 −0.005  =  0.0293608 −0.0667514 0.0130326   Delta-v at the conclusion of the maneuver is {/Delta1vf}={δv+ f}−{δv− f}=  0 00  −  −0.0258223 −0.000472444 −0.0222449  =  0.0258223 0.000472444 0.0222449  (km/s) The total delta-v requirement is /Delta1vtotal=/bardbl/Delta1v0/bardbl+/bardbl/Delta1vf/bardbl= 0.0740787 +0.03559465 =0.109673 km /s=109.7m/s 7.5 Two-impulse rendezvous maneuvers 335 From Equation 7.42a, we have, for 0 <t<tf,   δx(t) δy(t) δz(t)  = 4−3c o s nt 00 6(sin nt−nt)1 0 00 c o s nt   20 2020   + 1 nsinnt2 n(1−cosnt)0 2 n(c o s nt−1)1 n(4 sin nt−3nt)0 001 nsinnt   0.00936084 −0.0467514 0.00803263   xy z t = 8 hrFive ‘orbits’ of the target 20 km20 km 20 kmt = 0 Figure 7.7 Rendezvous trajectory of the chase vehicle relative to the target. Substituting nfrom (a), we obtain the relative posit ion vector as a function of time. It is plotted in Figure 7.7. Example 7.3A target and a chase vehicle are in the same 300 km circular earth orbit. The chaser is 2 km behind the target when the chaser initiates a two-impulse rendezvous maneuverso as to rendezvous with the target in 1.49 hours. Find the total delta-v requirement. For the circular orbit v=/radicalbigg µ r=/radicalbigg 398 600 6378+300=7.726 km /s( a) so that the mean motion is n=v r=7.726 6678=0.0011569 rad /s (b) 336 Chapter 7 Relative motion and rendezvous (Example 7.3 continued)For this mean motion and the rendezvous trajectory time t=1.49 hr =5364 s, the Clohessy–Wiltshire matrices are [/Phi1rr]= 1.0090 0 0 −37.699 1 0 00 0 .99700  [/Phi1rv]= −66.946 5 .1928 0 −5.1928 −16 360 0 00 −66.946  (c) [/Phi1vr]= −2.6881×10−400 −2.0851×10−500 00 8 .9603×10−5  [/Phi1vv]= 0.99700 −0.15490 0 0.15490 0 .98798 0 00 0 .99700  The initial and final positions of the chaser in the CW frame are {δr0}=  0 −2 0  (km) {δrf}=  0 00  (d) Thus, solving the first CW equation, {δrf}= [/Phi1rr]{δr0}+[/Phi1rv]{δv+ 0}, for{δv+ 0}, we get {δv+ 0}=− [/Phi1rv]−1[/Phi1rr]{δr0}=− −0.014937 −4.7412×10−60 4.7412×10−6−6.1124×10−50 00 −0.014937  × 1.0090 0 0 −37.699 1 0 00 0 .99700   0 −2 0   {δv+ 0}=  −9.4824×10−6 −1.2225×10−4 0  (km/s) (e) Therefore, the second CW equation, {δv− f}=[/Phi1vr]{δr0}+[/Phi1vv]{δv+ 0}, yields {δv− f}= −2.6881×10−400 −2.0851×10−500 00 8 .9603×10−5   0 −2 0   + 0.99700 −0.15490 0 0.15490 0 .98798 0 00 0 .99700   −9.4824×10−6 −1.2225×10−4 0   7.5 Two-impulse rendezvous maneuvers 337 {δv− f}=  9.4824 ×10−6 −1.2225 ×10−4 0  (km/s) (f) Since the chaser is in the same circular orbit as the target, its relative velocity is initially zero, i.e., {δv− 0}={ 0}. (See also Equation 7.58 at the end of the next section.) Thus, {/Delta1v 0}={δv+ 0}−{δv− 0}=  −9.4824 ×10−6 −1.2225 ×10−4 0  −  0 00  =  −9.4824 ×10 −6 −1.2225 ×10−4 0  (km/s) which implies /bardbl/Delta1v 0/bardbl=0.1226 m /s( g) At the end of the rendezvous maneuver, {δv+ f}={ 0}, so that {/Delta1v f}={δv+ f}−{δv− f}=  0 00  −  9.4824 ×10 −6 −1.2225 ×10−4 0  =  −9.4824 ×10−6 1.2225 ×10−4 0  (km/s) Therefore /bardbl/Delta1v f/bardbl=0.1226 m /s( h) The total delta-v required is /Delta1vtotal=/bardbl/Delta1v0/bardbl+/bardbl/Delta1vf/bardbl=0.2452 m /s (i) Observe that in this case the motion takes place entirely in the plane of the target orbit. There is no motion normal to the plane (in the zdirection). The copla- nar rendezvous trajectory relative to the CW frame is sketched in Figure 7.8. /H110020.5 /H110021 /H110021.5Chaser /H110022y /H110020.4/H110020.200.20.4x 2 kmPerigee of chaser's transfer orbitClohessy–Wiltshire frame Circular orbit of target Continuation if no rendezvousTarget Figure 7.8 Motion of the chaser relative to the target. 338 Chapter 7 Relative motion and rendezvous 7.6 Relative motion in close-proximity circular orbits Figure 7.9 shows two spacecraft in coplanar circular orbits. Let us calculate the velocity δvof the chase vehicle Brelative to the target Awhen they are in close proximity. ‘Close proximity’ means that δr r0<<1 T o solve this problem, we must use the relative velocity equation, vB=vA+/Omega1×δr+δv (7.50) where /Omega1is the angular velocity of the CW frame attached to A, /Omega1=nˆk nis the mean motion of the target vehicle, n=vA r0(7.51) where, by virtue of the circular orbit, vA=/radicalbiggµ r0(7.52) Chaser, B Target, A r0rx yCoplanar circular orbits Earthûr û⊥ dr Figure 7.9 Two spacecraft in close proximity. 7.6 Relative motion in close-proximity circular orbits 339 Solving Equation 7.50 for the relative velocity δvyields δv=vB−vA−(nˆk)×δr (7.53) Since the chase orbit is circular, we have for the first term on the right-hand side of Equation 7.53 vB=/radicalbiggµ rˆu⊥=/radicalbiggµ r(ˆk׈ur)=√µˆk×(1√rr r) (7.54) Since, as is apparent from Figure 7.9, r=r0+δr, we can write this expression for vBas follows: vB=√µˆk×r−3 2(r0+δr) (7.55) Now r−3 2=(r2)−3 4= See Equation 7.10/bracehtipdownleft/bracehtipupright/bracehtipupleft /bracehtipdownright r2 0/parenleftbigg 1+2r0·δr r2 0/parenrightbigg −3 4 =r−3 2 0/parenleftbigg 1+2r0·δr r2 0/parenrightbigg−3 4 (7.56) Using the binomial theorem (cf. Equation 5.44), and retaining terms at most linear inδr,w eg e t /parenleftbigg 1+2r0·δr r2 0/parenrightbigg−3 4 =1−3 2r0·δr r2 0 Substituting this into Equation 7.56 leads to r−3 2=r−3 2 0−3 2r0·δr r7 2 0 Upon substituting this result into Equation 7.55, we get vB=√µˆk×(r0+δr) r−3 2 0−3 2r0·δr r7 2 0  Retaining terms at most linear in δr, we can write this as vB=ˆk×/braceleftbigg/radicalbiggµ r0r0 r0+√µ/r 0 r0δr−3 2√µr0 r0/bracketleftbigg/parenleftbiggr0 r0/parenrightbigg ·δr/bracketrightbiggr0 r0/bracerightbigg Using Equations 7.51 and 7.52, together with the facts that δr=δxˆi+δyˆjand r0/r0=ˆi, this reduces to vB=ˆk×/braceleftbigg vAˆi+vA r0(δxˆi+δyˆj)−3 2vA r0[ˆi·(δxˆi+δyˆj)]ˆi/bracerightbigg =vAˆj+(−nδ yˆi+nδxˆj)−3 2nδxˆj 340 Chapter 7 Relative motion and rendezvous x yO Neighboring circular orbits Earthδv Figure 7.10 Circular orbits, with relative velocity directions, in the vicinity of the Clohessy–Wiltshire frame. so that vB=− nδyˆi+(vA−1 2nδx)ˆj (7.57) This is the absolute velocity of the chaser resolved into components in the target’s Clohessy–Wiltshire frame. Substituting Equation 7.57 into 7.53 and using the fact that vA=vAˆjyields δv=[−nδyˆi+(vA−1 2nδx)ˆj]−(vAˆj)−(nˆk)×(δxˆi+δyˆj) =− nδyˆi+vAˆj−1 2nδxˆj−vAˆj−nδxˆj+nδyˆi so that δv=−3 2nδxˆj (7.58) This is the velocity of the chaser as measured in the moving reference frame of the neighboring target. Keep in mind that circular orbits were assumed at the outset. In the Clohessy–Wiltshire frame, neighboring coplanar circular orbits appear to be straight lines parallel to the yaxis, which is the orbit of the origin. Figure 7.10 illus- trates this point, showing also the linear velocity variation according to Equation 7.58. Problems 7.1 Two manned spacecraft, AandB(see figure), are in circular, polar ( i=90◦) orbits around the earth. A’s orbital altitude is 300 km; B’s is 250 km. At the instant shown ( Aover the equator, Bover the North Pole), calculate (a) the position, (b) velocity, and (c) the acceleration of Brelative to A. Problems 341 A’syaxis points always in the flight direction, and its xaxis is directed radially outward at all times.{Ans.: (a) r rel)xyz=− 6678ˆi+6628ˆjkm; (b) vrel)xyz=− 0.08693 ˆikm/s; (c) arel)xyz= −1.140 ×10−6ˆjkm/s2} γXYZ AB y zx EarthN Figure P .7.1 7.2 Spacecraft AandBare in coplanar, circular geocentric orbits. The orbital radii are shown in the figure. When Bis directly below A, as shown, calculate B’s acceleration relative to A. {Ans.: (a rel)xyz=− 0.268ˆi(m/s2)} x y A B Earth8000 km7000 km Figure P .7.2 342 Chapter 7 Relative motion and rendezvous 7.3 Use the order of magnitude analysis in this chapter as a guide to answer the following questions. (a) If r=r0+δr, express√r(where r=√r·r) to the first order in δr(i.e., to the first order in the components of δr=δxˆi+δyˆj+δzˆk). In other words, find O(δr), such that√r=√r0+O(δr), where O(δr) is linear in δr. (b) For the special case r0=3ˆi+4ˆj+5ˆkandδr=0.01ˆi−0.01ˆj+0.03ˆk, calculate√r−√r0and compare that result with O(δr). (c) Repeat part (b) using δr=ˆi−ˆj+3ˆkand compare the results. {Ans.: (a) O(δr)=r0·δr//parenleftbigg 2r3 2 0/parenrightbigg ; (b) O(δr)//parenleftbig√r−√r0/parenrightbig =0.998; (c) O(δr)/ (√r−√r0)=0.903} r0r xyz dr Figure P .7.3 7.4 Write the expression r=a(1−e2) 1+ecosθas a linear function of e, valid for small values of e(e<<1). 7.5 Given ¨x+9x=10, with the initial conditions x=5 and ˙x=− 3a t t=0, find xand˙xat t=1.2. {Ans.: x(1.2)=− 1.934,˙x(1.2)=7.853} 7.6 Given that ¨x+10x+2˙y=0 ¨y+3˙x=0 with initial conditions x(0)=1,y(0)=2,˙x(0)=− 3 and ˙y(0)=4, find xand yatt=5. {Ans.: x(5)=− 6.460, y(5)=97.31} 7.7 A space station is in a 90-minute period earth orbit. At t=0, a satellite has the following position and velocity components relative to a Clohessy–Wiltshire frame attached to the Problems 343 space station: {δr}=⌊ 100 ⌋T(km), {δv}=⌊ 01 00 ⌋T(m/s). How far is the satellite from the space station 15 minutes later?{Ans.: 11.2 km} 7.8 A space station is in a circular earth orbit of radius 6600 km. An approaching space- craft executes a delta-v burn when its positio n vector relative to the space station is {δr 0}=⌊ 111 ⌋T(km). Just before the burn the relative velocity of the spacecraft was{δv− 0}=⌊ 005 ⌋T(m/s). Calculate the total delta-v required for the space shuttle to rendezvous with the station in one-third period of the space station orbit.{Ans.: 6.21 m/s} 7.9 A space station is in circular orbit 2 of radius r 0. A spacecraft is in coplanar circular orbit 1o fr a d i u s r0+δr.A tt=0 the spacecraft executes an impulsive maneuver to rendezvous with the space station at time tf=one-half the period T0of the space station. For a Hohmann transfer orbit ( δu+ 0=0), find (a) the initial position of the spacecraft relative to the space station, and (b) the relative velocity of the spacecraft when it arrives at the target. Sketch the rendezvous trajectory relative to the target. {Ans.: (a) {δr0}=⌊δr3π/(4δr)0⌋T, (b){δv− f}=⌊ 0πδr/(2T 0)0⌋T} XY t /H11005 01 2 fr0δr Figure P .7.9 7.10 Assuming a Hohmann transfer ( δu+ 0=0), calculate the total delta-v required for ren- dezvous if {δr0}=⌊ 0δy00⌋T,{δv− 0}=⌊ 000 ⌋Tand tf=the period of the circular target orbit. Sketch the rendezvous trajectory relative to the target.{Ans.: /Delta1v tot=2δy0/(3T )} 344 Chapter 7 Relative motion and rendezvous 7.11 Spacecraft Aand Bare in the same circular earth orbit with a period of 2 hours. B is 6 km ahead of A.A t t=0,Bapplies an in-track delta-v (retrofire) of 3 m/s. Using a Clohessy–Wiltshire frame attached to A, determine the distance between Aand Bat t=30 minutes and the velocity of Brelative to A. {Ans.: /bardblδr/bardbl=10.9k m ,/bardblδv/bardbl=10.8 m/s} 7.12 A GEO satellite strikes some orbiting debris and is found 2 hours afterwards to have drifted to the position {δr}=⌊−10 10 0 ⌋Tkm relative to its original location. At that time the only slightly damaged satellite initiates a two-impulse maneuver to return to itsoriginal location in 6 hours. Find the total delta-v for this maneuver.{Ans.: 3.5 m/s} 7.13 A space station is in a 245 km circular earth orbit inclined at 30 ◦. The right ascension of its node line is 40◦. Meanwhile, a space shuttle has been launched into a 280 km by 250 km orbit inclined at 30.1◦, with a nodal right ascension of 40◦and argument of perigee equal to 60◦. When the shuttle’s true anomaly is 40◦, the space station is 99◦beyond its node line. At that instant, the space shuttle executes a delta-v burn to rendezvous with the spacestation in (precisely) t fhours, where tfis selected by you or assigned by the instructor. Calculate the total delta-v required and sketch the projection of the rendezvous trajectoryon the xyplane of the space station coordinates. 7.14 The space station is in a circular earth orbit of radius 6600 km. The space shuttle is also in a circular orbit in the same plane as the space station’s. At the instant that the positionof the shuttle relative to the space station, in Clohessy–Wiltshire coordinates, is (5 km,0, 0). What is the relative velocity δvof the space shuttle in meters/s? {Ans.: 8.83 m/s} x yz Space stationShuttle CW frame5 km Figure P .7.14 7.15 The Clohessy–Wiltshire coordinates and velocities of a spacecraft upon entering a ren- dezvous trajectory with the target vehicle are shown. The spacecraft orbits are coplanar.Calculate the distance dof the spacecraft from the target when t=π/2n,w h e r e nis the mean motion of the target’s circular orbit.{Ans.: 0.900 δr} Problems 345 (δx, δy) d TargetSpacecraft at t /H11005 0/H11001 yx CW frameπ 16nδr nδr47 δr πδr Figure P .7.15 7.16 The target Tis in a circular earth orbit with mean motion n. The chaser Cis directly above Tin a slightly larger circular orbit having the same plane as T’s. What relative initial veloc- ityδv+ 0is required so that Carrives at the target Tat time tf=one-half the target’s period? {Ans.: δv+ 0=− 0.589nδ x0ˆi−1.75nδ x0ˆj} x y C TCW frameδx0 Figure P .7.16 7.17 The space shuttle and the International Space Station are in coplanar circular orbits. The space station has an orbital radius rand a mean motion n. The shuttle’s radius is 346 Chapter 7 Relative motion and rendezvous r−d(d<<r). If a two-impulse rendezvous maneuver with tf=π/(4n) is initiated with zero relative velocity in the xdirection ( δ˙x+ 0=0), calculate the initial relative ycoordinate of the shuttle.{Ans.: δy 0=− 1.98d} x yISS orbit Shuttle orbitd Figure P .7.17 8Chapter Interplanetary trajectories Chapter outline 8.1 Introduction 347 8.2 Interplanetary Hohmann transfers 348 8.3 Rendezvous opportunities 349 8.4 Sphere of influence 354 8.5 Method of patched conics 359 8.6 Planetary departure 360 8.7 Sensitivity analysis 366 8.8 Planetary rendezvous 368 8.9 Planetary flyby 375 8.10 Planetary ephemeris 387 8.11 Non-Hohmann interplanetary trajectories 391 Problems 398 8.1 Introduction In this chapter we consider some basic aspects of planning interplanetary missions. We begin by considering Hohmann transfers, which are the easiest to analyze and the most energy efficient. The orbits of the planets involved must lie in the sameplane and the planets must be positioned just right for a Hohmann transfer to beused. The time between such opportunities is derived. The method of patched conicsis employed to divide the mission into three parts: the hyperbolic departure trajectoryrelative to the home planet; the cruise ellipse relative to the sun; and the hyperbolicarrival trajectory, relative to the target planet. 347 348 Chapter 8 Interplanetary trajectories The use of patched conics is justified by calculating the radius of a planet’s sphere of influence and showing how small it is on the scale of the solar system. Matchingthe velocity of the spacecraft at the home planet’s sphere of influence to that requiredto initiate the outbound cruise phase and then specifying the periapse radius of thedeparture hyperbola determines the delta-v requirement at departure. The sensitivityof the target radius to the burnout conditions is discussed. Matching the velocitiesat the target planet’s sphere of influence and specifying the periapse of the arrivalhyperbola yields the delta-v at the target for a planetary rendezvous or the directionof the outbound hyperbola for a planetary flyby. Flyby maneuvers are discussed,including the effect of leading and trailing side flybys, and some noteworthy examplesof the use of gravity assist maneuvers are presented. The chapter concludes with an analysis of the situation in which the planets’ orbits are not coplanar and the transfer ellipse is tangent to neither orbit. This is akin to thechase maneuver in Chapter 6 and requires the solution of Lambert’s problem using Algorithm 5.2. 8.2 Interplanetary Hohmann transfers As can be seen from Table A.1, the orbits of most of the planets in the solar system lie very close to the earth’s orbital plane (the ecliptic plane). The innermost planet,Mercury, and the outermost planet, Pluto, differ most in inclination (7 ◦and 17◦, respectively). The orbital planes of the other planets lie within 3.5◦of the ecliptic. It is also evident from Table A.1 that most of the planetary orbits have small eccentricities,the exceptions once again being Mercury and Pluto. T o simplify the beginning of ourstudy of interplanetary trajectories, we will assume that all of the planets’ orbits arecircular and coplanar. Later on, in Section 8.10, we will relax this assumption. The most energy efficient way for a spacecraft to transfer from one planet’s orbit to another is to use a Hohmann transfer ellipse (Section 6.2). Consider Figure 8.1,which shows a Hohmann transfer from an inner planet 1 to an outer planet 2. Thedeparture point Dis at the periapse (perihelion) of the transfer ellipse and the arrival point is at the apoapse (aphelion). The cir cular orbital speed of planet 1 relative to the sun is given by Equation 2.53, V 1=/radicalbiggµsun R1(8.1) The specific angular momentum hof the transfer ellipse relative to the sun is found from Equation 6.2, so that the velocity of the space vehicle on the transfer ellipse atthe departure point Dis V(v) D=h R1=/radicalbig 2µsun/radicalBigg R2 R1(R1+R2)(8.2) This is greater than the speed of the planet. Therefore the required delta-v at Dis /Delta1VD=V(v) D−V1=/radicalbiggµsun R1/parenleftBigg/radicalBigg 2R2 R1+R2−1/parenrightBigg (8.3) 8.3 Rendezvous opportunities 349 SunPlanet 1 at departurePlanet 2 at arrivalPlanet 2 at departure Planet 1 at arrivalHeliocentric elliptical transfer trajectory R2 R112 DA VAVD V1 V2 Figure 8.1 Hohmann transfer from inner planet 1 to outer planet 2. Likewise, the delta-v at the arrival point Ais /Delta1V A=V2−V(v) A=/radicalbiggµsun R2/parenleftBigg 1−/radicalBigg 2R1 R1+R2/parenrightBigg (8.4) This velocity increment, like that at point D, is positive since planet 2 is traveling faster than the spacecraft at point A. For a mission from an outer planet to an inner planet, as illustrated in Figure 8.2, the delta-vs computed using Equations 8.3 and 8.4 will both be negative instead of positive. That is because the departure point and arrival point are now at aphelionand perihelion, respectively, of the tr ansfer ellipse. The speed of the spacecraft must be reduced for it to drop into the lower-energy transfer ellipse at the departure pointD, and it must be reduced again at point Ain order to arrive in the lower-energy circular orbit of planet 2. 8.3 Rendezvous opportunities The purpose of an interplanetary mission is for the spacecraft not only to intercept aplanet’s orbit but also to rendezvous with the planet when it gets there. For rendezvousto occur at the end of a Hohmann transfer, the location of planet 2 in its orbit at thetime of the spacecraft’s departure from planet 1 must be such that planet 2 arrivesat the apse line of the transfer ellipse at the same time the spacecraft does. Phasing 350 Chapter 8 Interplanetary trajectories SunPlanet 1 at departurePlanet 2 at arrival Planet 2 at departurePlanet 1 at arrivalHeliocentric elliptical transfer trajectory R1 R21 2 A DVDV1 VAV2 Figure 8.2 Hohmann transfer from outer planet 1 to inner planet 2. maneuvers (Section 6.7) are clearly not pr actical, especially for manned missions, due to the large periods of the heliocentric orbits. Consider planet 1 and planet 2 in circular orbits around the sun, as shown in Figure 8.3. Since the orbits are circular, we can choose a common horizontal apse linefrom which to measure the true anomaly θ. The true anomalies of planets 1 and 2, respectively, are θ 1=θ10+n1t (8.5) θ2=θ20+n2t (8.6) where n1and n2are the mean motions (angular velocities) of the planets and θ10 andθ20are their true anomalies at time t=0. The phase angle between the position vectors of the two planets is defined as φ=θ2−θ1 (8.7) φis the angular position of planet 2 relative t o planet 1. Substituting Equations 8.5 and 8.6 into 8.7 we get φ=φ0+(n2−n1)t (8.8) φ0is the phase angle at time zero. n2−n1is the orbital angular velocity of planet 2 relative to planet 1. If the orbit of planet 1 lies inside that of planet 2, as in Figure 8.3(a), then n1>n2. Therefore, the relative angular velocity n2−n1is negative, which means 8.3 Rendezvous opportunities 351 Sun12 /H9278 Sun1 2/H9278 (a) (b)θ2 θ1θ2 θ1 Figure 8.3 Planets in circular orbits around the sun. (a) Planet 2 outside the orbit of planet 1. (b) Planet 2 inside the orbit of planet 1. planet 2 moves clockwise relative to planet 1. On the other hand, if planet 1 is outside of planet 2 then n2−n1is positive, so that the relative motion is counterclockwise. The phase angle obviously varies linearly with time according to Equation 8.8. If the phase angle is φ0att=0 ,h o wl o n gw i l li tt a k et ob e c o m e φ0again? The answer: when the position vector of planet 2 rotates through 2 πradians relative to planet 1. The time required for the phase angle to return to its initial value is called thesynodic period, which is denoted T syn. For the case shown in Figure 8.3(a) in which the relative motion is clockwise, Tsynis the time required for φto change from φ0to φ0−2π. From Equation 8.8 we have φ0−2π=φ0+(n2−n1)Tsyn so that Tsyn=2π n1−n2(n1>n2) For the situation illustrated in Figure 8.3(b) ( n2>n1),Tsynis the time required for φto go from φ0toφ0+2π, in which case Equation 8.8 yields Tsyn=2π n2−n1(n2>n1) Both cases are covered by writing Tsyn=2π |n1−n2|(8.9) Recalling Equation 3.6, we can write n1=2π/T1and n2=2π/T2. Thus, in terms of the orbital periods of the two planets, Tsyn=T1T2 |T1−T2|(8.10) Observe that Tsynis the orbital period of planet 2 relative to planet 1. 352 Chapter 8 Interplanetary trajectories Example 8.1Calculate the synodic period of Mars relative to the earth In Table A.1 we find the orbital periods of earth and Mars: Tearth=365.26 days (1 year) TMars=1 year 321 .73 days =687.99 days Hence, Tsyn=TearthTMars |Tearth−TMars|=365.26×687.99 |365.26−687.99|=777.9d a y s These are earth days (1 day = 24 hours). Therefore it takes 2.13 years for a given configuration of Mars relative to the earth to occur again. Figure 8.4 depicts a mission from planet 1 to planet 2. Following a heliocentric Hohmann transfer, the spacecraft intercepts and rendezvous with planet 2. Later itreturns to planet 1 by means of another Hohmann transfer. The major axis of theheliocentric transfer ellipse is the sum of the radii of the two planets’ orbits, R 1+R2. The time t12required for the transfer is one-half the period of the ellipse. Hence, according to Equation 2.73, t12=π√µsun/parenleftbiggR1+R2 2/parenrightbigg3/2 (8.11) During the time it takes the spacecraft to fly from orbit 1 to orbit 2, through an angle of πradians, planet 2 must move around its circular orbit and end up at a point directly opposite planet 1’s position when the spacecraft departed. Since planet 2’s angularvelocity is n 2, the angular distance traveled by the planet during the spacecraft’s trip isn2t12. Hence, as can be seen from Figure 8.4(a), the initial phase angle φ0between the two planets is φ0=π−n2t12 (8.12) (a) (b)Planet 1 at arrival SunPlanet 1 at departure Planet 2 at arrivalPlanet 2 at departure2 1 n2t12 Planet 1 at arrivalPlanet 1 at departureSun Planet 2 at arrivalPlanet 2 at departure 21n2t12 φ0 φfφ/H110320 φ/H11032f Figure 8.4 Round-trip mission, with layover, to planet 2. (a) Departure and rendezvous with planet 2. (b) Return and rendezvous with planet 1. 8.3 Rendezvous opportunities 353 When the spacecraft arrives at planet 2, the phase angle will be φf, which is found using Equations 8.8 and 8.12: φf=φ0+(n2−n1)t12=(π−n2t12)+(n2−n1)t12 φf=π−n1t12 (8.13) For the situation illustrated in Figure 8.4, planet 2 ends up being behind planet 1 by an amount equal to the magnitude of φf. At the start of the return trip, illustrated in Figure 8.4(b), planet 2 must be φ/prime 0 radians ahead of planet 2. Since the spacecraft flies the same Hohmann transfer trajectory back to planet 1, the time of flight is t12, the same as the outbound leg. Therefore, the distance traveled by planet 1 during the return trip is the same as the outbound leg, which means φ/prime 0=−φf (8.14) In any case, the phase angle at the beginning of the return trip must be the negative of the phase angle at arrival from planet 1. The time required for the phase angle toreach its proper value is called the wait time, t wait. Setting time equal to zero at the instant we arrive at planet 2, Equation 8.8 becomes φ=φf+(n2−n1)t φbecomes −φ fafter the time twait. That is −φ f=φf+(n2−n1)twait or twait=−2φ f n2−n1(8.15) where φfis given by Equation 8.13. Equation 8.15 may yield a negative result, which means the desired phase relation occurred in the past. Therefore we must add orsubtract an integral multiple of 2 πto the numerator in order to get a positive value fort wait. Specifically, if N=0, 1, 2, ..., then twait=−2φ f−2πN n2−n1(n1>n2) (8.16) twait=−2φ f+2πN n2−n1(n1<n2) (8.17) where Nis chosen to make twaitpositive. twaitwould probably be the smallest positive number thus obtained. Example 8.2Calculate the minimum wait time for initiating a return trip from Mars to earth. From Tables A.1 and A.2 we have Rearth=149.6×106km RMars=227.9×106km µsun=132.71 ×109km3/s2 354 Chapter 8 Interplanetary trajectories (Example 8.2 continued)According to Equation 8.11, the time of flight from earth to Mars is t12=π√µsun/parenleftbiggRearth+RMars 2/parenrightbigg3/2 =π√ 132.71×109/parenleftbigg149.6×106+227.9×106 2/parenrightbigg3/2 =2.2362×107s or t12=258.82 days From Equation 3.6 and the orbital periods of earth and Mars (see Example 8.1 above) we obtain the mean motions of the earth and Mars. nearth=2π 365.26=0.017202 rad /day nMars=2π 687.99=0.0091327 rad /day The phase angle between earth and Mars when the spacecraft reaches Mars is given by Equation 8.13. φf=π−neartht12=π−0.017202 ·258.82=− 1.3107 (rad) Since nearth>nMars, we choose Equation 8.16 to find the wait time: twait=−2φf−2πN nMars−nearth=−2(−1.3107) −2πN 0.0091327 −0.017202=778.65N−324.85 (days) N=0 yields a negative value, which we cannot accept. Setting N=1, we get twait=453.8d a y s This is the minimum wait time. Obviously, we could set N=2, 3,...to obtain longer wait times. In order for a spacecraft to depart on a mission to Mars by means of a Hohmann (minimum energy) transfer, the phase angle between earth and Mars must be that given by Equation 8.12. Using the results of Example 8.2, we find it to be φ0=π−nMarst12=π−0.0091327 ·258.82=0.7778 rad =44.57◦ This opportunity occurs once every synodic period, which we found to be 2.13 years in Example 8.1. In Example 8.2 we found that the time to fly to Mars is 258.8 days,followed by a wait time of 453.8 days, followed by a return trip time of 258.8 days.Hence, the minimum total time for a manned Mars mission is t total=258.8+453.8+253.8=971.4d a y s =2.66 years 8.4 Sphere of influence The sun, of course, is the dominant celestial body in the solar system. It is over 1000 times more massive than the largest planet, Jupiter, and has a mass of over 300 000 8.4 Sphere of influence 355 2 4 6 80.20.40.60.8 r/r0Fg/Fg0 101.0 Figure 8.5 Decrease of gravitational force with distance from a planet’s surface. earths. The sun’s gravitational pull holds all of the planets in its grasp according to Newton’s law of gravity, Equation 2.6. However, near a given planet the influence of itsown gravity exceeds that of the sun. For example, at its surface the earth’s gravitationalforce is over 1600 times greater than the sun’s. The inverse-square nature of the lawof gravity means that the force of gravity F gdrops off rapidly with distance rfrom the center of attraction. If Fg0is the gravitational force at the surface of a planet with radius r0, then Figure 8.5 shows how rapidly the force diminishes with distance. At ten body radii, the force is 1 percent of its value at the surface. Eventually, the forceof the sun’s gravitational field overwhelms that of the planet. In order to estimate the radius of a planet’s gravitational sphere of influence, consider the three-body system comprising a planet pof mass m p, the sun sof mass msand a space vehicle vof mass mvillustrated in Figure 8.6. The position vectors of the planet and spacecraft relative to an inertial frame centered at the sun are Rand Rv, respectively. The position vector of the space vehicle relative to the planet is r. (Throughout this chapter we will use upper case letters to represent position, velocityand acceleration measured relative to the sun and lower case letters when they are measured relative to a planet.) The gravitational force exerted on the vehicle by the planet is denoted F(v) p, and that exerted by the sun is F(v) s. Likewise, the forces on the planet are F(p) sand F(p) v, whereas on the sun we have F(s) vand F(s) p. According to Newton’s law of gravitation (Equation 2.6), these forces are F(v) p=−Gmvmp r3r (8.18a) F(v) s=−Gmvms R3vRv (8.18b) F(p) s=−Gm pms R3R (8.18c) 356 Chapter 8 Interplanetary trajectories /H9258 msmv mp RRυ r Fυ(s) Fp(s)Fs(υ) Fp(υ) Fs(p)Fυ(p) Figure 8.6 Relative position and gravitational force vectors among the three bodies. Observe that Rv=R+r (8.19) From Figure 8.6 and the law of cosines we see that the magnitude of Rvis Rv=(R2+r2−2Rrcosθ)1 2=R/bracketleftbigg 1−2r Rcosθ+/parenleftBigr R/parenrightBig2/bracketrightbigg1 2 (8.20) We expect that within the planet’s sphere of influence, r/R/lessmuch1. In that case, the terms involving r/Rin Equation 8.20 can be neglected, so that, approximately, Rv=R (8.21) The equation of motion of the spacecraft relative to the sun-centered inertial frame is mv¨Rv=F(v) s+F(v) p Solving for ¨Rvand substituting the gravitational forces given by Equations 8.18a and 8.18b, we get ¨Rv=1 mv/parenleftbigg −Gmvms R3vRv/parenrightbigg +1 mv/parenleftbigg −Gmvmp r3r/parenrightbigg =−Gm s R3vRv−Gm p r3r (8.22) Let us write this as ¨Rv=As+Pp (8.23) where As=−Gm s R3vRv Pp=−Gm p r3r (8.24) 8.4 Sphere of influence 357 Asis the primary gravitational acceleration of the vehicle due to the sun, whereas Ppis the secondary or perturbing acceleration due to the planet. The magnitudes of Asand Ppare As=Gm s R2Pp=Gm p r2(8.25) where we made use of the approximation given by Equation 8.21. The ratio of the perturbing acceleration to the primary acceleration is, therefore, Pp As=Gm p r2 Gm s R2=mp ms/parenleftbiggR r/parenrightbigg2 (8.26) The equation of motion of the planet relative to the inertial frame is mp¨R=F(p) v+F(p) s Solving for ¨R, noting that F(p) v=− F(v) p, and using Equations 8.18b and 8.18c, yields ¨R=1 mp/parenleftbiggGmvmp r3r/parenrightbigg +1 mp/parenleftbigg −Gm pms R3R/parenrightbigg =Gmv r3r−Gm s R3R (8.27) Subtracting Equation 8.27 from 8.22 and collecting terms, we find ¨Rv−¨R=−Gm p r3r/parenleftbigg 1+mv mp/parenrightbigg −Gm s R3v/bracketleftBigg Rv−/parenleftbiggRv R/parenrightbigg3 R/bracketrightBigg Recalling Equation 8.19, we can write this as ¨r=−Gm p r3r/parenleftbigg 1+mv mp/parenrightbigg −Gm s R3v/braceleftBigg r+/bracketleftBigg 1−/parenleftbiggRv R/parenrightbigg3/bracketrightBigg R/bracerightBigg (8.28) This is the equation of motion of the vehic le relative to the planet. By using Equation 8.21 and the fact that mv/lessmuchmp, we can write this in approximate form as ¨r=ap+ps (8.29) where ap=−Gm p r3rp s=−Gm s R3r (8.30) In this case apis the primary gravitational acceleration of the vehicle due to the planet, and psis the perturbation caused by the sun. The magnitudes of these vectors are ap=Gm p r2ps=Gm s R3r (8.31) The ratio of the perturbing acceleration to the primary acceleration is ps ap=Gm sr R3 Gm p r2=ms mp/parenleftBigr R/parenrightBig3 (8.32) 358 Chapter 8 Interplanetary trajectories For motion relative to the planet, the ratio ps/apis a measure of the deviation of the vehicle’s orbit from the Keplerian orbit arising from the planet acting by itself(p s/ap=0). Likewise, Pp/Asis a measure of the planet’s influence on the orbit of the vehicle relative to the sun. If ps ap<Pp As(8.33) then the perturbing effect of the sun on the vehicle’s orbit around the planet is less than the perturbing effect of the planet on the vehicle’s orbit around the sun. Wesay that the vehicle is therefore within the planet’s sphere of influence. SubstitutingEquations 8.26 and 8.32 into 8.33 yields m s mp/parenleftBigr R/parenrightBig3 <mp ms/parenleftbiggR r/parenrightbigg2 which means /parenleftBigr R/parenrightBig5 </parenleftbiggmp ms/parenrightbigg2 or r R</parenleftbiggmp ms/parenrightbigg2 5 LetrSOIbe the radius of the sphere of influence. Within the planet’s sphere of influence, defined by rSOI R=/parenleftbiggmp ms/parenrightbigg2 5 (8.34) the motion of the spacecraft is determined b y its equations of motion relative to the planet (Equation 8.28). Outside of the sphere of influence, the path of the spacecraftis computed relative to the sun (Equation 8.22). The sphere of influence radius presented in Equation 8.34 is not an exact quantity. It is simply a reasonable estimate of the distance beyond which the sun’s gravitationalattraction dominates that of a planet. The spheres of influence of all of the planets and the earth’s moon are listed in Table A.2. Example 8.3Calculate the radius of the earth’s sphere of influence. In Table A.1 we find mearth=5.974×1024kg msun=1.989×1030kg Rearth=149.6×106km Substituting this data into Equation 8.34 yields rSOI=149.6×106/parenleftbigg5.974×1024 1.989×1024/parenrightbigg2 5 =925×106km 8.5 Method of patched conics 359 Since the radius of the earth is 6378 km, rSOI=145 earth radii Relative to the earth, its sphere of influence is very large. However, relative to the sun it is tiny, as illustrated in Figure 8.7. Sun Radius = 109 earth radiiEarth's SOI Radius = 145 earth radii 23 460 earth radii Figure 8.7 The earth’s sphere of influence and the sun, drawn to scale. 8.5 Method of patched conics ‘Conics’ refers to the fact that two-body or Keplerian orbits are conic sections with the focus at the attracting body. T o study an interplanetary trajectory we assumethat when the spacecraft is outside the sphere of influence of a planet it follows anunperturbed Keplerian orbit around the sun. Because interplanetary distances are sovast, for heliocentric orbits we may neglect the size of the spheres of influence andconsider them, like the planets they surround, to be just points in space coincidingwith the planetary centers. Within each planetary sphere of influence, the spacecrafttravels an unperturbed Keplerian path about the planet. While the sphere of influenceappears as a mere speck on the scale of the solar system, from the point of view of theplanet it is very large indeed and may be considered to lie at infinity. T o analyze a mission from planet 1 to planet 2 using the method of patched conics, we first determine the heliocentric trajectory – such as the Hohmann transfer ellipsediscussed in Section 8.2 – that will intersect the desired positions of the two planets intheir orbits. This trajectory takes the spacecraft from the sphere of influence of planet1 to that of planet 2. At the spheres of influence, the heliocentric velocities of thetransfer orbit are computed relative to the planet to establish the velocities ‘at infinity’ which are then used to determine planetocentric departure trajectory at planet 1 andarrival trajectory at planet 2. In this way we ‘patch’ together the three conics, onecentered at the sun and the other two centered at the planets in question. Whereas the method of patched conics is remarkably accurate for interplanetary trajectories, such is not the case for lunar rendezvous and return trajectories. The orbitof the moon is determined primarily by the earth, whose sphere of influence extendswell beyond the moon’s 384 400 km orbital radius. T o apply patched conics to lunartrajectories we ignore the sun and consider the motion of a spacecraft as influenced byjust the earth and moon, as in the restricted three-body problem discussed in Section2.12. The size of the moon’s sphere of influence is found using Equation 8.34, withthe earth playing the role of the sun: r SOI=R/parenleftbiggmmoon mearth/parenrightbigg2 5 360 Chapter 8 Interplanetary trajectories where Ris the radius of the moon’s orbit. Thus, using Table A.1, rSOI=384 400/parenleftbigg73.48×1021 5974×1021/parenrightbigg2 5 =66 200 km as recorded in Table A.2. The moon’s sphere of influence extends out to over one-sixth of the distance to the earth. We can hardly consider it to be a mere speck relative tothe earth. Another complication is the fact that the earth and the moon are somewhatcomparable in mass, so that their center of mass lies almost three-quarters of an earthradius from the center of the earth. The motion of the moon cannot be accuratelydescribed as rotating around the center of the earth. Complications such as these place the analysis of cislunar trajectories beyond our scope. Extensions of the patched conic technique to such orbits may be found in Bate,Mueller and White (1971), Kaplan (1976) and Battin (1999). 8.6 Planetary departure In order to escape the gravitational pull of a planet, the spacecraft must travel ahyperbolic trajectory relative to the planet, arriving at its sphere of influence witha relative velocity v ∞(hyperbolic excess velocity) greater than zero. On a parabolic trajectory, according to Equation 2.80, the spacecraft will arrive at the sphere ofinfluence ( r=∞ ) with a relative speed of zero. In that case the spacecraft remains in the same orbit as the planet and does not embar k upon a heliocentric elliptical path. Figure 8.8 shows a spacecraft departing on a Hohmann trajectory from planet 1 towards a target planet 2 which is farther away from the sun (as in Figure 8.1). At the sphere of influence crossing, the heliocentric velocity V(v) Dof the spacecraft is parallel to the asymptote of the departure hyperbola as well as to the planet’s heliocentric velocity vector V1.V(v) Dand V1must be parallel and in the same direction for a Hohmann transfer such that /Delta1VDin Equation 8.3 is positive. Clearly, /Delta1VDis the hyperbolic excess speed of the departure hyperbola, v∞=/radicalbiggµsun R1/parenleftBigg/radicalBigg 2R2 R1+R2−1/parenrightBigg (8.35) It would be well at this point for the reader to review Section 2.9 on hyperbolic trajectories and compare Figures 8.8 and 2.23. Recall that point C is the center of thehyperbola. A space vehicle is ordinarily launched into an interplanetary trajectory from a circular parking orbit. The radius of this parking orbit equals the periapse radiusr pof the departure hyperbola. According to Equation 2.40, the periapse radius is given by rp=h2 µ11 1+e(8.36) where his the angular momentum of the departure hyperbola (relative to the planet), eis the eccentricity of the hyperbola and µ1is the planet’s gravitational parameter. 8.6 Planetary departure 361 To the sun Planet 1's orbital trackSpacecraft parking orbit CSphere of influencev/H11009 V1 P Periapse of the departure hyperbolaApse line of thedeparture hyperbolaAsymptote/H9004 b Figure 8.8 Departure of a spacecraft on a mission from an inner planet to an outer planet. The hyperbolic excess speed is found in Equation 2.105, from which we obtain h=µ1√ e2−1 v∞(8.37) Substituting this expression for the angular momentum into Equation 8.36 and solving for the eccentricity yields e=1+rpv2 ∞ µ1(8.38) We place this result back into Equation 8.37 to obtain the following expression for the angular momentum: h=rp/radicalBigg v2∞+2µ1 rp(8.39) Since the hyperbolic excess speed is specified by the mission requirements (Equa- tion 8.35), choosing a departure periapse rpyields the parameters eand hof the 362 Chapter 8 Interplanetary trajectories v/H11009 Circle of injection pointsSurface of revolution of departure hyperbolas A A∆ Figure 8.9 Locus of possible departure trajectories for a given v∞and rp. departure hyperbola. From the angular momentum we get the periapse speed, vp=h rp=/radicalBigg v2∞+2µ1 rp(8.40) which can also be found from an energy approach using Equation 2.103. With Equation 8.40 and the speed of the circular parking orbit (Equation 2.53), vc=/radicalbiggµ1 rp(8.41) we can calculate the delta-v required to put the vehicle onto the hyperbolic departure trajectory, /Delta1v=vp−vc=vc /radicalBigg 2+/parenleftbiggv∞ vc/parenrightbigg2 −1  (8.42) The location of periapse, where the delta-v maneuver must occur, is found using Equations 2.89 and 8.38, β=cos−1/parenleftbigg1 e/parenrightbigg =cos−1 1 1+rpv2 ∞ µ1 (8.43) βgives the orientation of the apse line of the hyperbola to the planet’s heliocentric velocity vector. It should be pointed out that the only requirement on the orientation of the plane of the departure hyperbola is that it contains the center of mass of the planet as wellas the relative velocity vector v ∞. Therefore, as shown in Figure 8.11, the hyperbola can be rotated about a line A–Awhich passes through the planet’s center of mass and 8.6 Planetary departure 363 N Sv/H11009 G AA EquatorLatitude of launch site ab1 1'2 Departure trajectoriesimax imin Figure 8.10 Parking orbits and departure trajectories for a launch site at a given latitude. is parallel to v∞(or V1, which of course is parallel to v∞for Hohmann transfers). Rotating the hyperbola in this way sweeps out a s urface of revolution on which lie all possible departure hyperbolas. The periapse of the hyperbola traces out a circle which,for the specified periapse radius r p, is the locus of all possible points of injection into a departure trajectory towards the target planet. This circle is the base of a cone withvertex at the center of the planet. From Figure 3.23 we can determine that its radiusisr psinβ,w h e r e βis given just above in Equation 8.43. The plane of the parking orbit, or direct ascent trajectory, must contain the line A–A and the launch site at the time of launch. The possible inclinations of a prograde orbit range from a minimum of imin,w h e r e iminis the latitude of the launch site, to imax, which cannot exceed 90◦. Launch site safety considerations may place additional limits on that range. For example, orbits originating from the Kennedy Space Center in Florida, USA, (latitude 28.5◦) are limited to inclinations between 28.5◦and 52.5◦. For the scenario illustrated in Figure 8.12 the location of the launch site limits access to just the departure trajectories having periapses lying between a and b. The figure shows that there are two times per day – when the planet rotates the launch site through positions 1 and 1/prime– that a spacecraft can be launched into a parking orbit. These times are closer together (the launch window is smaller) thelower the inclination of the parking orbit. Once a spacecraft is established in its parking orbit, then an opportunity for launch into the departure trajectory occurs each orbital circuit. If the mission is to send a spacecraft from an outer planet to an inner planet, as in Figure 8.2, then the spacecraft’s heliocentric speed V(v) Dat departure must be less than that of the planet. That means the spacecraft must emerge from the back side of the sphere of influence with its relative velocity vector v∞directed opposite to V1,a s shown in Figure 8.11. Figures 8.9 and 8.10 apply to this situation as well. 364 Chapter 8 Interplanetary trajectories Planet 1's orbital trackSpacecraft parking orbitCSphere of influenceV1 PPeriapse of the departure hyperbola /H9252 Asymptote v/H11009∆ Figure 8.11 Departure of a spacecraft on a trajectory from an outer planet to an inner planet. Example 8.4A spacecraft is launched on a mission to Mars starting from a 300 km circular parking orbit. Calculate (a) the delta-v required; (b) the location of perigee of the departure hyperbola; (c) the amount of propellant required as a percentage of the spacecraftmass before the delta-v burn, assuming a specific impulse of 300 seconds. From Tables A.1 and A.2 we obtain the gravitational parameters for the sun and theearth, µ sun=1.327×1011km3/s2 µearth=398 600 km3/s2 and the orbital radii of the earth and Mars, Rearth=149.6×106km RMars=227.9×106km 8.6 Planetary departure 365 (a) According to Equation 8.35, the hyperbolic excess speed is v∞=/radicalbiggµsun Rearth/parenleftBigg/radicalBigg 2RMars Rearth+RMars−1/parenrightBigg =/radicalBigg 1.327×1011 149.6×106 /radicalBigg 2(227.9 ×106) 149.6×106+227.9×106−1  from which v∞=2.943 km /s The speed of the spacecraft in its 300 km circular parking orbit is given by Equation 8.41, vc=/radicalbiggµearth rearth+300=/radicalbigg 398 600 6678=7.726 km /s Finally, we use Equation 8.42 to calculate the delta-v required to step up to the departure hyperbola: /Delta1v=vp−vc=vc /radicalBigg 2+/parenleftbiggv∞ vc/parenrightbigg2 −1 =7.726 /radicalBigg 2+/parenleftbigg2.943 7.726/parenrightbigg2 −1  /Delta1v=3.590 km /s (b) Perigee of the departure hyperbola, relative to the earth’s orbital velocity vector, is found using Equation 8.43, β=cos−1 1 1+rpv2 ∞ µearth =cos−1 1 1+6678·2.9432 368 600  β=29.16◦ Figure 8.12 shows that the perigee can be located on either the sunlit or dark side of the earth. It is likely that the parking orbit would be a prograde orbit (west toeast), which would place the burnout point on the dark side. (c) From Equation 6.1 we have /Delta1m m=1−e−/Delta1v Ispgo Substituting /Delta1v=3.590 km /s,Isp=300 s and go=9.81×10−3km/s2, this yields /Delta1m m=0.705 That is, prior to the delta-v maneuver, over 70 percent of the spacecraft mass must be propellant. 366 Chapter 8 Interplanetary trajectories (Example 8.4 continued) 29.2/H11034Sun 29.2/H11034SunVearth Vearth Parking orbitDeparture hyperbola PerigeeParking orbit Perigee (a) (b) Figure 8.12 Departure trajectory to Mars initiated from (a) the dark side and (b) the sunlit side of the earth. 8.7 Sensitivity analysis The initial maneuvers required to place a spacecraft on an interplanetary trajectory occur well within the sphere of influence of the departure planet. Since the sphereof influence is just a point on the scale of the solar system, one may ask what effectssmall errors in position and velocity at the maneuver point have on the trajectory. Assuming the mission is from an inner to an outer planet, let us consider the effect which small changes in the burnout velocity v pand radius rphave on the target radius R2of the heliocentric Hohmann transfer ellipse (see Figures 8.1 and 8.8). R2is the radius of aphelion, so we use Equation 2.60 to obtain R2=h2 µsun1 1−e Substituting h=R1V(v) Dand e=(R2−R1)/(R2+R1), and solving for R2, yields R2=R2 1[V(v) D]2 2µsun−R1[V(v) D]2(8.44) (This expression holds as well for a mission from an outer to inner planet.) The change δR2inR2due to a small variation δV(v) DofV(v) Dis δR2=dR2 dV(v) DδV(v) D=4R2 1µsun/braceleftBig 2µsun−R1[V(v) D]2/bracerightBig2V(v) DδV(v) D Dividing this equation by Equation 8.44 leads to δR2 R2=2 1−R1[V(v) D]2 2µsunδV(v) D V(v) D(8.45) 8.7 Sensitivity analysis 367 The departure speed V(v) Dof the space vehicle is the sum of the planet’s speed V1and excess speed v∞: V(v) D=V1+v∞ We can solve Equation 8.40 for v∞, v∞=/radicalBigg v2 p−2µ1 rp Hence V(v) D=V1+/radicalBigg v2 p−2µ1 rp(8.46) The change in V(v) Ddue to variations δrpandδvpof the burnout position (periapse) rpand speed vpis given by δV(v) D=∂V(v) D ∂rpδrp+∂V(v) D ∂vpδvp (8.47) From Equation 8.46 we obtain ∂V(v) D ∂rp=µ1 v∞r2 p∂V(v) D ∂vp=vp v∞ Therefore δV(v) D=µ1 v∞r2 pδrp+vp v∞δvp Once again making use of Equation 8.40, this can be written as follows δV(v) D V(v) D=µ1 V(v) Dv∞rpδrp rp+v∞+2µ1 rp V(v) Dδvp vp(8.48) Substituting this into Equation 8.45 finally yields the desired result, an expression for the variation of R2due to variations in rpandvp: δR2 R2=2 1−R1[V(v) D]2 2µsun µ1 V(v) Dv∞rpδrp rp+v∞+2µ1 rp V(v) Dδvp vp (8.49) Consider a mission from earth to Mars, starting from a 300 km parking orbit. We have µsun=1.327×1011km3/s2 µpl1=µearth=398 600 km3/s2 R1=149.6×106km R2=227.9×106km rp=6678 km 368 Chapter 8 Interplanetary trajectories In addition, from Equations 8.1 and 8.2, V1=Vearth=/radicalbiggµsun R1=/radicalBigg 1.327×1011 149.6×106=29.78 km/s V(v) D=/radicalbig 2µsun/radicalBigg R2 R1(R1+R2) =/radicalbig 2·1.327×1011/radicalBigg 227.9×106 149.6×106(149.6×106+227.9×106)=32.73 km/s Therefore v∞=V(v) D−Vearth=2.943 km /s and, from Equation 8.40, vp=/radicalBigg v2∞+2µearth rp=/radicalbigg 2.9432+2·398 600 6678=11.32 km/s Substituting these values into Equation 8.49 yields δR2 R2=3.127δrp rp+6.708δvp vp This expression shows that a 0.01 percent variation (1.1 m/s) in the burnout speed vpchanges the target radius R2by 0.067 percent or 153 000 km! Likewise, an error of 0.01 percent (0.67 km) in burnout radius rpproduces an error of over 70 000 km. Thus small errors which are likely to occur in the launch phase of the mission mustbe corrected by midcourse maneuvers during the coasting flight along the ellipticaltransfer trajectory. 8.8 Planetary rendezvous A spacecraft arrives at the sphere of influence of the target planet with a hyperbolic excess velocity v∞relative to the planet. In the case illustrated in Figure 8.1, a mission from an inner planet 1 to an outer planet 2 (e.g., earth to Mars), the spacecraft’s heliocentric approach velocity V(v) Ais smaller in magnitude than that of the planet, V2. Therefore, it crosses the forward portion of the sphere of influence, as shown in Figure 8.13. For a Hohmann transfer, V(v) Aand V2are parallel, so the magnitude of the hyperbolic excess velocity is, simply, v∞=V2−V(v) A(8.50) If the mission is as illustrated in Figure 8.2, from an outer planet to an inner one (e.g., earth to Venus), then V(v) Ais greater than V2, and the spacecraft must cross the rear portion of the sphere of influence, as shown in Figure 8.14. In that case v∞=V(v) A−V2 (8.51) 8.8 Planetary rendezvous 369 To the sunPlanet 2's orbital track Capture orbitFlyby trajectoryCSphere of influence v/H11009/H9254 V2P/H9252 Apse line of the arrival hyperbolaAsymptoteAsymptote ∆ Figure 8.13 Spacecraft approach trajectory for a Hohmann transfer to an outer planet from an inner one. Pis the periapse of the approach hyperbola. What happens after crossing the sphere of influence depends on the nature of the mission. If the goal is to impact the planet (or its atmosphere), the aiming radius /Delta1of the approach hyperbola must be such that hyperbola’s periapse rpequals essentially the radius of the planet. If the intent is to go into orbit around the planet, then /Delta1 must be chosen so that the delta-v burn at periapse will occur at the correct altitudeabove the planet. If there is no impact with the planet and no drop into a capture orbitaround the planet, then the spacecraft will simply continue past periapse on a flybytrajectory, exiting the sphere of influence with the same relative speed v ∞it entered, but with the velocity vector rotated through the turn angle δ, given by Equation 2.90, δ=2 sin−1/parenleftbigg1 e/parenrightbigg (8.52) With the hyperbolic excess speed v∞and the periapse radius rpspecified, the eccentricity of the approach hyperbola is found from Equation 8.38, e=1+rpv2 ∞ µ2(8.53) where µ2is the gravitational parameter of planet 2. Hence, the turn angle is δ=2 sin−1 1 1+rpv2 ∞ µ2 (8.54) 370 Chapter 8 Interplanetary trajectories To the sunCapture orbit CSphere of influence Flyby trajectory Planet 2's orbital trackv∞ δ V2P βApse line of the arrival hyperbolaAsymptote Asymptote Asymptote∆ Figure 8.14 Spacecraft approach trajectory for a Hohmann transfer to an inner planet from an outer one. Pis the periapse of the approach hyperbola. We can combine Equations 2.93 and 2.97 to obtain the following expression for the aiming radius, /Delta1=h2 µ21√ e2−1(8.55) The angular momentum of the approach hyperbola relative to the planet is found using Equation 8.39, h=rp/radicalBigg v2∞+2µ2 rp(8.56) Substituting Equations 8.53 and 8.56 into 8.55 yields the aiming radius in terms of the periapse radius and the hyperbolic excess speed, /Delta1=rp/radicalBigg 1+2µ2 rpv2∞(8.57) Just as we observed when discussing departure trajectories, the approach hyperbola does not lie in a unique plane. We can rotate the hyperbolas illustrated in Figures 8.11and 8.12 about a line A–Aparallel to v ∞and passing through the target planet’s center 8.8 Planetary rendezvous 371 v∞ Locus of periapsesυ∞ Target circle A A∆ Figure 8.15 Locus of approach hyperbolas to the target planet. Periapses of approach hyperbolas v∞ Target planet υ∞SOI ∆i∆ Figure 8.16 Family of approach hyperbolas having the same v∞but different /Delta1. of mass, as shown in Figure 8.15. The approach hyperbolas in that figure terminate at the circle of periapses. Figure 8.16 is a plane through the solid of revolution revealingthe shape of hyperbolas having a common v ∞but varying /Delta1. Let us suppose that the purpose of the mission is to enter an elliptical orbit of eccentricity earound the planet. This will require a delta-v maneuver at periapse P(Figures 8.13 and 8.14), which is also periapse of the ellipse. The speed in the hyperbolic trajectory at periapse is given by Equation 8.40 vp)hyp=/radicalBigg v2∞+2µ2 rp(8.58) 372 Chapter 8 Interplanetary trajectories The velocity at periapse of the capture orbit is found by setting h=rpvpin Equation 2.40 and solving for vp: vp)capture=/radicalBigg µ2(1+e) rp(8.59) Hence, the required delta-v is /Delta1v=vp)hyp−vp)capture=/radicalBigg v2∞+2µ2 rp−/radicalBigg µ2(1+e) rp(8.60) F o rag i v e n v∞,/Delta1vclearly depends upon the choice of periapse radius rpand capture orbit eccentricity e. Requiring the maneuver point to remain the periapse of the capture orbit means that /Delta1vis maximum for a circular capture orbit and decreases with increasing eccentricity until /Delta1v=0, which, of course, means no capture (flyby). In order to determine optimal capture radius, let us write Equation 8.60 in non- dimensional form as /Delta1v v∞=/radicalBigg 1+2 ξ−/radicalBigg 1+e ξ(8.61) where ξ=rpv2 ∞ µ2(8.62) The first and second derivatives of /Delta1v/v ∞with respect to ξare d dξ/Delta1v v∞=/parenleftbigg −1√ξ+2+√1+e 2/parenrightbigg1 ξ3 2(8.63) d2 dξ2/Delta1v v∞=/bracketleftBigg 2ξ+3 (ξ+2)3 2−3 4√ 1+e/bracketrightBigg 1 ξ5 2(8.64) Setting the first derivative equal to zero and solving for ξyields ξ=21−e 1+e(8.65) Substituting this value of ξinto Equation 8.64, we get d2 dξ2/Delta1v v∞=√ 2 64(1+e)3 (1−e)3 2(8.66) This expression is positive for elliptical orbits (0 ≤e<1), which means that when ξis given by Equation 8.65 /Delta1vis a minimum. Therefore, from Equation 8.62, the optimal periapse radius as far as fuel expenditure is concerned is rp=2µ2 v2∞1−e 1+e(8.67) 8.8 Planetary rendezvous 373 We can combine Equations 2.40 and 2.60 to get 1−e 1+e=rp ra(8.68) where rais the apoapse radius. Thus, Equation 8.67 implies ra=2µ2 v2∞(8.69) That is, the apoapse of this capture ellipse is independent of the eccentricity and equals the radius of the optimal circular orbit. Substituting Equation 8.65 back into Equation 8.61 yields the minimum /Delta1v /Delta1v=v∞/radicalbigg 1−e 2(8.70) Finally, placing the optimal rpinto Equation 8.57 leads to an expression for the aiming radius required for minimum /Delta1v, /Delta1=2√ 2√1−e 1+eµ2 v2∞=/radicalbigg 2 1−erp (8.71) Clearly, the optimal /Delta1v(and periapse height) are reduced for highly eccentric elliptical capture orbits ( e→1). However, it should be pointed out that the use of optimal /Delta1v may have to be sacrificed in favor of a variety of other mission requirements. Example 8.5After a Hohmann transfer from earth, calculate the minimum delta-v required to place a spacecraft in Mars orbit with a period of seven hours. Also calculate the periapse radius, the aiming radius and the angle between periapse and Mars’ velocityvector. The following data is required from Tables A.1 and A.2: µsun=1.327×1011km3/s2 µMars=42 830 km3/s2 Rearth=149.6×106km RMars=227.9×106km rMars=3396 km The hyperbolic excess speed is found using Equation 8.4, v∞=/Delta1V A=/radicalbiggµsun RMars/parenleftBigg 1−/radicalBigg 2Rearth Rearth+RMars/parenrightBigg =/radicalBigg 1.327×1011 227.9×106 1−/radicalBigg 2·149.6×106 149.6×106+227.9×106  v∞=2.648 km /s 374 Chapter 8 Interplanetary trajectories (Example 8.5 continued)We can use Equation 2.73 to express the semimajor axis aof the capture orbit in terms of its period T, a=/parenleftbiggT√µMars 2π/parenrightbigg2 3 Substituting T=7·3600 s yields a=/parenleftBigg 25 200√ 42 830 2π/parenrightBigg2 3 =8832 km From Equation 2.63 we obtain a=rp 1−e Upon substituting the optimal periapse radius, Equation 8.67, this becomes a=2µMars v2∞1 1+e from which e=2µMars av2∞−1=2·42 830 8832·2.6482−1=0.3833 Thus, using Equation 8.70, we find /Delta1v=v∞/radicalbigg 1−e 2=2.648/radicalbigg 1−0.3833 2=1.470 km /s From Equations 8.66 and 8.71 we obtain the periapse radius rp=2µMars v2∞1−e 1+e=2·42 830 2.64821−0.3833 1+0.3833=5447 km and the aiming radius /Delta1=rp/radicalbigg 2 1−e=5447/radicalbigg 2 1−0.3833=9809 km Finally, using Equation 8.43, we get the angle to periapse β=cos−1 1 1+rpv2 ∞ µMars =cos−1 1 1+5447·2.6482 42 830 =58.09◦ 8.9 Planetary flyby 375 Mars, the approach hyperbola, and the capture orbit are shown to scale in Figure 8.17. The approach could also be made from the dark side of the planet instead of the sunlit side. The approach hyperbola and capture ellipse would be the mirror image of thatshown, as is the case in Figure 8.12. PC58.1/H11034 To the sun 9809 km VMars = 24.13 km/s /H9271∞ = 2.648 km/s12 217 km5447 km Figure 8.17 An optimal approach to a Mars capture orbit with a seven hour period. rMars=3396 km . 8.9 Planetary flyby A spacecraft which enters a planet’s sphere of influence and does not impact the planet or go into orbit around it will continue in its hyperbolic trajectory through periapsePand exit the sphere of influence. Figure 8.18 shows a hyperbolic flyby trajectory along with the asymptotes and apse line of the hyperbola. It is a leading-side flybybecause the periapse is on the side of the planet facing into the direction of motion.Likewise, Figure 8.19 illustrates a trailing-side flyby. At the inbound crossing point, the heliocentric velocity V (v) 1of the spacecraft equals the planet’s heliocentric velocity Vplus the hyperbolic excess velocity v∞)1of the spacecraft (relative to the planet), V(v) 1=V+v∞1 (8.72) Similarly, at the outbound crossing we have V(v) 2=V+v∞2 (8.73) The change /Delta1V(v)in the spacecraft’s heliocentric velocity is /Delta1V(v)=V(v) 2−V(v) 1=(V+v∞2)−(V+v∞1) 376 Chapter 8 Interplanetary trajectories v∞2 δTo the sun PC δβ β V(υ)V(υ) V VVûS v∞2v∞1 v∞1Apse line α1φ1α2φ2 2 1∆V(υ)uVˆ Figure 8.18 Leading-side planetary flyby. which means /Delta1V(v)=v∞2−v∞1=/Delta1v∞ (8.74) The excess velocities v∞1and v∞2lie along the asymptotes of the hyperbola and are therefore inclined at the same angle βto the apse line (see Figure 2.23), with v∞1 pointing towards and v∞2pointing away from the center C. They both have the same magnitude v∞, with v∞2having rotated relative to v∞1by the turn angle δ.H e n c e , /Delta1v∞– and therefore /Delta1V(v)– is a vector which lies along the apse line and always points away from periapse, as illustrated in Figures 8.18 and 8.19. From those figuresit can be seen that, in a leading-side flyby, the component of /Delta1V (v)in the direction of the planet’s velocity is negative, whereas for the trailing-side flyby it is positive. Thismeans that a leading-side flyby results in a decrease in the spacecraft’s heliocentricspeed. On the other hand, a trailing-side flyby increases that speed. In order to analyze a flyby problem, we proceed as follows. First, let ˆu Vbe the unit vector in the direction of the planet’s heliocentric velocity Vand let ˆuSbe the unit vector pointing from the planet to the sun. At the inbound crossing of the sphere of influence, the heliocentric velocity V(v) 1of the spacecraft is V(v) 1=[V(v) 1]VˆuV+[V(v) 1]SˆuS (8.75) 8.9 Planetary flyby 377 α1δTo the sun PC δβ β V1(υ)V2(υ) VVV uVuS v∞1 v∞1v∞2 v∞2Apse line φ1α2 φ2 ∆V(υ)ˆˆ Figure 8.19 Trailing-side planetary flyby. where the scalar components of V(v) 1are [V(v) 1]V=V(v) 1cosα1 [V(v) 1]S=V(v) 1sinα1 (8.76) α1is the angle between V(v) 1and V. All angles are measured positive counterclockwise. Referring to Figure 2.11, we see that the magnitude of α1is the flight path angle γ of the spacecraft’s heliocentric trajectory when it encounters the planet’s sphere ofinfluence (a mere speck) at the planet’s distance Rfrom the sun. Furthermore, [V(v) 1]V=V⊥1 [V(v) 1]S=− Vr1 (8.77) V⊥1and Vr1are furnished by Equations 2.38 and 2.39 V⊥1=µsun h11 1+e1cosθ1Vr1=µsun h1e1sinθ1 (8.78) in which e1,h1andθ1are the eccentricity, angular momentum and true anomaly of the heliocentric approach trajectory. The velocity of the planet relative to the sun is V=VˆuV (8.79) 378 Chapter 8 Interplanetary trajectories where V=√µsun/R. At the inbound crossing of the planet’s sphere of influence, the hyperbolic excess velocity of the spacecraft is obtained from Equation 8.72 v∞1=V(v) 1−V Using this we find v∞1=(v∞1)VˆuV+(v∞1)SˆuS (8.80) where the scalar components of v∞1are (v∞1)V=V(v) 1cosα1−V (v∞1)S=V(v) 1sinα1 (8.81) v∞is the magnitude of v∞1, v∞=√v∞1·v∞1=/radicalbigg/bracketleftBig V(v) 1/bracketrightBig2 +V2−2V(v) 1Vcosα1 (8.82) At this point v∞is known, so that upon specifying the periapse radius rpwe can compute the angular momentum and eccentricity of the flyby hyperbola (relative tothe planet), using Equations 8.38 and 8.39: h=r p/radicalBigg v2∞+2µ rpe=1+rpv2 ∞ µ(8.83) where µis the gravitational parameter of the planet. The angle between v∞1and the planet’s heliocentric velocity is φ1. It is found using the components of v∞1in Equation 8.81, φ1=tan−1(v∞1)S (v∞1)V=tan−1 V(v) 1sinα1 V(v) 1cosα1−V(8.84) At the outbound crossing the angle between v∞2and Visφ2,w h e r e φ2=φ1+δ (8.85) For the leading-side flyby in Figure 8.18, the turn angle is δpositive (counterclockwise) whereas in Figure 8.19 it is negative. Since the magnitude of v∞2isv∞, we can express v∞2in components as v∞2=v∞cosφ2ˆuV+v∞sinφ2ˆuS (8.86) Therefore, the heliocentric velocity of the spacecraft at the outbound crossing is V(v) 2=V+v∞2=[V(v) 2]VˆuV+[V(v) 2]SˆuS (8.87) where the components of V(v) 2are [V(v) 2]V=V+v∞cosφ2 [V(v) 2]S=v∞sinφ2 (8.88) From this we obtain the radial and transverse heliocentric velocity components, V⊥2=[V(v) 2]V Vr2=− [V(v) 2]S (8.89) 8.9 Planetary flyby 379 Finally, we obtain the three elements e2,h2andθ2of the new heliocentric departure trajectory by means of Equation 2.21, h2=RV⊥2 (8.90) Equation 2.35, R=h2 2 µsun1 1+e2cosθ2(8.91) and Equation 2.39, Vr2=µsun h2e2sinθ2 (8.92) Notice that the flyby is considered to be an impulsive maneuver during which the heliocentric radius of the spacecraft, which is confined within the planet’s sphereof influence, remains fixed at R. The heliocentric velocity analysis is similar to that described in Section 6.7. Example 8.6A spacecraft departs earth with a velocity perpendicular to the sun line on a flyby mission to Venus. Encounter occurs at a true anomaly in the approach trajectory of−30◦. Periapse altitude is to be 300 km. (a) For an approach from the dark side of the planet, show that the post-flyby orbit is as illustrated in Figure 8.20. (b) For an approach from the sunlit side of the planet, show that the post-flyby orbit is as illustrated in Figure 8.21. Earth at departureEarth at arrival Venus at departureVenus at arrival SunPre-flyby ellipse Post-flyby ellipseAphelion Perihelion30° 44.32°1 2 Figure 8.20 Spacecraft orbits before and after a flyby of Venus, approaching from the dark side. 380 Chapter 8 Interplanetary trajectories (Example 8.6 continued) Earth at departureEarth at arrival Venus at departureVenus at arrival SunPre-flyby ellipse Post-flyby ellipseAphelionPerihelion 30°66.76 °1 2 Figure 8.21 Spacecraft orbits before and after a flyby of Venus, approaching from the sunlit side. The following data is found in Tables A.1 and A.2: µsun=1.3271×1011km3/s2 µVenus=324 900 km3/s2 Rearth=149.6×106km RVenus=108.2×106km rVenus=6052 km Pre-flyby ellipse (orbit 1) Evaluating the orbit formula, Equation 2.35, at perihelion of orbit 1 yields Rearth=h2 1 µsun1 1−e1 Thus h2 1=µsunRearth(1−e1)( a ) At intercept RVenus=h2 1 µsun1 1+e1cos(θ1) 8.9 Planetary flyby 381 Substituting Equation (a) and θ1=− 30◦and solving the resulting expression for e1 leads to e1=Rearth−RVenus Rearth+RVenus cos(θ1)=149.6×106−108.2×106 149.6×106+108.2×106cos(−30◦)=0.1702 With this result, Equation (a) yields h1=/radicalbig 1.327×1011·149.6×106(1−0.1702) =4.059×109km2/s Now we can use Equations 8.78 to calculate the radial and transverse components of the spacecraft’s heliocentric velocity at the inbound crossing of Venus’s sphere of influence: V⊥1=h1 RVenus=4.059×109 108.2×106=37.51 km /s Vr1=µsun h1e1sin(θ 1)=1.327×1011 4.059×109·0.1702 ·sin(−30◦)=− 2.782 km /s The flight path angle, from Equation 2.41, is γ1=tan−1Vr1 V⊥1=tan−1/parenleftbigg−2.782 37.51/parenrightbigg =− 4.241◦ The negative sign is consistent with the fact that the spacecraft is flying towards perihelion of the pre-flyby elliptical trajectory (orbit 1). The speed of the space vehicle at the inbound crossing is V(v) 1=/radicalBig V2r1+V2 ⊥1=/radicalbig (−2.782)2+37.512=37.62 km /s (b) Flyby hyperbola From Equations 8.75 and 8.77 we obtain V(v) 1=37.51ˆuV+2.782ˆuS(km/s) The velocity of Venus in its presumed circular orbit around the sun is V=/radicalbiggµsun RVenusˆuV=/radicalBigg 1.327×1011 108.2×106ˆuV=35.02ˆuV(km/s) (c) Hence v∞1=V(v) 1−V=(37.51 ˆuV+2.782ˆuS)−35.02ˆuV=2.490ˆuV+2.782ˆuS(km/s) (d) It follows that v∞=√v∞1·v∞1=3.733 km /s The periapse radius is rp=rVenus+300=6352 km 382 Chapter 8 Interplanetary trajectories (Example 8.6 continued)Equations 8.38 and 8.39 are used to compute the angular momentum and eccentricityof the planetocentric hyperbola: h=6352/radicalbigg v2∞+2µVenus 6352=6352/radicalbigg 3.7332+2·324 900 6352=68 480 km2/s e=1+rpv2 ∞ µVenus=1+6352·3.7332 324 900=1.272 The turn angle and true anomaly of the asymptote are δ=2 sin−1/parenleftbigg1 e/parenrightbigg =2 sin−1/parenleftbigg1 1.272/parenrightbigg =103.6◦ θ∞=cos−1/parenleftbigg −1 e/parenrightbigg =cos−1/parenleftbigg −1 1.272/parenrightbigg =141.8◦ From Equations 2.40, 2.93 and 2.97, the aiming radius is /Delta1=rp/radicalbigg e+1 e−1=6352/radicalbigg 1.272+1 1.272−1=18 340 km (e) Finally, from Equation (d) we obtain the angle between v∞1and V, φ1=tan−12.782 2.490=48.17◦(f) There are two flyby approaches, as shown in Figure 8.22. In the dark side approach, the turn angle is counterclockwise ( +102.9◦) whereas for the sunlit side approach it is clockwise ( −102.9◦). 48.2°48.2°Venus’ orbital trac kTo the sun Dark side approachSunlit side approach SOI 3.73 km/s 3.73 km/s ∆ = 18 340 km Figure 8.22 Initiation of a sunlit side approach and dark side approach at the inbound crossing. 8.9 Planetary flyby 383 Dark side approach According to Equation 8.85, the angle between v∞and VVenus at the outbound crossing is φ2=φ1+δ=48.17◦+103.6◦=151.8◦ Hence, by Equation 8.86, v∞2=3.733(cos 151.8◦ˆuV+sin 151.8◦ˆuS)=− 3.289ˆuV+1.766ˆuS(km/s) Using this and Equation (c) above, we compute the spacecraft’s heliocentric velocity at the outbound crossing: V(v) 2=V+v∞2=31.73ˆuV+1.766ˆuS(km/s) It follows from Equation 8.89 that V⊥2=31.73 km /s Vr2=− 1.766 km /s( g) The speed of the spacecraft at the outbound crossing is V(v) 2=/radicalBig V2r2+V2 ⊥2=/radicalbig (−1.766)2+31.732=31.78 km /s This is 5.83 km/s less than the inbound speed. Post-flyby ellipse (orbit 2) for the dark side approach For the heliocentric post flyby trajectory, labeled orbit 2 in Figure 8.20, the angular momentum is found using Equation 8.90 h2=RVenus V⊥2=(108.2 ×106)·31.73=3.434×109(km2/s) (h) From Equation 8.91, ecosθ2=h2 2 µsunRVenus−1=(3.434 ×106)2 1.327×1011·108.2×106−1=− 0.1790 (i) and from Equation 8.92 esinθ2=Vr2h2 µsun=−1.766 ·3.434×109 1.327×1011=− 0.04569 (j) Thus tanθ2=esinθ2 ecosθ2=−0.04569 −0.1790=0.2553 (k) which means θ2=14.32◦or 194.32◦(l) Butθ2must lie in the third quadrant since, according to Equations (i) and (j), both the sine and cosine are negative. Hence, θ2=194.32◦(m) 384 Chapter 8 Interplanetary trajectories With this value of θ2, we can use either Equation (i) or (j) to calculate the eccentricity, e2=0.1847 (n) Perihelion of the departure orbit lies 194.32◦clockwise from the encounter point (so that aphelion is 14.32◦therefrom), as illustrated in Figure 8.20. The perihelion radius is given by Equation 2.40, Rperihelion =h2 2 µsun1 1+e2=(3.434×109)2 1.327×10111 1+0.1847=74.98×106km which is well within the orbit of Venus. Sunlit side approach In this case the angle between v∞and VVenus at the outbound crossing is φ2=φ1−δ=48.17◦−103.6◦=− 55.44◦ Therefore, v∞2=3.733[cos( −55.44◦)ˆuV+sin(−55.44◦)ˆuS]=2.118ˆuV−3.074ˆuS(km/s) The spacecraft’s heliocentric velocity at the outbound crossing is V(v) 2=VVenus+v∞2=37.14ˆuV−3.074ˆuS(km/s) which means V⊥2=37.14 km/s Vr2=3.074 km /s The speed of the spacecraft at the outbound crossing is V(v) 2=/radicalBig 3.0742+V2 ⊥2=/radicalbig 3.0502+37.142=37.27 km/s This speed is just 0.348 km/s less than the inbound crossing speed. The relatively small speed change is due to the fact that the apse line of this hyperbola is nearly perpendicular to Venus’s orbital track, as shown in Figure 8.23. Nevertheless, theperiapses of both hyperbolas are on the leading side of the planet. Post-flyby ellipse (orbit 2) for the sunlit side approach T o determine the heliocentric post-flyby trajectory, labeled orbit 2 in Figure 8.21, werepeat steps (h) through (n) above: h 2=RVenus V⊥2=(108.2×106)·37.14=4.019×109(km2/s) ecosθ2=h2 2 µsunRVenus−1=(4.019×109)2 1.327×1011·108.2×106−1=0.1246 (o) esinθ2=Vr2h2 µsun=3.074·4.019×109 1.327×1011=0.09309 (p) 8.9 Planetary flyby 385 48.2° 48.2°3.73 km/s55.4°3.73 km/s 3.73 km/s3.73 km/s152° To the sun Venus’ orbital track iii9.2° 3.6° Apse line of iiApse line of i Figure 8.23 Hyperbolic flyby trajectories for (i) the dark side approach and (ii) the sunlit side approach. tanθ2=esinθ2 ecosθ2=0.09309 0.1246=0.7469 θ2=36.08◦or 216.08◦ θ2must lie in the first quadrant since both the sine and cosine are positive. Hence, θ2=36.76◦(q) With this value of θ2, we can use either Equation (o) or (p) to calculate the eccentricity, e2=0.1556 Perihelion of the departure orbit lies 36.76◦clockwise from the encounter point as illustrated in Figure 8.21. The perihelion radius is Rperihelion =h2 2 µsun1 1+e2=(4.019 ×109)2 1.327×10111 1+0.1556=105.3×106km which is just within the orbit of Venus. Aphelion lies between the orbits of earth and Venus. Gravity assist maneuvers are used to add momentum to a spacecraft over and above that available from a spacecraft’s o n-board propulsion system. A sequence of flybys of planets can impart the delta-v needed to reach regions of the solar systemthat would be inaccessible using only existing propulsion technology. The technique 386 Chapter 8 Interplanetary trajectories can also reduce the flight time. Interplanetary missions using gravity assist flybys must be carefully designed in order to take advantage of the relative positions of planets. The 260 kg spacecraft Pioneer 11, launched in April 1973, used a December 1974 flyby of Jupiter to gain the momentum required to carry it to the first ever flybyencounter with Saturn on 1 September 1979. Following its September 1977 launch, Voyager 1 likewise used a flyby of Jupiter (March 1979) to reach Saturn in November 1980. In August 1977 Voyager 2 waslaunched on its ‘grand tour’ of the outer planets and beyond. This involved gravityassist flybys of Jupiter (July 1979), Saturn (August 1981), Uranus (January 1986) andNeptune (August 1989), after which the spacecraft departed at an angle of 30 ◦to the ecliptic. With a mass nine times that of Pioneer 11, the dual-spin Galileo spacecraft departed on 18 October 1989 for an extensive international exploration of Jupiterand its satellites lasting until September 2003. Galileo used gravity assist flybys of Venus (February 1990), earth (December 1990) and earth again (December 1992) before arriving at Jupiter in December 1995. The international Cassini mission to Saturn also made extensive use of gravity assist flyby maneuvers. The Cassini spacecraft was launched on 15 October 1997from Cape Canaveral, Florida, and arrived at Saturn nearly seven years later, on 1 July2004. The mission involved four flybys, as illustrated in Figure 8.24. A little over eightmonths after launch, on 26 April 1998, Cassini flew by Venus at a periapse altitudeof 284 km and received a speed boost of about 7 km/s. This placed the spacecraft inan orbit which sent it just outside the orbit of Mars (but well away from the planet)and returned it to Venus on 24 June 1999 for a second flyby, this time at an altitudeof 600 km. The result was a trajectory that vectored Cassini toward the earth for an18 August 1999 flyby at an altitude of 1171 km. The 5.5 km/s speed boost at earthsent the spacecraft toward Jupiter for its next flyby maneuver. This occurred on 30December 2000 at a distance of 9.7 million km from Jupiter, boosting Cassini’s speedby about 2 km/s and adjusting its trajectory so as to rendezvous with Saturn aboutthree and a half years later. Jupiter gravity assist flyby 30 Dec 2000 Earth gravity assist flyby 18 Aug 1999Earth at launch 15 Oct 1997First Venus gravity assist flyby 26 Apr 1998Second Venus gravity assist flyby 24 Jun 1999Mars orbit /H9253SunArrival at Saturn 1 Jul 2004 Figure 8.24 Cassini seven-year mission to Saturn. 8.10 Planetary ephemeris 387 8.10 Planetary ephemeris The state vector R,Vof a planet is defined relative to the heliocentric ecliptic frame of reference illustrated in Figure 8.25. This is very similar to the geocentric equatorialframe of Figure 4.5. The sun replaces the earth as the center of attraction, and the planeof the ecliptic replaces the earth’s equatorial plane. The vernal equinox continues todefine the inertial Xaxis. In order to design realistic interplanetary missions we must be able to determine the state vector of a planet at any given time. Table 8.1 provides the orbital elementsof the planets and their rates of change per century with respect to the J2000 epoch(1 January 2000, 12 hr UT). The table, covering the years 1800 to 2050, is sufficientlyaccurate for our needs. From the orbital elements we can infer the state vector using Algorithm 4.2. In order to interpret Table 8.1, observe the following: 1 astronomical unit (1 AU) is 1 .49597871 ×10 8km, the average distance between the earth and the sun. 1 arcsecond (1/prime/prime) is 1/3600 of a degree. ais the semimajor axis. eis the eccentricity. iis the inclination to the ecliptic plane. Y XPerihelionZNorth ecliptic pole Ascending node ˆ I ˆ K R /H9253Ecliptic plane ˆ J Node lineSunˆ w ˆ n ii /H9258 /H9275VPlanetary orbit Ω Figure 8.25 Planetary orbit in the heliocentric ecliptic frame. 388 Chapter 8 Interplanetary trajectories T able 8.1 Planetary orbital elements and their centennial rates. From Standish et al. (1992). Used with permission a,A U ei ,d e g /Omega1,d e g ˜ω,d e g L,d e g ˙a, AU/Cy ˙e,1/Cy ˙i,/prime/prime/Cy ˙/Omega1,/prime/prime/Cy ˙˜ω,/prime/prime/Cy ˙L,/prime/prime/Cy Mercury 0.38709893 0.20563069 7.00487 48.33167 77.45645 252.25084 0.00000066 0.00002527 −23.51 −446.30 573.57 538 101 628.29 Venus 0.72333199 0.00677323 3.39471 76.68069 131.53298 181.97973 0.00000092 −0.00004938 −2.86 −996.89 −108.80 210 664 136.06 Earth 1.00000011 0.01671022 0.00005 −11.26064 102.94719 100.46435 −0.00000005 −0.00003804 −46.94−18228.25 1198.28 129 597 740.63 Mars 1.52366231 0.09341233 1.85061 49.57854 336.04084 355.45332 −0.00007221 0.00011902 −25.47 −1020.19 1560.78 68 905 103.78 Jupiter 5.20336301 0.04839266 1.30530 100.55615 14.75385 34.40438 0.00060737 −0.00012880 −4.15 1217.17 839.93 10 925 078.35 Saturn 9.53707032 0.05415060 2.48446 113.71504 92.43194 49.94432 −0.00301530 −0.00036762 6.11 −1591.05−1948.89 4 401 052.95 Uranus 19.19126393 0.04716771 0.76986 74.22988 170.96424 313.23218 0.00152025 −0.00019150 −2.09 −1681.4 1312.56 1 542 547.79 Neptune 30.06896348 0.00858587 1.76917 131.72169 44.97135 304.88003 −0.00125196 0.00002514 −3.64 −151.25 −844.43 786 449.21 Pluto 39.48168677 0.24880766 17.14175 110.30347 224.06676 238.92881 −0.00076912 0.00006465 11.07 −37.33 −132.25 522 747.90 /Omega1is the right ascension of the ascending node (relative to the J2000 vernal equinox). ˜ω, the longitude of perihelion, is defined as ˜ω=ω+/Omega1,w h e r e ωis the argument of perihelion. L, the mean longitude, is defined as L=˜ω+M,w h e r e Mis the mean anomaly. ˙a,˙e,˙/Omega1, etc., are the rates of change of the above orbital elements per Julian century. 1 century (Cy) equals 36 525 days. Algorithm 8.1Determine the state vector of a planet at a g iven date and time. All angular calculations must be adjusted so that they lie in the range 0◦to 360◦. Recall that the gravitational parameter of the sun is µ=1.327×1011km3/s2. This procedure is implemented in MATLAB in Appendix D.17. 1. Use Equations 5.47 and 5.48 to calculate the Julian day number JD. 2. Calculate T0, the number of Julian centuries between J2000 and the date in question T0=JD−2 451 545 36 525(8.104a) 3. If Qis any one of the six planetary orbital elements listed in Table 8.1, then calculate its value at JDby means of the formula Q=Q0+˙QT 0 (8.104b) 8.10 Planetary ephemeris 389 where Q0is the value listed for J2000 and ˙Qis the tabulated rate. All angular quantities must be adjusted to lie in the range 0◦to 360◦. 4. Use the semimajor axis aand the eccentricity eto calculate the angular momentum hatJDfrom Equation 2.61 h=/radicalbig µa(1−e2) 5. Obtain the argument of perihelion ωand mean anomaly MatJDfrom the results of step 3 by means of the definitions ω=˜ω−/Omega1 M=L−˜ω 6. Substitute the eccentricity eand the mean anomaly MatJDinto Kepler’s equation (Equation 3.11) and calculate the eccentric anomaly E. 7. Calculate the true anomaly θusing Equation 3.10. 8. Use h,e,/Omega1,i,ωandθto obtain the heliocentric position vector Rand velocity Vby means of Algorithm 4.2, with the heliocentric ecliptic frame replacing the geocentric equatorial frame. Example 8.7Find the distance between the earth and Mars at 12 hr UT on 27 August 2003. Use Algorithm 8.1. Step 1: According to Equation 5.56, the Julian day number J0for midnight (0 hr UT) of this date is J0=367·2003−INT  7/bracketleftbigg 2003+INT/parenleftbigg8+9 12/parenrightbigg/bracketrightbigg 4   +INT/parenleftbigg275·8 9/parenrightbigg +27+1 721 013 .5 =735 101 −3507+244+27+1 721 013 .5 =2 452 878 .5 AtUT=12, the Julian day number is JD=2 452 878 .5+12 24=2 452 879 .0 Step 2: The number of Julian centuries between J2000 and this date is T0=JD−2 451 545 36 525=2 452 879 −2 451 545 36 525=0.036523 Cy 390 Chapter 8 Interplanetary trajectories (Example 8.7 continued)Step 3: Table 8.1 and Equation 8.104 yield the orbital elements of earth and Mars at 12 hr UT on 27 August 2003. a,k m ei ,d e g /Omega1,d e g ˜ω,d e g L,d e g Earth 1.4960 ×1080.016709 0.00042622 348.55 102.96 335.27 Mars 2.2794 ×1080.093417 1.8504 49.568 336.06 334.51 Step 4: hearth=4.4451×109km2/s hMars=5.4760×109km2/s Step 5: ωearth=(˜ω−/Omega1)earth=102.96−348.55=− 245.59◦(114.1◦) ωMars=(˜ω−/Omega1)Mars=336.06−49.568=286.49◦ Mearth=(L−˜ω)earth=335.27−102.96=232.31◦ MMars=(L−˜ω)Mars=334.51−336.06=− 1.55◦(358.45◦) Step 6: Eearth−0.016709 sin Eearth=232.31◦(π/180)⇒Eearth=231.56◦ EMars−0.093417 sin EMars=358.45◦(π/180)⇒EMars=358.30◦ Step 7: θearth=2 tan−1/parenleftBigg/radicalbigg 1−0.016709 1+0.016709tan231.56◦ 2/parenrightBigg =− 129.19◦⇒θearth=230.81◦ θMars=2 tan−1/parenleftBigg/radicalbigg 1−0.093417 1+0.093417tan358.30◦ 2/parenrightBigg =− 1.8669◦⇒θMars=358.13◦ Step 8: From Algorithm 4.2, Rearth=(135.59ˆI−66.803ˆJ−0.00028691 ˆK)×106(km) Vearth=12.680ˆI+26.61ˆJ−0.00021273 ˆK(km/s) RMars=(185.95ˆI−89.916ˆJ−6.4566ˆK)×106(km) VMars=11.474ˆI+23.884ˆJ+0.21826 ˆK(km/s) The distance dbetween the two planets is therefore d=/bardblRMars−Rearth/bardbl =/radicalBig (185.95−135.59)2+[−89.916−(−66.803) ]2+(−6.4566−0.00028691)2×106 8.11 Non-Hohmann interplanetary trajectories 391 or d=55.79×106km The positions of earth and Mars are illustrated in Figure 8.26. It is a rare event for Mars to be in opposition (lined up with earth on the same side of the sun) when Marsis at or near perihelion. The two planets had not been this close in recorded history. 103°49.6° 24° 26° Earth MarsEarth perihelion Mars perihelionMars ascending node /H9253 Mars descending nodeSun Figure 8.26 Earth and Mars on 27 August 2003. Angles shown are heliocentric latitude, measured in the plane of the ecliptic counterclockwise from the vernal equinox of J2000. 8.11 Non-Hohmann interplanetary trajectories T o implement a systematic patched conic procedure for three-dimensional trajecto- ries, we will use vector notation and the procedures described in Sections 4.4 and 4.6(Algorithms 4.1 and 4.2), together with the solution of Lambert’s problem presentedin Section 5.3 (Algorithm 5.2). The mission is to send a spacecraft from planet 1 toplanet 2 in a specified time t 12. As previously in this chapter, we break the mission down into three parts: the departure phase, the cruise phase and the arrival phase. We start with the cruise phase. The frame of reference that we use is the heliocentric ecliptic frame shown in Figure 8.27. The first step is to obtain the state vector of planet 1 at departure (time t) and the state vector of planet 2 at arrival (time t+t12). That is accomplished by means of Algorithm 8.1. The next step is to determine the spacecraft’s transfer trajectory from planet 1 to planet 2. We first observe that, according to the patched conic procedure, the 392 Chapter 8 Interplanetary trajectories Y X/H9258PerihelionZNorth ecliptic pole Spacecraft trajectory Ascending node InJK wR1 /H9253Ecliptic plane Node lineR2 SunPlanet 1 at departure Planet 2 at arrival /H9251itr itrvtr Ωtr∆u Figure 8.27 Heliocentric orbital elements of a three-dimensional transfer trajectory from planet 1 to planet 2. heliocentric position vector of the spacecraft at time tis that of planet 1 ( R1) and at time t+t12its position vector is that of planet 2 ( R2). With R1,R2and the time of flight t12we can use Algorithm 5.2 (Lambert’s problem) to obtain the spacecraft’s departure and arrival velocities V(v) Dand V(v) Arelative to the sun. Either of the state vectors R1,V(v) DorR2,V(v) Acan be used to obtain the transfer trajectory’s six orbital elements by means of Algorithm 4.1. The spacecraft’s hyperbolic excess velocity upon exiting the sphere of influence of planet 1 is v∞)Departure =V(v) D−V1 (8.102a) and its excess speed is v∞)Departure =/vextenddouble/vextenddouble/vextenddoubleV(v) D−V1/vextenddouble/vextenddouble/vextenddouble (8.102b) Likewise, at the sphere of influence crossing at planet 2, v ∞)Arrival=V(v) A−V2 (8.103a) v∞)Arrival=/vextenddouble/vextenddouble/vextenddoubleV(v) A−V2/vextenddouble/vextenddouble/vextenddouble (8.103b) 8.11 Non-Hohmann interplanetary trajectories 393 Algorithm 8.2Given the departure and arrival dates (and, therefore, the time of flight), determine the trajectory for a mission from planet 1 to planet 2. This procedure is implementedin MATLAB in Appendix D.19. 1. Use Algorithm 8.1 to determine the state vector R1,V1of planet 1 at departure and the state vector R2,V2of planet 2 at arrival. 2. Use R1,R2and the time of flight in Algorithm 5.2 to find the spacecraft velocity V(v) Dat departure from planet 1’s sphere of influence and its velocity V(v) Aupon arrival at planet 2’s sphere of influence. 3. Calculate the hyperbolic excess velocities at departure and arrival using Equations 8.102 and 8.103. Example 8.8A spacecraft departs earth’s sphere of influence on 7 November 1996 (0 hr UT) on a prograde coasting flight to Mars, arriving at Mars’sphere of influence on 12 September 1997 (0 hr UT). Use Algorithm 8.2 to determine the trajectory and then compute thehyperbolic excess velocities at departure and arrival. Step 1: Algorithm 8.1 yields the state vectors for earth and Mars: Rearth=1.0500 ×108ˆI+1.0466 ×108ˆJ +988.33 ˆK(km) (Rearth=1.482×108km) Vearth=− 21.516 ˆI+20.987 ˆJ+0.00013228 ˆK(km/s) ( Vearth=30.06 km /s) RMars=− 2.0833 ×107ˆI−2.1840 ×108ˆJ −4.0629 ×106ˆK(km) (RMars=2.194×108km) VMars=25.047 ˆI−0.22029 ˆJ−0.62062 ˆK(km/s) ( VMars=25.05 km /s) Step 2: The position vector R1of the spacecraft at crossing the earth’s sphere of influence is just that of the earth, R1=Rearth=1.0500 ×108ˆI+1.0466 ×108ˆJ+988.33 ˆK(km) Upon arrival at Mars’ sphere of influence the spacecraft’s position vector is R2=RMars=− 2.0833 ×107ˆI−2.1840 ×108ˆJ−4.0629 ×106ˆK(km) According to Equations 5.47 and 5.48 JDDeparture =2 450 394 .5 JDArrival=2 450 703 .5 Hence, the time of flight is t12=2 450 703 .5−2 450 394 .5=309 days 394 Chapter 8 Interplanetary trajectories (Example 8.8 continued)Entering R1,R2and t12into Algorithm 5.2 yields V(v) D=− 24.427ˆI+21.781ˆJ+0.94803 ˆK(km/s)/bracketleftBig V(v) D=32.741 km /s/bracketrightBig V(v) A=22.158ˆI−0.19668ˆJ−0.45785 ˆK(km/s)/bracketleftBig V(v) A=22.164 km /s/bracketrightBig Using the state vector R1,V(v) Dwe employ Algorithm 4.1 to find the orbital elements of the transfer trajectory. h=4.8456×106km2/s e=0.20579 /Omega1=44.895◦ i=1.6621◦ ω=19.969◦ θ1=340.04◦ a=1.8474×108km Step 3: At departure the hyperbolic excess velocity is v∞)Departure =V(v) D−Vearth=− 2.913ˆI+0.7958ˆJ+0.9480ˆK(km/s) Therefore, the hyperbolic excess speed is v∞)Departure =/vextenddouble/vextenddoublev∞)Departure/vextenddouble/vextenddouble=3.1651 km /s (a) Likewise, at arrival v∞)Arrival=V(v) A−VMars=− 2.8804ˆI+0.023976 ˆJ+0.16277 ˆK(km/s) so that v∞)Arrival=/vextenddouble/vextenddoublev∞)Arrival/vextenddouble/vextenddouble=2.8851 km /s (b) For the previous example, Figure 8.28 shows the orbits of earth, Mars and the space- craft from directly above the ecliptic plane. Dotted lines indicate the portions of anorbit which are below the plane. λis the heliocentric longitude measured counter- clockwise from the vernal equinox of J2000. Also shown are the position of Mars atdeparture and the position of earth at arrival. The transfer orbit resembles that of the Mars Global Surveyor, which departed earth on 7 November 1996 and arrived at Mars 309 days later, on 12 September 1997. Example 8.9In Example 8.8, calculate the delta-v required to launch the spacecraft onto its cruise trajectory from a 180 km circular parking orbit. Sketch the departure trajectory. Recall that rearth=6378 km µearth=398 600 km3/s2 8.11 Non-Hohmann interplanetary trajectories 395 Mars ascending node (λ /H11005 49.58°) Mars descending node (λ /H11005 229.6°)Mars at arrival (λ /H11005 264.6°)Mars perihelion (λ /H11005 336.0°)Earth at launch and spacecraf t ascending node (λ /H1100544.91°)Earth perihelion (λ /H11005 102.9°)Spacecraft perihelion (λ /H11005 64.85°) Spacecraft descending node (λ /H11005 224.9°)SunMars at launch (λ /H11005 119.3°) Earth at arrival (λ /H11005 349.3°)/H9253/H9261 Figure 8.28 The transfer trajectory, together with the orbits of earth and Mars, as viewed from directly above the plane of the ecliptic. The radius to periapse of the departure hyperbola is the radius of the earth plus the altitude of the parking orbit, rp=6378+180=6558 km Substituting this and Equation (a) from Example 8.8 into Equation 8.40 we get the speed of the spacecraft at periapse of the departure hyperbola, vp=/radicalBigg [v∞)Departure ]2+2µearth rp=/radicalbigg 3.16512+2·398 600 6558=11.47 km /s The speed of the spacecraft in its circular parking orbit is vo=/radicalbiggµearth rp=/radicalbigg 398 600 6558=7.796 km /s Hence, the delta-v requirement is /Delta1v=vp−v0=3.674 km /s The eccentricity of the hyperbola is given by Equation 8.38, e=1+rpv2 ∞ µearth=1+6558·3.16512 398 600=1.165 If we assume that the spacecraft is launched from a parking orbit of 28◦inclination, then the departure appears as shown in the three-dimensional sketch in Figure 8.29. 396 Chapter 8 Interplanetary trajectories (Example 8.9 continued) VearthZ XTo the sun /H9020v/H11009 Parking orbit /H9253Earth's equatorial planePerigee Figure 8.29 The departure hyperbola, assumed to be at 28◦inclination to earth’s equator. Example 8.10In Example 8.8, calculate the delta-v required to place the spacecraft in an elliptical capture orbit around Mars with a periapse altitude of 300 km and a period of 48hours. Sketch the approach hyperbola. From Tables A.1 and A.2 we know that rMars=3380 km µMars=42 830 km3/s2 The radius to periapse of the arrival hyperbola is the radius of Mars plus the periapse of the elliptical capture orbit, rp=3380+300=3680 km According to Equation 8.40 and Equation (b) of Example 8.8, the speed of the spacecraft at periapse of the arrival hyperbola is vp)hyp=/radicalBigg [v∞)Arrival]2+2µMars rp=/radicalbigg 2.88512+2·42 830 3680=5.621 km /s T o find the speed vp)ellat periapse of the capture ellipse, we use the required period (48 hours) to determine the ellipse’s semimajor axis, using Equation 2.73 aell=/parenleftbiggT√µMars 2π/parenrightbigg3 2 =/parenleftBigg 48·3600·√ 42 830 2π/parenrightBigg3 2 =31 880 km 8.11 Non-Hohmann interplanetary trajectories 397 From Equation 2.63 we obtain eell=1−rp aell=1−3680 31 880=0.8846 Then Equation 8.59 yields vp)ell=/radicalbiggµMars rp(1+eell)=/radicalbigg 42 830 3680(1+0.8846) =4.683 km /s Hence, the delta-v requirement is /Delta1v=vp)hyp−vp)ell=0.9382 km /s The eccentricity of the approach hyperbola is given by Equation 8.38, e=1+rpv2 ∞ µMars=1+3680·2.88512 42 830=1.715 Assuming that the capture ellipse is a polar orbit of Mars, then the approach hyperbola is as illustrated in Figure 8.30. Note that Mars’ equatorial plane is inclined 25◦to the plane of its orbit around the sun. Furthermore, the vernal equinox of Mars lies at anangle of 85 ◦from that of the earth. /H9020 VMarsTo the sunX v/H11009Z3680 by 60 070 km polar capture orbit (48 hour period) PeriapseApoapse gMars Mars equatorial plane Figure 8.30 The approach hyperbola and capture ellipse. 398 Chapter 8 Interplanetary trajectories Problems 8.1 On 6 February 2006, when the earth is 147 .4×106km from the sun, a spacecraft parked in a 200 km altitude circular earth orbit is to be launched directly into an ellipticalorbit around the sun with perihelion of 120 ×10 6km and aphelion equal to the earth’s distance from the sun on the launch date. Calculate the delta-v required and v∞of the departure hyperbola.{Ans.: v ∞=30 km/s,/Delta1v=3.34 km/s} 8.2 Estimate the total delta-v requirement for a Hohmann transfer from earth to Mercury, assuming a 150 km circular parking orbit at earth and a 150 km circular capture orbit atMercury. Furthermore, assume that the planets have coplanar circular orbits with radiiequal to the semimajor axes listed in Table A.1.{Ans.: 15 km/s} 8.3 Calculate the radius of the spheres of influence of Mercury, Venus, Mars and Jupiter. {Ans.: See Table A.2} 8.4 Calculate the radius of the spheres of influence of Saturn, Uranus, Neptune and Pluto. {Ans.: See Table A.2} 8.5 Suppose a spacecraft approaches Jupiter on a Hohmann transfer ellipse from earth. If the spacecraft flies by Jupiter at an altitude of 200 000 km on the sunlit side of the planet,determine the orbital elements of the post-flyby trajectory and the delta-v imparted tothe spacecraft by Jupiter’s gravity. Assume that all of the orbits lie in the same (ecliptic)plane.{Ans.: /Delta1V=10.6k m/s,a=4.79×10 6km, e=0.8453} 8.6 Use Table 8.1 to verify that the orbital elements for earth and Mars presented in Example 8.7. 8.7 Use Table 8.1 to determine the day of the year 2005 when the earth is farthest from the sun.{Ans.: 4 July} 8.8 On 1 December 2005 a spacecraft leaves a 180 km altitude circular orbit around the earth on a mission to Venus. It arrives at Venus 121 days later on 1 April 2006, entering a 300 kmby 9000 km capture ellipse around the planet. Calculate the total delta-v requirementfor this mission.{Ans.: 6.75 km/s} 8.9 On 15 August 2005 a spacecraft in a 190 km, 52 ◦inclination circular parking orbit around the earth departs on a mission to Mars, arriving at the red planet on 15 March2006, whereupon retro rockets place it into a highly elliptic orbit with a periapse of300 km and a period of 35 hours. Determine the total delta-v required for this mission.{Ans.: 4.86 km/s} 8.10 Calculate the propellant mass required to launch a 2000 kg spacecraft from a 180 km circular earth orbit on a Hohmann transfer trajectory to Saturn. Calculate the timerequired for the mission and compare it to that of Cassini. Assume the propulsionsystem has a specific impulse of 300 s.{Ans.: 6.03 y; 21 810 kg} 9Chapter Rigid-body dynamics Chapter outline 9.1 Introduction 399 9.2 Kinematics 400 9.3 Equations of translational motion 408 9.4 Equations of rotational motion 410 9.5 Moments of inertia 414 9.5.1 Parallel axis theorem 428 9.6 Euler’s equations 435 9.7 Kinetic energy 441 9.8 The spinning top 443 9.9 Euler angles 448 9.10 Yaw, pitch and roll angles 459 Problems 463 9.1 Introduction Just as Chapter 1 provides a foundation for the development of the equations of orbital mechanics, this chapter serves as a basis for developing the equations of satellite attitude dynamics. Chapter 1 deals with particles, whereas here we areconcerned with rigid bodies. Those familiar with rigid body dynamics can move onto the next chapter, perhaps returning from time to time to review concepts. The kinematics of rigid bodies is presented first. The subject depends on a theorem of the French mathematician Michel Chasles (1793–1880). Chasles’ theorem states 399 400 Chapter 9 Rigid-body dynamics that the motion of a rigid body can be described by the displacement of any point of the body (the base point) plus a rotation abo ut a unique axis through that point. The magnitude of the rotation does not depend on the base point. Thus, at any instant arigid body in a general state of motion has an angular velocity vector whose directionis that of the instantaneous axis of rotation. Describing the rotational componentof the motion a rigid body in three dimensions requires taking advantage of thevector nature of angular velocity and knowing how to take the time derivative ofmoving vectors, which is explained in Chapter 1. Several examples illustrate how thisis done. We then move on to study the interaction between the motion of a rigid body and the forces acting on it. Describing the translational component of the motion requiressimply concentrating all of the mass at a point, the center of mass, and applying the methods of particle mechanics to determine its motion. Indeed, our study of the two-body problem up to this point has focused on the motion of their centers of masswithout regard to the rotational aspect. Analyzing the rotational dynamics requirescomputing the body’s angular momentum, and that in turn requires accounting forhow the mass is distributed throughout the body. The mass distribution is describedby the six components of the moment of inertia tensor. Writing the equations of rotational motion r elative to coordinate axes embedded in the rigid body and aligned with the principal axes of inertia yields the non-linearEuler equations of motion, which are applied to a study of the dynamics of a spinningtop (or one-axis gyro). The expression for the kinetic energy of a rigid body is derived because it will be needed in the following chapter. The chapter concludes with a description of two sets of three angles commonly employed to specify the orientation of a body in three-dimensional space. One ofthese are the Euler angles, which are the same as the right ascension of the node ( /Omega1), argument of periapse ( ω) and inclination ( i)i n t r o d u c e di nC h a p t e r4t oo r i e n to r b i t s in space. The other set comprises the yaw, pitch and roll angles, which are suitable fordescribing the orientation of an airplane. Both the Euler angles and yaw–pitch–rollangles will be employed in Chapter 10. 9.2 Kinematics Figure 9.1 shows a moving rigid body and its instantaneous axis of rotation, which defines the direction of the absolute angular velocity vector ω.T h e XYZ axes are a fixed, inertial frame of reference. The position vectors RAand RBof two points on the rigid body are measured in the inertial frame. The vector RB/Adrawn from point Ato point Bis the position vector of Brelative to A. Since the body is rigid, RB/Ahas a constant magnitude even though its direction is continuously changing. Clearly, RB=RA+RB/A Differentiating this equation thr ough with respect to time, we get ˙RB=˙RA+dRB/A dt(9.1) 9.2 Kinematics 401 XYZ A RB RAB RB/A/H9275 Figure 9.1 Rigid body and its instantaneous axis of rotation. ˙RAand˙RBare the absolute velocities vAand vBof points Aand B. Because the magnitude of RB/Adoes not change, its time derivative is given by Equation 1.24, dRB/A dt=ω×RB/A Thus, Equation 9.1 becomes vB=vA+ω×RB/A (9.2) Taking the time derivative of Equation 9.1 yields ¨RB=¨RA+d2RB/A dt2(9.3) ¨RAand¨RBare the absolute accelerations aAand aBof the two points of the rigid body, while from Equation 1.25 we have d2RB/A dt2=α×RB/A+ω×(ω×RB/A) in which αis the angular acceleration, α=dω/dt . Therefore, Equation 9.3 can be written aB=aA+α×RB/A+ω×(ω×RB/A) (9.4) Equations 9.2 and 9.4 are the relative velocity and acceleration formulas. Note that all quantities in these expressions are measured in the same inertial frame of reference. 402 Chapter 9 Rigid-body dynamics When the rigid body under consideration is connected to and moving relative to another rigid body, computation of its inertial angular velocity ωand angular acceleration αmust be done with care. The key is to remember that angular velocity is a vector. It may be found as the vector sum of a sequence of angular velocities, eachmeasured relative to another, starting with one measured relative to an absolute frame,as illustrated in Figure 9.2. In that case, the absolute angular velocity ωof body 4 is ω=ω 1+ω2/1+ω3/2+ω4/3 (9.5) Each of these angular velocities is resolved into components along the axes of the moving frame of reference xyzshown in Figure 9.2, so that ω=ωxˆi+ωyˆj+ωzˆk (9.6) The moving frame is chosen for convenience of the analysis, and its inertial angu- lar velocity is denoted /Omega1, as discussed in Section 1.5. According to Equation 1.30, the absolute angular acceleration αis obtained from Equation 9.6 by means of the following calculation, α=dω dt/parenrightbigg rel+/Omega1×ω (9.7) where dω dt/parenrightbigg rel=dωx dtˆi+dωy dtˆj+dωz dtˆk (9.8) XYZ12 34/H92751 /H92752/1 /H92753/2/H92754/3x yz ˆIˆJˆKˆi ˆjˆk/H9024 Figure 9.2 Angular velocity is the vector sum of the relative angular velocities starting with ω1, measured relative to the inertial frame. 9.2 Kinematics 403 Example 9.1An airplane flies at constant speed vwhile simultaneously undergoing a constant yaw rateωyawabout a vertical axis and describing a circular loop in the vertical plane with ar a d i u s /rho1. The constant propeller spin rate is ωspinrelative to the airframe. Find the velocity and acceleration of the tip Pof the propeller relative to the hub H, when Pis directly above H.T h ep r o p e l l e rr a d i u si sl . xy zP Hvyaw vspin υl/rho1 Figure 9.3 Airplane with attached xyzbody frame. The xyz axes are rigidly attached to the airplane. The xaxis is aligned with the propeller’s spin axis. The yaxis is vertical, and the zaxis is in the spanwise direction, so that xyzforms a right-handed triad. Although the xyzframe is not inertial, we can imagine it to instantaneously coincide with an inertial frame. The absolute angular velocity of the airplane has two components, the yaw and the counterclockwise pitch angular velocity v//rho1 of its rotation in the circular loop, ωairplane =ωyawˆj+ωpitchˆk=ωyawˆj+v /rho1ˆk The angular velocity of the body-fixed moving frame is that of the airplane, /Omega1= ωairplane , so that /Omega1=ωyawˆj+v /rho1ˆk The absolute angular velocity of the propeller is that of the airplane plus the angular velocity propeller relative to the airplane, ωprop=ωairplane +ωspinˆi which means ωprop=ωspinˆi+ωyawˆj+v /rho1ˆk (a) From Equation 9.2, the velocity of point Pon the propeller relative to Hon the hub, vP/H,i sg i v e nb y vP/H=vP−vH=ωprop×rP/H where rP/His the position vector of Prelative to H. Thus, using (a), vP/H=/parenleftbigg ωspinˆi+ωyawˆj+v /rho1ˆk/parenrightbigg ×(lˆj) 404 Chapter 9 Rigid-body dynamics (Example 9.1 continued)from which vP/H=−v /rho1lˆi+ωspinlˆk The absolute angular acceleration of the propeller is found from Equation 9.7, αprop=dωprop dt/parenrightbigg rel+/Omega1×ωprop=/parenleftbiggdωspin dtˆi+dωyaw dtˆj+d(v//rho1) dtˆk/parenrightbigg +/parenleftbigg ωyawˆj+v /rho1ˆk/parenrightbigg ×/parenleftbigg ωspinˆi+ωyawˆj+v /rho1ˆk/parenrightbigg Sinceωspin,ωyaw,vand/rho1are all constant, this reduces to αprop=/parenleftbigg ωyawˆj+v /rho1ˆk/parenrightbigg ×/parenleftbigg ωspinˆi+ωyawˆj+v /rho1ˆk/parenrightbigg (b) Carrying out the cross product yields αprop=v /rho1ωspinˆj−ωyawωspinˆk (c) From Equation 9.4, the acceleration of Prelative to H,aP/H,i sg i v e nb y aP/H=aP−aH=αprop×rP/H+ωprop×/parenleftbig ωprop×rP/H/parenrightbig Substituting (a) and (c) into this expression yields aP/H=/parenleftbiggv /rho1ωspinˆj−ωyawωspinˆk/parenrightbigg ×(lˆj)+/parenleftbigg ωspinˆi+ωyawˆj+v /rho1ˆk/parenrightbigg ×/bracketleftbigg/parenleftbigg ωspinˆi+ωyawˆj+v /rho1ˆk/parenrightbigg ×rP/H/bracketrightbigg From this we find aP/H=/parenleftBig ωyawωspinlˆi/parenrightBig +/parenleftbigg ωspinˆi+ωyawˆj+v /rho1ˆk/parenrightbigg ×/parenleftbigg −v /rho1lˆi+ωspinlˆk/parenrightbigg =/parenleftBig ωyawωspinlˆi/parenrightBig +/bracketleftbigg ωyawωspinlˆi−/parenleftbiggv2 /rho12+ω2 spin/parenrightbigg lˆj+ωyawv /rho1lˆk/bracketrightbigg so that finally, aP/H=2ωyawωspinlˆi−/parenleftbiggv2 /rho12+ω2 spin/parenrightbigg lˆj+ωyawv /rho1lˆk Example 9.2The satellite is rotating about the zaxis at a constant rate N.T h e xyzaxes are attached to the spacecraft, and the zaxis has a fixed orientation in inertial space. The solar panels rotate at a constant rate ˙θin the direction shown. Calculate the absolute velocity and acceleration of point Aon the panel relative to point Owhich lies at the center of the spacecraft and on the centerline of the panels. 9.2 Kinematics 405 z x y dN A w/2 w/2θ Figure 9.4 Rotating solar panel on a rotating satellite. The position vector of Arelative to Ois rA/O=−w 2sinθˆi+dˆj+w 2cosθˆk (a) The absolute angular velocity of the panel is the absolute angular velocity of the spacecraft plus the angular velocity of the panel relative to the spacecraft, ωpanel=−˙θˆj+Nˆk (b) According to Equation 9.2, the velocity of Arelative to Ois vA/O=vA−vO=ωpanel×rA/O=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆi ˆj ˆk 0 −˙θ N −w 2sinθ dw 2cosθ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle from which vA/O=−/parenleftBigw 2˙θcosθ+Nd/parenrightBig ˆi−w 2Nsinθˆj−w 2˙θsinθˆk Since the moving xyzframe is attached to the body of the spacecraft, its angular velocity is /Omega1=Nˆk The absolute angular acceleration of the panel is obtained from Equation 9.7, αpanel=dωpanel dt/parenrightbigg rel+/Omega1×ωpanel =/parenleftbiggd(−˙θ) dtˆj+dN dtˆk/parenrightbigg +(Nˆk)×/parenleftBig −˙θˆj+Nˆk/parenrightBig Since Nand˙θare constants, this reduces to αpanel=˙θNˆi (c) 406 Chapter 9 Rigid-body dynamics (Example 9.2 continued)The acceleration of Arelative to Ois found using Equation 9.4, aA/O=aA−aO=αpanel×rA/O+ωpanel×/parenleftbig ωpanel×rA/O/parenrightbig =/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆi ˆj ˆk ˙θN 00 −w 2sinθdw 2cosθ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle+(−˙θˆj+Nˆk)×/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆi ˆj ˆk 0 −˙θ N −w 2sinθ dw 2cosθ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =/parenleftBig −w 2N˙θcosθˆj+N˙θdˆk/parenrightBig +/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆi ˆj ˆk 0 −˙θ N −w 2˙θcosθ−Nd−Nw 2sinθ−w 2˙θsinθ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle which leads to aA/O=w 2(N2+˙θ2)sinθˆi−N(Nd+w˙θcosθ)ˆj−w 2˙θ2cosθˆk Example 9.3The gyro rotor shown has a constant spin rate ωspinaround axis b–ain the direction shown. The XYZ axes are fixed. The xyzaxes are attached to the gimbal ring, whose angleθwith the vertical is increasing at the constant rate ˙θin the direction shown. The assembly is forced to precess at the constant rate Naround the vertical, as shown. Calculate the absolute angular velocity and acceleration of the rotor in the position shown, expressing the results in both the XYZ and the xyzframes of reference. X YZ z xy Gimbal ringRotorωspin a bd Nc Gθ Figure 9.5 Rotating, precessing, nutating gyro. 9.2 Kinematics 407 We will need the instantaneous relationship between the unit vectors of the inertial XYZ axes and the co-moving xyzframe, which by inspection are ˆI=− cosθˆj+sinθˆk ˆJ=ˆi (a) ˆK=sinθˆj+cosθˆk so that the matrix of the transformation from xyztoXYZ is [Q]xX= 0−cosθsinθ 10 00 sin θ cosθ  (b) The absolute angular velocity of the gimbal ring is that of the base plus the angular velocity of the gimbal relative to the base, ωgimbal=NˆK+˙θˆi=N(sinθˆj+cosθˆk)+˙θˆi=˙θˆi+Nsinθˆj+Ncosθˆk (c) where we made use of (a) 3. Since the moving xyzframe is attached to the gimbal, /Omega1=ωgimbal , so that /Omega1=˙θˆi+Nsinθˆj+Ncosθˆk (d) The absolute angular velocity of the rotor is its spin relative to the gimbal, plus the angular velocity of the gimbal, ωrotor=ωgimbal+ωspinˆk (e) From (c) it follows that ωrotor=˙θˆi+Nsinθˆj+(Ncosθ+ωspin)ˆk (f) Because ˆi,ˆjandˆkmove with the gimbal, this expression is valid for any time, not just the instant shown in Figure 9.5. Alternatively, applying the vector transformation {ωrotor}XYZ=[Q]xX{ωrotor}xyz (g) we obtain the angular velocity of the rotor in the inertial frame, but only at the instant shown in the figure, i.e., when the xaxis aligns with the Yaxis   ωX ωY ωZ  = 0−cosθsinθ 10 00 sin θ cosθ   ˙θ Nsinθ Ncosθ+ω spin   =  −N sinθcosθ+Nsinθcosθ+ωspinsinθ ˙θ Nsin2θ+Ncos2θ+ωspincosθ   or ωrotor=ωspinsinθˆI+˙θˆJ+(N+ωspincosθ)ˆK (h) 408 Chapter 9 Rigid-body dynamics (Example 9.3 continued)The angular acceleration of the rotor is obtained from Equation 9.7, recalling that N, ˙θ, andωspinare independent of time: αrotor=dωrotor dt/parenrightbigg rel+/Omega1×ωrotor=/bracketleftbiggd(˙θ) dtˆi+d(Nsinθ) dtˆj+d(Ncosθ+ωspin) dtˆk/bracketrightbigg +/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆi ˆj ˆk ˙θNsinθ Ncosθ ˙θNsinθNcosθ+ω spin/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =(N˙θcosθˆj−N˙θsinθˆk)+[ˆi(Nω spinsinθ)−ˆj(ωspin˙θ)+ˆk(0)] Upon simplification, this becomes αrotor=Nωspinsinθˆi+˙θ(Ncosθ−ωspin)ˆj−N˙θsinθˆk (i) This expression, like (f), is valid at any time. The components of αrotor along the XYZ axes are found in the same way as for ωrotor, {αrotor}XYZ=[Q]xX{αrotor}xyz which means   αX αY αZ  = 0−cosθsinθ 10 00 sin θ cosθ   Nω spinsinθ ˙θ(Ncosθ−ωspin) −N˙θsinθ   =  −N˙θcos2θ+˙θωspincosθ−N˙θsin2θ Nωspinsinθ N˙θsinθcosθ−˙θωspinsinθ−N˙θsinθcosθ   or αrotor=˙θ(ωspincosθ−N)ˆI+NωspinsinθˆJ−˙θωspinsinθˆK (j) Note carefully that (j) is not simply the time derivative of (h). Equations (h) and (j) are valid only at the instant that the xyzand XYZ axes have the alignments shown in Figure 9.4. 9.3 Equations of translational motion Figure 9.6 again shows an arbitrary, continuous, three-dimensional body of mass m. ‘Continuous’ means that as we zoom in on a point it remains surrounded by a con- tinuous distribution of matter having the infinitesimal mass dmin the limit. The point never ends up in a void. In particular, we ignore the actual atomic and molec-ular microstructure in favor of this continuum hypothesis, as it is called. Molecularmicrostructure does not bear upon the overall dynamics of a finite body. We will useGto denote the center of mass. Position vectors of points relative to the origin of the inertial frame will be designated by capital letters. Thus, the position of the center of 9.3 Equations of translational motion 409 XYZ Gdm R RGdFnet dfnet Figure 9.6 Forces on the mass element dmof a continuous medium. mass is RG, defined as mR G=/integraldisplay mRdm (9.9) Ris the position of a mass element dmwithin the continuum. Each element of mass i sa c t e du p o nb yan e te x t e r n a lf o r c e dFnetand a net internal force dfnet. The external force comes from direct contact with other objects and from action at a distance, suchas gravitational attraction. The internal forces are those exerted from within the bodyby neighboring particles. These are the forces which hold the body together. For eachmass element, Newton’s second law, Equation 1.10, is written dF net+dfnet=dm¨R (9.10) Writing this equation for the infinite number of mass elements of which the body is composed and then summing them all together leads to the integral, /integraldisplay dFnet+/integraldisplay dfnet=/integraldisplay m¨Rdm Because the internal forces occur in action–reaction pairs,/integraltext dfnet=0. (External forces on the body are those without an internal reactant; the reactant lies outside the bodyand, hence, outside our purview.) Thus F net=/integraldisplay m¨Rdm (9.11) where Fnetis the resultant external force on the body, Fnet=/integraltext dFnet.F r o m Equation 9.9 /integraldisplay m¨Rdm=m¨RG 410 Chapter 9 Rigid-body dynamics where ¨RG=aG, the absolute acceleration of the center of mass. Therefore, Equation 9.11 can be written Fnet=m¨RG (9.12) We are therefore reminded that the motion of the center of mass of a body is deter- mined solely by the resultant of the external forces acting on it. So far our study oforbiting bodies has focused exclusively on the motion of their centers of mass. In thischapter we will turn our attention to rotational motion around the center of mass. T osimplify things, we will ultimately assume that the body is not only continuous, butthat it is also rigid. That means all points of the body remain a fixed distance fromeach other and there is no flexing, bending or twisting deformation. 9.4 Equations of rotational motion Our development of the rotational dynamics equations does not require at the outset that the body under consideration be rigid. It may be a solid, liquid or gas. Point Pin the Figure 9.7 is arbitrary; it need not be fixed in space nor attached to a point on the body. Then the moment about Pof the forces on mass element dm (cf. Figure 9.6) is dMP=r×dFnet+r×dfnet where ris the position vector of the mass element dmrelative to the point P. Writing the right-hand side as r×(dFnet+dfnet), substituting Equation 9.10, and integrating XYZ G Pdm R r RG RPrG/P/rho1 Figure 9.7 Position vectors of a mass element in a continuum from several key reference points. 9.4 Equations of rotational motion 411 over all of the mass elements of the body yields MPnet=/integraldisplay mrרRdm (9.13) where ¨Ris the absolute acceleration of dmrelative to the inertial frame and MPnet=/integraldisplay r×dFnet+/integraldisplay r×dfnet But/integraltext r×dfnet=0because the internal forces occur in action–reaction pairs. Thus, MPnet=/integraldisplay r×dFnet which means the net moment includes only the moment of all of the external forces on the body. Observe that rרR=d dt(r×˙R)−˙r×˙R (9.14) Since r=R−RP,w h e r e RPis the absolute position vector of P, it is true that ˙r×˙R=(˙R−˙RP)×˙R=−˙RP×˙R (9.15) Substituting Equation 9.15 into Equation 9.14, then moving that result into Equation 9.13 yields MPnet=d dt/integraldisplay mr×˙Rdm+˙RP×/integraldisplay m˙Rdm (9.16) Now, r×˙Rdmis the absolute angular momentum of mass element dmabout P. The angular momentum of the entire body is the integral of this cross product overall of its mass elements. That is, the absolute angular momentum of the body relativeto point Pis H P=/integraldisplay mr×˙Rdm (9.17) Observing from Figure 9.7 that r=rG/P+/rho1, we can write Equation 9.17 as HP=/integraldisplay m(rG/P+/rho1)×˙Rdm=rG/P×/integraldisplay m˙Rdm+/integraldisplay m/rho1×˙Rdm (9.18) The last term is the absolute angular momentum relative to the center of mass G, HG=/integraldisplay /rho1×˙Rdm (9.19) Furthermore, by the definition of center of mass, Equation 9.9, /integraldisplay m˙Rdm=m˙RG (9.20) Equations 9.19 and 9.20 allow us to write Equation 9.18 as HP=HG+rG/P×mvG (9.21) 412 Chapter 9 Rigid-body dynamics This useful relationship shows how to obtain the absolute angular momentum about any point Ponce HGis known. For calculating the angular momentum about the center of mass, Equation 9.19 can be cast in a much more useful form by making the substitution (cf. Figure 9.7)R=R G+/rho1, so that HG=/integraldisplay m/rho1×(˙RG+˙/rho1)dm=/integraldisplay m/rho1×˙RGdm+/integraldisplay m/rho1×˙/rho1dm In the two integrals on the right, the variable is /rho1.˙RGis fixed and can therefore be factored out of the first integral to obtain HG=/parenleftbigg/integraldisplay m/rho1dm/parenrightbigg ×˙RG+/integraldisplay m/rho1×˙/rho1dm By definition of the center of mass,/integraltext m/rho1dm=0(the position vector of the center of mass relative to itself is zero), which means HG=/integraldisplay m/rho1×˙/rho1dm (9.22) Since /rho1and˙/rho1are the position and velocity relative to the center of mass G,/integraltext m/rho1×˙/rho1dm is the total moment about the center of mass of the linear momentum relative to thecenter of mass, H Grel. In other words, HG=HGrel (9.23) This is a rather surprising fact, hidden in Equation 9.19 and true in general for no other point of the body. Another useful angular momentum formula, similar to Equation 9.21, may be found by substituting R=RP+rinto Equation 9.17, HP=/integraldisplay mr×(˙RP+˙r)dm=/parenleftbigg/integraldisplay mrdm/parenrightbigg ×˙RP+/integraldisplay mr×˙rdm (9.24) The term on the far right is the net moment of relative linear momentum about P, HPrel=/integraldisplay mr×˙rdm (9.25) Also,/integraltext mrdm=mrG/P,w h e r e rG/Pis the position of the center of mass relative to P. Thus, Equation 9.24 can be written HP=HPrel+rG/P×mvP (9.26) Finally, substituting this into Equation 9.21, solving for HPrel, and noting that vG−vP=vG/Pyields HPrel=HG+rG/P×mvG/P (9.27) This expression is useful when the absolute velocity vGof the center of mass, which is required in Equation 9.21, is not available. 9.4 Equations of rotational motion 413 So far we have written down some formulas for calculating the angular momen- tum about an arbitrary point in space and about the center of mass of the bodyitself. Let us now return to the problem of relating angular momentum to the appliedtorque. Substituting Equation 9.17 into 9.16 and noting that by definition of thecenter of mass, /integraldisplay m˙Rdm=m˙RG we obtain MPnet=˙HP+˙RP×m˙RG Thus, for an arbitrary point P, MPnet=˙HP+vP×mvG (9.28) where vPand vGare the absolute velocities of points Pand G, respectively. This expression is applicable to two important special cases. If the point Pis at rest in inertial space ( vP=0), then Equation 9.28 reduces to MPnet=˙HP (9.29) This equation holds as well if vPand vGare parallel, e.g., if Pis the point of contact of a wheel rolling while slipping in the plane. Note that the validity of Equation 9.29depends neither on the body’s being rigid nor on its being in pure rotation about P. If point Pis chosen to be the center of mass, then, since v G×vG=0, Equation 9.28 becomes MGnet=˙HG (9.30) This equation is valid for any state of motion. If Equation 9.30 is integrated over a time interval, then we obtain the angular impulse–momentum principle, /integraldisplayt2 t1MGnetdt=HG2−HG1 (9.31) A similar expression follows from Equation 9.29./integraltext Mdtis the angular impulse. If the net angular impulse is zero, then /Delta1H=0, which is a statement of the conservation of angular momentum. Keep in mind that the angular impulse–momentum principleis not valid for just any reference point. Additional versions of Equations 9.29 and 9.30 can be obtained which may prove useful in special circumstances. For example, substituting the expression for H P (Equation 9.21) into Equation 9.28 yields MPnet=/bracketleftbigg ˙HG+d dt(rG/P×mvG)/bracketrightbigg +vP×mvG =˙HG+d dt[(rG−rP)×mvG]+vP×mvG =˙HG+(vG−vP)×mvG+rG/P×maG+vP×mvG 414 Chapter 9 Rigid-body dynamics or, finally, MPnet=˙HG+rG/P×maG (9.32) This expression is useful when it is convenient to compute the net moment about a point other than the center of mass. Alternatively, by simply differentiating Equation9.27 we get ˙H Prel=˙HG+=0/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright vG/P×mvG/P+rG/P×maG/P Solving for ˙HG, invoking Equation 9.30, and using the fact that aP/G=− aG/P leads to MGnet=˙HPrel+rG/P×maP/G (9.33) Finally, if the body is rigid, the magnitude of the position vector /rho1of any point relative to the center of mass does not change with time. Therefore, Equation 1.24 requiresthat˙/rho1=ω×/rho1, leading us to conclude H G=/integraldisplay m/rho1×(ω×/rho1)dm (9.34) Again, the absolute angular momentum about the center of mass depends only on the absolute angular velocity and not on the absolute translational velocity of any pointof the body. No such simplification of Equation 9.17 exists for an arbitrary reference point P. However, if the point Pis fixed in inertial space and the rigid body is rotating about P, then the magnitude of the position vector rfrom Pto any point of the body is constant. It follows from Equation 1.24 that ˙r=ω×r. According to Figure 9.7, R=R p+r Differentiating with respect to time gives ˙R=˙Rp+˙r=0+ω×r=ω×r Substituting this into Equation 9.17 yields the formula for angular momentum in this special case, HP=/integraldisplay mr×(ω×r)dm (9.35) Although Equations 9.34 and 9.35 are mathematically identical, one must keep in mind the notation of Figure 9.7. Equation 9.35 applies only if the rigid body is inpure rotation about a stationary point in ine rtial space, whereas Equation 9.34 applies unconditionally to any situation. 9.5 Moments of inertia T o use Equation 9.29 or 9.30 to solve problems, the vectors within them have to be resolved into components. T o find the components of angular momentum, we must 9.5 Moments of inertia 415 XYZ Gdm xy z/H9275 ˆiˆj ˆk/rho1 Figure 9.8 Co-moving xyzframe used to compute the moments of inertia. appeal to its definition. We will focus on the formula for angular momentum of a rigid body about its center of mass, Equation 9.34, because the expression for fixed-pointrotation (Equation 9.35) is mathematically the same. The integrand of Equation 9.34can be rewritten using the bac−cabvector identity presented in Equation 2.23, /rho1×(ω×/rho1)=ω/rho1 2−/rho1(ω·/rho1) (9.36) Let the origin of a co-moving xyzcoordinate system be attached to G, as shown in Figure 9.8. The unit vectors of this frame are ˆi,ˆjandˆk. The vectors /rho1andω can be resolved into components in the xyzd i r e c t i o n st og e t /rho1=xˆi+yˆj+zˆkand ω=ωxˆi+ωyˆj+ωzˆk. Substituting these vector expressions into the right side of Equation 9.36 yields /rho1×(ω×/rho1)=(ωxˆi+ωyˆj+ωzˆk)(x2+y2+z2)−(xˆi+yˆj+zˆk)(ω xx+ωyy+ωzz) Expanding the right side and collecting terms having the unit vectors ˆi,ˆjandˆkin common, we get /rho1×(ω×/rho1)=[(y2+z2)ωx−xyωy−xzωz]ˆi +[−yxωx+(x2+z2)ωy−yzωz]ˆj +[−zxωx−zyωy+(x2+y2)ωz]ˆk (9.37) We put this result into the integrand of Equation 9.34 to obtain HG=Hxˆi+Hyˆj+Hzˆk (9.38) where  Hx Hy Hz  = IxIxy Ixz Iyx IyIyz Izx Izy Iz   ωx ωy ωz  (9.39a) 416 Chapter 9 Rigid-body dynamics or, in matrix notation, {H}=[I]{ω} (9.39b) The components of the moment of inertia matrix [ I] about the center of mass are Ix=/integraltext (y2+z2)dm I xy=−/integraltext xydm I xz=−/integraltext xzdm Iyx=Ixy Iy=/integraltext (x2+z2)dm I yz=−/integraltext yzdm Izx=Ixz Izy=Iyz Iz=/integraltext (x2+y2)dm(9.40) [I] is clearly a symmetric matrix: [ I]T=[I]. Observe that, whereas the products of inertia Ixy,Ixzand Iyzcan be positive, negative or zero, the moments of inertia Ix, Iyand Izare always positive (never zero or negative) for bodies of finite dimensions. For this reason, [ I] is a positive-definite matrix. Keep in mind that Equations 9.38 and 9.39 are valid as well for axes attached to a fixed point Pabout which the body is rotating. The moments of inertia reflect how the mass of a rigid body is distributed. They manifest a body’s rotational inertia, its resistance to being set into rotary motion orstopped once rotation is under way. It is not an object’s mass alone but how that massis distributed which determines how the body will respond to applied torques. It is easy to show that the following statements are true: If the xyplane is a plane of symmetry of the body, then I xz=Iyz=0. If the xzplane is a plane of symmetry of the body, then Ixy=Iyz=0. If the yzplane is a plane of symmetry of the body, then Ixy=Ixz=0. Obviously, if the body has just two planes of symmetry relative to the xyzframe of reference, then all three products of inertia vanish, and [ I] becomes a diagonal matrix, [I]= A00 0B0 00 C  (9.41) where A,BandCare the principal moments of inertia (all positive), and the xyzaxes are the principal axes of inertia. In this case, relative to either the center of mass or afixed point of rotation, we have H x=AωxHy=BωyHz=Cωz (9.42) In general, the angular velocity ωand the angular momentum Hare not parallel. However, if (for example) ω=ωˆi, then according to Equations 9.42, {H}=A{ω}.I n other words, if the angular velocity points in a principal direction, so does the angularmomentum. In that case the two vectors are indeed parallel. Each of the three principal moments of inertia can be expressed as follows: A=mk 2 xB=mk2 yC=mk2 z (9.43) where mis the mass of the body and kx,kyand kzare the three radii of gyration. One may imagine the mass of a body to be concentrated around a principal axis at adistance equal to the radius of gyration. The moments of inertia for several common shapes are listed in Figure 9.9. By symmetry, their products of inertia vanish for the coordinate axes used. Formulas 9.5 Moments of inertia 417 l/2l/2 zx r Grl/2l/2 Gzx b al/2l/2 G xyz (a) (b) (c)Ix /H110051 4mr2 /H110011 12ml2Ix /H110051 2mr2 /H110011 12ml2Ix /H11005 m(a2 /H11001 l2)1 12 Iy /H11005 m(b2 /H11001 l2)1 12 Iz /H11005 m(a2 /H11001 b2)1 12Iz /H110051 2mr2Iz /H11005 mr2 Figure 9.9 Moments of inertia for three common homogeneous solids of mass m. (a) Solid circular cylinder. (b) Circular cylindrical shell . (c) Rectangular parallelepiped. for other solid geometries can be found in engineering handbooks and in dynamics textbooks. For a mass concentrated at a point, the moments of inertia in Equation 9.40 are just the mass times the integrand evaluated at the point. That is, the moment ofinertia matrix [ I m] of a point mass mis given by [Im]= m(y2+z2)−mxy −mxz −mxy m(x2+z2)−myz −mxz −myz m (x2+y2)  (9.44) Example 9.4The following table lists mass and coordinates of seven point masses. Find the center of mass of the system and the moments of inertia about the origin. Point, i Mass, mi(kg) xi(m) yi(m) zi(m) 13 −0.5 0.2 0.3 2 7 0.2 0.75 −0.4 35 1 −0.8 0.9 4 6 1.2 −1.3 1.25 52 −1.3 1.4 −0.8 64 −0.3 1.35 0.75 7 1 1.5 −1.7 0.85 The total mass of this system is m=7/summationdisplay i=1mi=28 kg 418 Chapter 9 Rigid-body dynamics (Example 9.4 continued)For concentrated masses the integral in Equation 9.9 is replaced by the mass times itsposition vector. Therefore, in this case the three components of the position vector of the center of mass are xG=7/summationtext i=1mixi m=0.35 m yG=7/summationtext i=1miyi m=0.01964 m zG=7/summationtext i=1mizi m=0.4411 m The total moment of inertia is the sum over all of the particles of Equation 9.44 evaluated at each point. Thus, [I]=(1)/bracehtipdownleft /bracehtipupright/bracehtipupleft /bracehtipdownright 0.39 0 .30 .45 0.31.02−0.18 0.45−0.18 0 .87 +(2)/bracehtipdownleft /bracehtipupright/bracehtipupleft /bracehtipdownright 5.0575 −1.05 0 .56 −1.05 1 .42 .1 0.56 2 .14.2175 +(3)/bracehtipdownleft /bracehtipupright/bracehtipupleft /bracehtipdownright 7.25 4 −4.5 49.05 3 .6 −4.53.68.2  +(4)/bracehtipdownleft /bracehtipupright/bracehtipupleft /bracehtipdownright 19.515 9 .36 −9 9.36 18 .015 9 .75 −99 .75 18 .78 +(5)/bracehtipdownleft /bracehtipupright/bracehtipupleft /bracehtipdownright 5.23 .64−2.08 3.64 4 .66 2 .24 −2.08 2.24 7 .3  +(6)/bracehtipdownleft /bracehtipupright/bracehtipupleft /bracehtipdownright 9.54 1 .62 0 .9 1.62 2 .61−4.05 0.9−4.05 7 .65 +(7)/bracehtipdownleft /bracehtipupright/bracehtipupleft /bracehtipdownright 3.6125 2 .55−1.275 2.55 2 .9725 1 .445 −1.275 1 .445 5 .14  or [I]= 50.56 20 .42−14.94 20.42 39 .73 14 .90 −14.94 14 .90 52 .16 (kg·m2) Example 9.5Calculate the moments of inertia of a slender, homogeneous straight rod of length l and mass m. One end of the rod is at the origin and the other has coordinates ( a,b,c). l (0, 0, 0) xyz AB (a, b, c) Figure 9.10 Uniform slender bar of mass mand length l. 9.5 Moments of inertia 419 A slender rod is one whose cross-sectional dimensions are negligible compared with its length. The mass is concentrated along its centerline. Since the rod is homogeneous, the mass per unit length /rho1is uniform and given by /rho1=m l(a) The length of the rod is l=/radicalbig a2+b2+c2 Starting with Ix, we have from Equations 9.40, Ix=/integraldisplayl 0(y2+z2)/rho1ds in which we replaced the element of mass dmby/rho1ds,w h e r e dsis the element of length along the rod. sis measured from end Aof the rod, so that the x,yand zcoordinates of any point along it are found in terms of sby the following relations, x=s lay=s lbz=s lc Thus Ix=/integraldisplayl 0/parenleftbiggs2 l2b2+s2 l2c2/parenrightbigg /rho1ds=/rho1b2+c2 l2/integraldisplayl 0s2ds=1 3/rho1(b2+c2)l Substituting (a) yields Ix=1 3m(b2+c2) In precisely the same way we find Iy=1 3m(a2+c2)Iz=1 3m(a2+b2) ForIxywe have Ixy=−/integraldisplayl 0xy/rho1ds=−/integraldisplayl 0s la·s lb/rho1ds=−/rho1ab l2/integraldisplayl 0s2ds=−1 3/rho1abl Once again using (a), Ixy=−1 3mab Likewise, Ixz=−1 3mac I yz=−1 3mbc 420 Chapter 9 Rigid-body dynamics Example 9.6The gyro rotor in Example 9.3 has a mass mof 5 kg, radius rof 0.08 m, and thickness tof 0.025 m. If N=2.1 rad/s, ˙θ=4 rad/s, ω=10.5 rad/s , and θ=60◦, calculate the angular momentum of the rotor about its center of mass Gin the xyzframe. What is the angle between the rotor’s angular velocity vector and its angular momentum vector? X YZ z xy G Rotor rt N ωspinθ Figure 9.11 Rotor of the gyroscope in Figure 9.4. In Example 9.3, Equation (f) gives the components of the absolute angular velocity of the rotor in the moving xyzframe, ωx=˙θ=4r a d/s ωy=Nsinθ=2.1·sin 60◦=1.819 rad /s( a ) ωz=ωspin+Ncosθ=10.5+2.1·cos 60◦=11.55 rad/s or ω=4ˆi+1.819ˆj+11.55ˆk(rad/s) (b) All three coordinate planes of the xyzframe contain the center of mass Gand all are planes of symmetry of the circular cylindrical rotor. Therefore, Ixy=Izx=Iyz=0 From Figure 9.9(a), we see that the non-zero diagonal entries in the moment of inertia tensor are A=B=1 12mt2+1 4mr2=1 125·0.0252+1 45·0.082=0.008260 kg ·m2 C=1 2mr2=1 25·0.082=0.0160 kg ·m2(c) 9.5 Moments of inertia 421 We can use Equation 9.42 to calculate the angular momentum, because the origin of thexyzframe is the rotor’s center of mass (which in this case also happens to be a fixed point of rotation, which is another reason we can use Equation 9.42). Substituting (a) and (c) into Equation 9.42 yields Hx=Aωx=0.008260 ·4=0.03304 kg ·m2/s Hy=Bωy=0.008260 ·1.819=0.0150 kg ·m2/s( d) Hz=Cωz=0.0160 ·11.55=0.1848 kg ·m2/s or H=0.03304 ˆi+0.0150 ˆj+0.1848 ˆk(kg·m2/s) (e) The angle φbetween Handωis found by using the dot product operation, φ=cos−1/parenleftbiggH·ω Hω/parenrightbigg =cos−1/parenleftbigg2.294 0.1883 ·12.36/parenrightbigg =9.717◦(f) As this illustrates, the angular momentum and the angular velocity are in general not collinear. Consider a coordinate system x/primey/primez/primewith the same origin as xyz, but different orientation. Let [ Q] be the orthogonal matrix ([ Q]−1=[Q]T) which transforms the components of a vector from thexyzsystem tothex/primey/primez/primeframe. Recall from Section 4.5 that the rows of [ Q] are the direction cosines of the x/primey/primez/primeaxes relative to xyz.T h e components of the angular momentum vector transform as follows {H/prime}=[Q ]{H} From Equation 9.39 we can write this as {H/prime}=[Q ][I]{ω} (9.45) Like the angular momentum vector, the components of the angular velocity vector in thexyzsystem are related to those in the primed system by the expression {ω/prime}=[Q ]{ω} The inverse relation is simply {ω}=[Q ]−1{ω/prime}=[Q ]T{ω/prime} (9.46) Substituting this into Equation 9.45, we get {H/prime}=[Q ][I][Q ]T{ω/prime} (9.47) But the components of angular momentum and angular velocity in the x/primey/primez/primeframe are related by an equation of the same form as Equation 9.39, so that {H/prime}=[I/prime]{ω/prime} (9.48) where [ I/prime] comprises the components of the inertia matrix in the primed system. Comparing the right-hand sides of Equations 9.47 and 9.48, we conclude that [I/prime]=[Q][I][Q ]T(9.49a) 422 Chapter 9 Rigid-body dynamics that is,  Ix/primeIx/primey/primeIx/primez/prime Iy/primex/primeIy/primeIy/primez/prime Iz/primex/primeIz/primey/primeIz/prime = Q11 Q12 Q13 Q21 Q22 Q23 Q31 Q32 Q33  IxIxy Ixz Iyx IyIyz Izx Izy Iz  Q11 Q21 Q31 Q12 Q22 Q32 Q13 Q32 Q33  (9.49b) This shows how to transform the components of the inertia matrix from the xyz coordinate system to any other orthogonal system with a common origin. Thus, forexample, I x/prime=⌊Row 1 ⌋/bracehtipdownleft/bracehtipupright/bracehtipupleft /bracehtipdownright/floorleftbigQ11 Q12 Q13/floorrightbig IxIxy Ixz Iyx IyIyz Izx Izy Iz ⌊Row 1 ⌋T /bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright  Q11 Q12 Q13   (9.50) Iy/primez/prime=⌊Row 2 ⌋/bracehtipdownleft/bracehtipupright/bracehtipupleft /bracehtipdownright/floorleftbigQ21 Q22 Q23/floorrightbig IxIxy Ixz Iyx IyIyz Izx Izy Izz ⌊Row 3 ⌋T /bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright  Q31 Q32 Q33   etc. Any object represented by a square matrix whose components transform according to Equation 9.49 is called a second order tensor. Example 9.7Find the mass moment of inertia of the system in Example 9.4 about an axis from the origin through the point with coordinates (2 m, −3 m, 4 m). From Example 9.4 the moment of inertia tensor for the system of point masses is [I]= 50.56 20 .42−14.94 20.42 39 .73 14 .90 −14.94 14 .90 52 .16 (kg·m2) The vector connecting the origin with (2 m, −3 m, 4 m) is V=2ˆi−3ˆj+4ˆk The unit vector in the direction of Vis ˆuV=V /bardblV/bardbl=0.3714ˆi−0.5571ˆj+0.7428ˆk We may consider ˆuVas the unit vector along the x/primeaxis of a rotated cartesian coordinate system. Then, from Equation 9.50, IV/prime=/floorleftbig0.3714 −0.5571 0 .7428/floorrightbig 50.56 20 .42−14.94 20.42 39 .73 14 .90 −14.94 14 .90 52 .16   0.3714 −0.5571 0.7428   =/floorleftbig0.3714 −0.5571 0 .7428/floorrightbig  −3.695 −3.482 24.90  =19.06 kg·m2 9.5 Moments of inertia 423 Example 9.8For the satellite of Example 9.2, reproduced in Figure 9.12, the data is as follows. N=0.1r a d /s and ˙θ=0.01 rad /s, in the directions shown. θ=40◦.d0=1.5m .T h e length, width and thickness of the panel are l=6m , w=2 m and t=0.025 m. The uniformly distributed mass of the panel is 50 kg. Find the angular momentum of the panel relative to the center of mass Oof the satellite. zx y lθ d0G x' y'z'N O w/2w/2 Figure 9.12 Satellite and solar panel. We can treat the panel as a thin parallelepiped. The panel’s xyzaxes have their origin at the center of mass Gof the panel and are parallel to its three edge directions. According to Figure 9.9(c), the moments of inertia relative to the xyzcoordinate system are IGx=1 12m(l2+t2)=1 12·50·(62+0.0252)=150.0k g ·m2 IGy=1 12m(w2+t2)=1 12·50·(22+0.0252)=16.67 kg ·m2(a) IGz=1 12m(w2+l2)=1 12·50·(22+62)=166.7k g ·m2 IGxy=IGxz=IGyz=0 In matrix notation, [IG]= 150.00 0 01 6 .67 0 0 0 166.7 (kg·m2) (b) The unit vectors of the satellite’s x/primey/primez/primesystem are related to those of panel’s xyzframe by inspection, ˆi/prime=− sinθˆi+cosθˆk=− 0.6428 ˆi+0.7660 ˆk ˆj/prime=−ˆj (c) ˆk/prime=cosθˆi+sinθˆk=0.7660 ˆi+0.6428 ˆk 424 Chapter 9 Rigid-body dynamics (Example 9.8 continued)The matrix [ Q] of the transformation from xyztox/primey/primez/primecomprises the direction cosines of ˆi/prime,ˆj/primeandˆk/prime: [Q]= −0.6428 0 0 .7660 0 −10 0.7660 0 0 .6428  (d) In Example 9.2 we found that the absolute angular velocity of the panel, in the satellite’s x/primey/primez/primeframe of reference, is ω=−˙θˆj/prime+Nˆk/prime=− 0.01ˆj/prime+0.1ˆk/prime(rad/s) That is, {ω/prime}=  0 −0.01 0.1  (rad/s) (e) T o find the absolute angular momentum {H/prime G}in the satellite system requires using Equation 9.39, {H/prime G}=[I/prime G]{ω/prime} (f) Before doing so, we must transform the components of the moments of inertia tensor in (b) from the unprimed system to the primed system, by means of Equation 9.49, [I/prime G]=[Q][IG][Q]T = −0.6428 0 0 .7660 0 −10 0.7660 0 0 .6428  150.00 0 01 6 .67 0 0 0 166 .7  −0.6428 0 0 .7660 0 −10 0.7660 0 0 .6428  so that [I/prime G]= 159.808 .205 01 6 .67 0 8.205 0 156 .9 (kg·m2)( g ) Then (f) yields {H/prime G}= 159.808 .205 01 6 .67 0 8.205 0 156 .9   0 −0.01 0.1  =  0.8205 −0.1667 15.69  (kg·m2/s) or, in vector notation, HG=0.8205ˆi/prime−0.1667ˆj/prime+15.69ˆk/prime(kg·m2/s) (h) This is the absolute angular momentum of the panel about its own center of mass, and it is used in Equation 9.27 to calculate the angular momentum HOrelrelative to the satellite’s center of mass O, HOrel=HG+rG/O×mvG/O (i) 9.5 Moments of inertia 425 rG/Ois the position vector from OtoG, rG/O=/parenleftbigg d0+l 2/parenrightbigg ˆj/prime=/parenleftbigg 1.5+6 2/parenrightbigg ˆj/prime=4.5ˆj/prime(m) (j) The velocity of Grelative to O,vG/O, is found from Equation 9.2, vG/O=ωsatellite ×rG/O=Nˆk/prime×rG/O=0.1ˆk/prime×4.5ˆj/prime=− 0.45ˆi/prime(m/s) (k) Substituting (h), (j) and (k) into (i) finally yields HOrel=(0.8205 ˆi/prime−0.1667 ˆj/prime+15.69ˆk/prime)+4.5ˆj/prime×[50(−0.45 ˆi/prime)] =0.8205 ˆi/prime−0.1667 ˆj/prime+116.9ˆk/prime(kg·m2/s) (l) Note that we were unable to use Equation 9.21 to find the absolute angular momentum HObecause that requires knowing the absolute velocity vG, which in turn depends on the absolute velocity of O, which was not provided. How can we find that transformation matrix [ Q] such that Equation 9.49 will yield a moment of inertia matrix [ I/prime] which is diagonal, i.e., of the form given by Equation 9.41? In other words, how do we find the principal directions of the moment of inertiatensor? Let the angular velocity vector {ω}be parallel to the principal direction defined by the vector {v}, so that {ω}=β {v},w h e r e βis a scalar. Since {ω}points in a principal direction of the inertia tensor, so must {H}, which means {H} is also parallel to {v}. Therefore, {H}=α {v},w h e r e αis a scalar. From Equation 9.39 it follows that α{v}=[I](β {v}) or [I]{v}=λ{v } where λ=α/β (a scalar). That is,  I xIxy Ixz Ixy IyIyz Ixz Iyz Iz   vx vy vz  =λ  vx vy vz   This can be written  Ix−λ Ixy Ixz Ixy Iy−λ Iyz Ixz Iyz Iz−λ   vx vy vz  =  0 00  (9.51) The trivial solution of Equation 9.51 is {v}={ 0}, which is of no interest. The only way that Equation 9.51 will not yield the trivial solution is if the coefficient matrix onthe left is singular. That will occur if its determinant vanishes, that is, if /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleI x−λ Ixy Ixz Ixy Iy−λ Iyz Ixz Iyz Iz−λ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=0 (9.52) 426 Chapter 9 Rigid-body dynamics Expanding the determinant, we find /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleI x−λ Ixy Ixz Ixy Iy−λ Iyz Ixz Iyz Iz−λ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=−λ 3+I1λ2−I2λ+I3 (9.53) where I1=Ix+Iy+Iz I2=/vextendsingle/vextendsingle/vextendsingle/vextendsingleI xIxy Ixy Iy/vextendsingle/vextendsingle/vextendsingle/vextendsingle+/vextendsingle/vextendsingle/vextendsingle/vextendsingleI xIxz Ixz Iz/vextendsingle/vextendsingle/vextendsingle/vextendsingle+/vextendsingle/vextendsingle/vextendsingle/vextendsingleI yIyz Iyz Iz/vextendsingle/vextendsingle/vextendsingle/vextendsingle(9.54) I 3=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleI xIxy Ixz Ixy IyIyz Ixz Iyz Iz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle Equations 9.52 and 9.53 yield the characteristic equation of the tensor [ I] λ 3−I1λ2+I2λ−I3=0 (9.55) The three roots λp(p=1, 2, 3) of this cubic equation are real, since [ I] is symmetric; furthermore they are all positive, since [ I] is a positive-definite matrix. Each root, or eigenvalue, λpis substituted back into Equation 9.51 to obtain  Ix−λp Ixy Ixz Ixy Iy−λp Iyz Ixz Iyz Iz−λp   v(p) x v(p) y v(p) z  =  0 00  ,p=1, 2, 3 (9.56) Solving this system yields the three eigenvectors {v (p)}corresponding to each of the three eigenvalues λp. The three eigenvectors are orthogonal, also due to the symmetry of the matrix [ I]. Each eigenvalue is a principal moment of inertia, and its corresponding eigenvector is a principal direction. Example 9.9Find the principal moments of inertia and the principal axes of inertia of the inertia tensor [I]= 100−20−100 −20 300 −50 −100−50 500 kg·m2 We seek the non-trivial solutions of the system  100−λ−20 −100 −20 300 −λ−50 −100 −50 500 −λ   vx vy vz  =  0 0 0  (a) 9.5 Moments of inertia 427 From Equation 9.54, I1=100+300+500=900 I2=/vextendsingle/vextendsingle/vextendsingle/vextendsingle100−20 −20 300/vextendsingle/vextendsingle/vextendsingle/vextendsingle+/vextendsingle/vextendsingle/vextendsingle/vextendsingle100−100 −100 500/vextendsingle/vextendsingle/vextendsingle/vextendsingle+/vextendsingle/vextendsingle/vextendsingle/vextendsingle300−50 −50 500/vextendsingle/vextendsingle/vextendsingle/vextendsingle=217 100 (b) I 3=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle100−20 −100 −20 300 −50 −100 −50 500/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=11 350 000 Thus, the characteristic equation is λ3−900λ2+217 100λ −11 350 000 =0( c) The three roots are the principal moments of inertia, which are found to be λ1=532.052 λ2=295.840 λ3=72.1083 (d) Each of these is substituted, in turn, back into (a) to find its corresponding principal direction. Substituting λ1=532.052 kg ·m2into (a) we obtain  −432.052 −20.0000 −100.0000 −20.0000 −232.052 −50.0000 −100.0000 −50.0000 −32.0519   v(1) x v(1) y v(1) z  =  0 00  (e) Since the determinant of the coefficient matrix is zero, at most two of the three equations in (e) are independent. Thus, at most two of the three components of the vector v(1)can be found in terms of the third. We can therefore arbitrarily set v(1) x=1 and solve for v(1) yandv(1) zusing any two of the independent equations in (e). With v(1) x=1, the first two of Equations (e) become −20.0000v(1) y−100.000v(1) z=432.052 −232.052v(1) y−50.000v(1) z=20.0000 (f) Solving these two equations for v(1) yandv(1) zyields, together with the assumption onv(1) x, v(1) x=1.00000 v(1) y=0.882793 v(1) z=− 4.49708 (g) T o obtain the unit vector in the direction of v(1) ˆi1=v(1) /bardblv(1)/bardbl=1.00000 ˆi+0.882793 ˆj−4.49708 ˆk/radicalbig 1.000002+0.8827932+(−4.49708)2 =0.213186 ˆi+0.188199 ˆj−0.958714 ˆk (h) Substituting λ2=295.840 into (a) and proceeding as above we find ˆi2=0.176732 ˆi−0.972512 ˆj−0.151609 ˆk (i) 428 Chapter 9 Rigid-body dynamics (Example 9.9 continued)The two unit vectors ˆi1andˆi2define two of the principal directions of the inertia tensor. Observe that ˆi1·ˆi2=0, as must be the case for symmetric matrices. T o obtain the third principal direction ˆi3, we can substitute λ3=72.1083 into (a) and proceed as above. However, since the inertia tensor is symmetric, we know that the three principal directions are mutually orthogonal. That means ˆi3=ˆi1׈i2. Substituting Equations (h) and (i) into the cross product, we find that ˆi3=− 0.960894 ˆi−0.137114 ˆj−0.240587 ˆk (j) We can check our work by substituting λ3andˆi3into (a) and verify that it is indeed satisfied:  100−72.1083 −20 −100 −20 300 −72.1083 −50 −100 −50 500 −72.1083   −0.960894 −0.137114 −0.240587  =  0 00  (k) The components of ˆi1,ˆi2andˆi3define the rows of the orthogonal transformation [ Q] from the xyzsystem into the x/primey/primez/primesystem aligned along the three principal directions: [Q]= 0.213186 0 .188199 −0.958714 0.176732 −0.972512 −0.151609 −0.960894 −0.137114 −0.240587  (l) If we apply the transformation in Equation 9.49, [ I/prime]=[Q][I][Q]T, we find [I/prime]= 0.213186 0 .188199 −0.958714 0.176732 −0.972512 −0.151609 −0.960894 −0.137114 −0.240587  100−20−100 −20 300 −50 −100−50 500  × 0.213186 0 .176732 −0.960894 0.188199 −0.972512 −0.137114 −0.958714 −0.151609 −0.240587  = 532.052 0 0 0 295 .840 0 00 7 2 .1083 (kg·m2) 9.5.1 Parallel axis theorem Suppose the rigid body in Figure 9.13 is in pure rotation about point P. Then, according to Equation 9.39, {HPrel}=[IP]{ω} (9.57) where [ IP] is the moment of inertia about P, given by Equations 9.40 with x=xG/P+ξ y=yG/P+η z=zG/P+ζ 9.5 Moments of inertia 429 G P xyz yPO xPzPdm ηζ xG/P yG/PzG/Pξ Figure 9.13 The moments of inertia are to be computed at P, given their values at G. On the other hand, we have from Equation 9.27 that HPrel=HG+rG/P×mvG/P (9.58) The vector rG/P×mvG/Pis the angular momentum about Pof the concentrated mass mlocated at G. Using matrix notation, it is computed as follows, {rG/P×mvG/P}≡{ HmPrel}=[I mP]{ω} (9.59) where [ ImP], the moment of inertia of mabout P, is obtained from Equation 9.43, with x=xG/P,y=yG/Pand z=zG/P. That is, [ImP]= m(y2 G/P+z2 G/P)−mx G/PyG/P −mx G/PzG/P −mx G/PyG/P m(x2 G/P+z2 G/P)−my G/PzG/P −mx G/PzG/P −my G/PzG/P m(x2 G/P+y2 G/P)  (9.60) Of course, Equation 9.39 requires {HG}=[I G]{ω} Substituting this together with Equations 9.57 and 9.59 into Equation 9.58 yields [IP]{ω}=[I G]{ω}+[I mP]{ω}=([I G]+[ImP]){ω} From this we may infer the parallel axis theorem, [IP]=[IG]+[ImP] (9.61) 430 Chapter 9 Rigid-body dynamics The moment of inertia about Pis the moment of inertia about parallel axes through the center of mass plus the moment of inertia of the center of mass about P. That is, IPx=IGx+m(y2 G/P+z2 G/P)IPy=IGy+m(x2 G/P+z2 G/P)IPz=IGz+m(x2 G/P+y2 G/P) IPxz=IGxz−mxG/PzG/PIPxy=IGxy−mxG/PyG/PIPyz=IGyz−myG/PzG/P (9.62) Example 9.10Find the moments of inertia of the rod in Example 9.5 (Figure 9.14) about its center of mass G. l/2 xyz A(0, 0, 0)B (a, b, c) G (a/2, b/2, c/2)l/2 Figure 9.14 Uniform slender rod. From Example 9.5, [IA]= 1 3m(b2+c2)−1 3mab −1 3mac −1 3mab1 3m(a2+c2)−1 3mbc −1 3mac −1 3mbc1 3m(a2+b2)  Using Equation 9.62 1, and noting the coordinates of the center of mass in Figure 9.14, IGx=IAx−m[(yG−0)2+(zG−0)2] =1 3m(b2+c2)−m/bracketleftBigg/parenleftbiggb 2/parenrightbigg2 +/parenleftBigc 2/parenrightBig2/bracketrightBigg =1 12m(b2+c2) Equation 9.62 4yields IGxy=IAxy+m(xG−0)(yG−0)=−1 3mab+m·a 2·b 2=−1 12mab The remaining four moments of inertia are found in a similar fashion, so that [IG]= 1 12m(b2+c2)−1 12mab −1 12mac −1 12mab1 12m(a2+c2)−1 12mbc −1 12mac −1 12mbc1 12m(a2+b2)  (9.63) 9.5 Moments of inertia 431 Example 9.11Calculate the principal moments of inertia about the center of mass and the corre- sponding principal directions for the bent rod in Figure 9.15. Its mass is uniformlydistributed at 2 kg/m. xyz 4321 0.4 m 0.5 m 0.3 m 0.2 m O Figure 9.15 Bent rod for which the principal moments of inertia are to be determined. The mass of each of the rod segments is m(1)=2·0.4=0.8k g m(2)=2·0.5=1k g m(3)=2·0.3=0.6k g m(4)=2·0.2=0.4k g ( a ) The total mass of the system is m=4/summationdisplay i=1m(i)=2.8 kg (b) The coordinates of each segment’s center of mass are xG1=0 yG1=0 zG1=0.2m xG2=0 yG2=0.25 m zG2=0.2m xG3=0.15 m yG3=0.5m zG3=0 xG4=0.3m yG4=0.4m zG4=0(c) If the slender rod of Figure 9.14 is aligned with, say, the xaxis, then a=landb=c=0, so that according to Equation 9.63, [IG]= 00 0 01 12ml20 001 12ml2  That is, the moment of inertia of a slender rod about axes normal to the rod at its center of mass is1 12ml2,w h e r e mandlare the mass and length of the rod, respectively. 432 Chapter 9 Rigid-body dynamics (Example 9.11 continued)Since the mass of a slender bar is assumed to be concentrated along the axis of thebar (its cross-sectional dimensions are infinitesimal), the moment of inertia about the centerline is zero. By symmetry, the products of inertia about axes through the center of mass are all zero. Using this information and the parallel axis theorem, we find the moments and products of inertia of each rod segment about the origin Oof thexyzsystem as follows. Rod 1: I(1) x=I(1) G1/parenrightBig x+m(1)(y2 G1+z2 G1)=1 12·0.8·0.42+0.8(0+0.22)=0.04267 kg ·m2 I(1) y=I(1) G1/parenrightBig y+m(1)(x2 G1+z2 G1)=1 12·0.8·0.42+0.8(0+0.22)=0.04267 kg ·m2 I(1) z=I(1) G1/parenrightBig z+m(1)(x2 G1+y2 G1)=0+0.8(0+0)=0 I(1) xy=I(1) G1/parenrightBig xy−m(1)xG1yG1=0−0.8(0)(0) =0 I(1) xz=I(1) G1/parenrightBig xz−m(1)xG1zG1=0−0.8(0)(0 .2)=0 I(1) yz=I(1) G1/parenrightBig yz−m(1)yG1zG1=0−0.8(0)(0) =0 Rod 2: I(2) x=I(2) G2/parenrightBig x+m(2)(y2 G2+z2 G2)=1 12·1.0·0.52+1.0(0+0.252)=0.08333 kg ·m2 I(2) y=I(2) G2/parenrightBig y+m(2)(x2 G2+z2 G2)=0+1.0(0+0)=0 I(2) z=I(2) G2/parenrightBig z+m(2)(x2 G2+y2 G2)=1 12·1.0·0.52+1.0(0+0.52)=0.08333 kg ·m2 I(2) xy=I(2) G2/parenrightBig xy−m(2)xG2yG2=0−1.0(0)(0 .5)=0 I(2) xz=I(2) G2/parenrightBig xz−m(2)xG2zG2=0−1.0(0)(0) =0 I(2) yz=I(2) G2/parenrightBig yz−m(2)yG2zG2=0−1.0(0.5)(0)=0 Rod 3: I(3) x=I(3) G3/parenrightBig x+m(3)(y2 G3+z2 G3)=0+0.6(0.52+0)=0.15 kg·m2 I(3) y=I(3) G3/parenrightBig y+m(3)(x2 G3+z2 G3)=1 12·0.6·0.32+0.6(0.152+0)=0.018 kg ·m2 I(3) z=I(3) G3/parenrightBig z+m(3)(x2 G3+y2 G3)=1 12·0.6·0.32+0.6(0.152+0.52)=0.1680 kg ·m2 I(3) xy=I(3) G3/parenrightBig xy−m(3)xG3yG3=0−0.6(0.15)(0.5)=− 0.045 kg ·m2 I(3) xz=I(3) G3/parenrightBig xz−m(3)xG3zG3=0−0.6(0.15)(0) =0 I(3) yz=I(3) G3/parenrightBig yz−m(3)yG3zG3=0−0.6(0.5)(0)=0 9.5 Moments of inertia 433 Rod 4: I(4) x=I(4) G4/parenrightBig x+m(4)(y2 G4+z2 G4)=1 12·0.4·0.22+0.4(0.42+0)=0.06533 kg ·m2 I(4) y=I(4) G4/parenrightBig y+m(4)(x2 G4+z2 G4)=0+0.4(0.32+0)=0.0360 kg ·m2 I(4) z=I(4) G4/parenrightBig z+m(4)(x2 G4+y2 G4)=1 12·0.4·0.22+0.4(0.32+0.42)=0.1013 kg ·m2 I(4) xy=I(4) G4/parenrightBig xy−m(4)xG4yG4=0−0.4(0.3)(0.4) =− 0.0480 kg ·m2 I(4) xy=I(4) G4/parenrightBig xy−m(4)xG4yG4=0−0.4(0.3)(0.4) =− 0.0480 kg ·m2 I(4) xz=I(4) G4/parenrightBig xz−m(4)xG4zG4=0−0.4(0.3)(0) =0 I(4) yz=I(4) G4/parenrightBig yz−m(4)yG4zG4=0−0.4(0.4)(0) =0 The total moments of inertia for all four rods are Ix=4/summationdisplay i=1I(i) x=0.3413 kg ·m2Iy=4/summationdisplay i=1I(i) y=0.09667 kg ·m2 Iz=4/summationdisplay i=1I(i) z=0.3527 kg ·m2Ixy=4/summationdisplay i=1I(i) xy=− 0.0930 kg ·m2(d) Ixz=4/summationdisplay i=1I(i) xz=0 Iyz=4/summationdisplay i=1I(i) yz=0 The coordinates of the center of mass of the system of four rods are, from (a), (b) and (c), xG=1 m4/summationdisplay i=1m(i)xGi=1 2.8·0.21=0.075 m yG=1 m4/summationdisplay i=1m(i)yGi=1 2.8·0.71=0.2536 m (e) zG=1 m4/summationdisplay i=1m(i)zGi=1 2.8·0.16=0.05714 m We use the parallel axis theorems to shift the moments of inertia in (d) to the center of mass Gof the system: IGx=Ix−m(y2 G+z2 G)=0.3413 −0.1892 =0.1522 kg ·m2 IGy=Iy−m(x2 G+z2 G)=0.09667 −0.02489 =0.07177 kg ·m2 IGz=Iz−m(x2 G+y2 G)=0.3527 −0.1958 =0.1569 kg ·m2 IGxy=Ixy+mxGyG=− 0.093+0.05325 =− 0.03975 kg ·m2 IGxz=Ixz+mxGzG=0+0.012=0.012 kg ·m2 IGyz=Iyz+myGzG=0+0.04057 =0.04057 kg ·m2 434 Chapter 9 Rigid-body dynamics (Example 9.11 continued)Therefore the inertia tensor, relative to the center of mass, is [I]= IGxIGxyIGxz IGxyIGyIGyz IGxzIGyzIGz = 0.1522 −0.03975 0 .012 −0.03975 0 .07177 0 .04057 0.012 0 .04057 0 .1569 (kg·m2)( f ) T o find the three principal moments of inertia, we set  0.1522−λ−0.03975 0 .012 −0.03975 0 .07177 −λ 0.04057 0.012 0 .04057 0 .1569−λ =0 from which we obtain the characteristic equation −λ3+0.3808λ2−0.04268λ+0.001166 =0 The three roots are the principal moments of inertia, λ1=0.04023 kg ·m2λ2=0.1658 kg ·m2λ3=0.1747 kg ·m2(g) We substitute each of these principal values, in turn, into the equation,  0.1522−λp−0.03975 0 .012 −0.03975 0 .07177 −λp 0.04057 0.012 0 .04057 0 .1569−λp   v(p) x v(p) y v(p) z  =  0 00  (h) in order to determine the components of the three principal vectors v(p),p=1, 2, 3. λ1=0.04023: Equation (h) becomes  0.1119 −0.03975 0 .012 −0.03975 0 .03154 0 .04057 0.012 0 .04057 0 .1166   v(1) x v(1) y v(1) z  =  0 00   We can arbitrarily set v(1) x=1, so that the first two equations become −0.03975v(1) y+0.012v(1) z=− 0.1119 0.03154v(1) y+0.04057v(1) z=0.03975 Solving for v(1) yandv(1) zyields v(1) y=2.520 and v(1) z=− 0.9794, so that v(1)=ˆi+2.520ˆj−0.9794ˆk /bardblv(1)/bardbl= 2.883 Normalizing this vector and calling it ˆi1,w eg e t ˆi1=v(1) /bardblv(1)/bardbl=0.3470ˆi+0.8742ˆj−0.3397ˆk (i) 9.6 Euler’s equations 435 λ2=0.1658 : Equation (h) becomes  −0.0137 −0.03975 0.012 −0.03975 −0.09408 0.04057 0.012 0.04057 −0.008968   v(2) x v(2) y v(2) z  =  0 00   Repeating the above procedure, we obtain v(2)=ˆi−0.1625 ˆj+0.6030 ˆk, so that ˆi2=v(2) /bardblv(2)/bardbl=0.8482 ˆi−0.1378 ˆj+0.5115 ˆk λ3=0.1747 : The third principal vector ˆi3is the cross product of the first two: ˆi3=ˆi1׈i2=0.4003 ˆi−0.4656 ˆj−0.7893 ˆk Check this result to see that it satisfies Equation (h):  0.1522 −0.1747 −0.03975 0.012 −0.03975 0.07177 −0.1747 0.04057 0.012 0.04057 0.1569 −0.1747   0.4003 −0.4656−0.7893  √ =  0 00   9.6 Euler’s equations For either the center of mass Go rafi x e dp o i n tP about which the body is in pure rotation, we know from Equations 9.29 and 9.30 that Mnet=˙H (9.64) Using a co-moving coordinate system, with angular velocity /Omega1and its origin located at the point (G orP), the angular momentum has the analytical expression H=Hxˆi+Hyˆj+Hzˆk (9.65) We shall henceforth assume, for simplicity, that (a) The moving xyzaxes are the principal axes of inertia, and (9.66a) (b) The moments of inertia relative to xyzare constant in time. (9.66b) Equations 9.42 and 9.66a imply that H=Aωxˆi+Bωyˆj+Cωzˆk (9.67) where A,Band Care the principal moments of inertia. 436 Chapter 9 Rigid-body dynamics According to Equation 1.28, the time derivative of His˙H=˙H)rel+/Omega1×H,s o that Equation 9.64 can be written Mnet=˙H/parenrightbig rel+/Omega1×H (9.68) Keep in mind that, whereas /Omega1(the angular velocity of the moving xyzcoordinate sys- tem) and ω(the angular velocity of the rigid body itself) are both absolute kinematic quantities, Equation 9.68 contains their components as projected onto the axes of thenon-inertial xyzframe, ω=ω xˆi+ωyˆj+ωzˆk /Omega1=/Omega1xˆi+/Omega1yˆj+/Omega1zˆk The absolute angular acceleration αis obtained using Equation 1.28, α=˙ω=αrel/bracehtipdownleft /bracehtipupright/bracehtipupleft /bracehtipdownright dωx dtˆi+dωy dtˆj+dωz dtˆk+/Omega1×ω that is, α=(˙ωx+/Omega1yωz−/Omega1zωy)ˆi+(˙ωy+/Omega1zωx−/Omega1xωz)ˆj+(˙ωz+/Omega1xωy−/Omega1yωx)ˆk(9.69) Clearly, it is generally true that αx/negationslash=˙ωxαy/negationslash=˙ωyαz/negationslash=˙ωz From Equations 1.29 and 9.67 ˙H/parenrightbig rel=d(Aωx) dtˆi+d(Bωy) dtˆj+d(Cωz) dtˆk Since A,Band Care constant, this becomes ˙H/parenrightbig rel=A˙ωxˆi+B˙ωyˆj+C˙ωzˆk (9.70) Substituting Equations 9.67 and 9.70 into Equation 9.68 yields Mnet=A˙ωxˆi+B˙ωyˆj+C˙ωzˆk+/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆi ˆj ˆk /Omega1 x/Omega1y/Omega1z AωxBωyCωz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle Expanding the cross product and collecting terms leads to M xnet=A˙ωx+C/Omega1yωz−B/Omega1zωy Mynet=B˙ωy+A/Omega1zωx−C/Omega1xωz (9.71) Mznet=C˙ωz+B/Omega1xωy−A/Omega1yωx If the co-moving frame is a rigidly attached body frame, then its angular velocity is the same as that of the body, i.e., /Omega1=ω. In that case, Equations 9.68 reduce to Euler’s equations of motion, Mnet=˙H/parenrightbig rel+ω×H (9.72a) 9.6 Euler’s equations 437 the three components of which are obtained from Equation 9.71, Mxnet=A˙ωx+(C−B)ωyωz Mynet=B˙ωy+(A−C)ωzωx (9.72b) Mznet=C˙ωz+(B−A)ω xωy Equation 9.68 is sometimes referred to as the modified Euler equation. When /Omega1=ω, it follows from Equation 9.69 that ˙ωx=αx ˙ωy=αy ˙ωz=αz (9.73) That is, the relative angular acceleration equals the absolute angular acceleration when /Omega1=ω. Rather than calculating the time derivatives ˙ωx,˙ωyand˙ωzfor use in Equation 9.72, we may in this case first compute αin the absolute XYZ frame α=dω dt=dωX dtˆI+dωY dtˆJ+dωZ dtˆK, and then project these components onto the xyzbody frame, so that   ˙ωx ˙ωy ˙ωz  =[Q]Xx  dωX/dt dωY/dt dωZ/dt  (9.74) where [ Q]Xxis the time-dependent orthogonal transformation from the inertial XYZ frame to the non-inertial xyzframe. Example 9.12Calculate the net moment on the solar panel of Examples 9.2 and 9.8. zx y lθ O d0G x' y'z'N MGnet Fnetw/2 w/2 Figure 9.16 Free-body diagram of the solar panel in Examples 9.2 and 9.8. Since the co-moving frame is rigidly attach ed to the panel, Euler’s equation (Equation 9.72) applies to this problem, MGnet=˙HG/parenrightbig rel+ω×HG (a) where HG=Aωxˆi+Bωyˆj+Cωzˆk (b) 438 Chapter 9 Rigid-body dynamics (Example 9.12 continued)and ˙HG/parenrightbig rel=A˙ωxˆi+B˙ωyˆj+C˙ωzˆk (c) In Example 9.2, the angular velocity of the panel in the satellite’s x/primey/primez/primeframe was found to be ω=−˙θˆj/prime+Nˆk/prime (d) In Example 9.8, the transformation from the panel’s xyzframe to that of the satellite was shown to be represented by the matrix [Q]= −sinθ 0c o s θ 0 −10 cosθ 0 sin θ  (e) We use the transpose of [ Q] to transform the components of ωinto the panel frame of reference, {ω}xyz=[Q]T{ω}x/primey/primez/prime= −sinθ 0c o s θ 0 −10 cosθ 0 sin θ   0 −˙θ N  =  Ncosθ ˙θ Nsinθ   or ωx=Ncosθω y=˙θω z=Nsinθ (f) In Example 9.2, Nand˙θwere said to be constant. Therefore, the time derivatives of (f) are ˙ωx=d(Ncosθ) dt=− N˙θsinθ ˙ωy=d˙θ dt=0( g ) ˙ωz=d(Nsinθ) dt=N˙θcosθ In Example 9.8 the moments of inertia in the panel frame of reference were listed as A=1 12m(l2+t2) B=1 12m(w2+t2) C=1 12m(w2+l2) (IGxy=IGxz=IGyz=0) (h) Substituting (b), (c), (f), (g) and (h) into (a) yields, MGnet=1 12m(l2+t2)(−N˙θsinθ)ˆi+1 12m(w2+t2)·0·ˆj +1 12m(w2+l2)(N˙θcosθ)ˆk +/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆi ˆj ˆk Ncosθ ˙θ Nsinθ 1 12m(l2+t2)(Ncosθ)1 12m(w2+t2)˙θ1 12m(w2+l2)(Nsinθ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle 9.6 Euler’s equations 439 Upon expanding the cross product and collecting terms, this reduces to MGnet=−1 6mt2N˙θsinθˆi+1 24m(t2−w2)N2sin 2θˆj+1 6mw2N˙θcosθˆk Using the numerical data of Example 9.8 ( m=50 kg, N=0.1 rad/s, θ=40◦, ˙θ=0.01 rad/s, l=6m , w=2 m and t=0.025 m), we get MGnet=− 3.348×10−6ˆi−0.08205 ˆj+0.02554 ˆk(N·m) Example 9.13Calculate the net moment on the gyro rotor of Examples 9.3 and 9.6. Figure 9.17 is a free-body diagram of the rotor. Since in this case the co-moving frame is not rigidly attached to the rotor, we must use Equation 9.68 to find the net moment about G, MGnet=˙HG/parenrightbig rel+/Omega1×HG (a) where HG=Aωxˆi+Bωyˆj+Cωzˆk (b) and ˙HG/parenrightbig rel=A˙ωxˆi+B˙ωyˆj+C˙ωzˆk (c) Z z xy Gθ r NMGnet Fnettωspin Figure 9.17 Free-body diagram of the gyro rotor of Examples 9.3 and 9.6. From Equation (h) of Example 9.3 we know that the components of the angular velocity of the rotor in the moving reference frame are ωx=˙θ ωy=Nsinθ (d) ωz=ωspin+Ncosθ 440 Chapter 9 Rigid-body dynamics (Example 9.13 continued)Since, as specified in Example 9.3, ˙θ,Nandωspinare all constant, it follows that ˙ωx=d˙θ dt=0 ˙ωy=d(Nsinθ) dt=N˙θcosθ (e) ˙ωz=d(ωspin+Ncosθ) dt=− N˙θsinθ The angular velocity /Omega1of the co-moving xyzframe is that of the gimbal ring, which equals the angular velocity of the rotor minus its spin. Therefore, /Omega1x=˙θ /Omega1y=Nsinθ (f) /Omega1z=Ncosθ In Example 9.6 we found that A=B=1 12mt2+1 4mr2 (g) C=1 2mr2 Substituting (b) through (g) into (a), we get MGnet=/parenleftbig1 12mt2+1 4mr2/parenrightbig ·0ˆi+/parenleftbig1 12mt2+1 4mr2/parenrightbig (N˙θcosθ)ˆj +1 2mr2(−N˙θsinθ)ˆk +/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆi ˆj ˆk ˙θ Nsinθ Ncosθ/parenleftbig 1 12mt2+1 4mr2/parenrightbig˙θ/parenleftbig1 12mt2+1 4mr2/parenrightbig Nsinθ1 2mr2(ωspin+Ncosθ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle Expanding the cross product, collecting terms, and simplifying leads to MGnet=/bracketleftbigg1 2ωspin+1 12/parenleftbigg 3−t2 r2/parenrightbigg Ncosθ/bracketrightbigg mr2Nsinθˆi +/parenleftbigg1 6t2 r2Ncosθ−1 2ωspin/parenrightbigg mr2˙θˆj−1 2N˙θsinθmr2ˆk (h) In Example 9.3 the following numerical data was provided: m=5k g , r=0.08 m, t=0.025 m, N=2.1 rad/s, θ=60◦,˙θ=4 rad/s and ωspin=105 rad/s. For this set of numbers, (h) becomes MGnet=0.3203ˆi−0.6698ˆj−0.1164ˆk(N·m) 9.7 Kinetic energy 441 9.7 Kinetic energy The kinetic energy Tof a rigid body is the integral of the kinetic energy1 2v2dmof its individual mass elements, T=/integraldisplay m1 2v2dm=/integraldisplay m1 2v·vdm (9.75) where vis the absolute velocity ˙Rof the element of mass dm. From Figure 9.7 we infer that ˙R=˙RG+˙/rho1. Furthermore, Equation 1.24 requires that ˙/rho1=ω×/rho1. Thus, v=vG+ω×/rho1, which means v·v=[vG+ω×/rho1]·[vG+ω×/rho1]=v2 G+2vG·(ω×/rho1)+(ω×/rho1)·(ω×/rho1) We can apply the vector identity introduced in Equation 2.32, A·(B×C)=B·(C×A) (9.76) to the last term to get v·v=v2 G+2vG·(ω×/rho1)+ω·[/rho1×(ω×/rho1)] Therefore, Equation 9.75 becomes T=/integraldisplay m1 2v2 Gdm+vG·/parenleftbigg ω×/integraldisplay m/rho1dm/parenrightbigg +1 2ω·/integraldisplay m/rho1×(ω×/rho1)dm Since/rho1is measured from the center of mass,/integraltext m/rho1dm=0. Recall that, according to Equation 9.34, /integraldisplay m/rho1×(ω×/rho1)dm=HG It follows that the kinetic energy may be written T=1 2mv2 G+1 2ω·HG (9.77) The second term is the rotational kinetic energy TR, TR=1 2ω·HG (9.78) If the body is rotating about a point Pwhich is at rest in inertial space, we have from Equation 9.2 and Figure 9.7 that vG=vP+ω×rG/P=0+ω×rG/P=ω×rG/P It follows that v2 G=vG·vG=(ω×rG/P)·(ω×rG/P) 442 Chapter 9 Rigid-body dynamics Making use once again of the vector identity in Equation 9.76, we find v2 G=ω·[rG/P×(ω×rG/P)]=ω·(rG/P×vG) Substituting this into Equation 9.77 yields T=1 2ω·[HG+rG/P×mvG] Equation 9.21 shows that this can be written T=1 2ω·HP (9.79) In this case, of course, all of the kinetic energy is rotational. In terms of the components of ωand H, whether it is HPorHG, the rotational kinetic energy expression becomes, with the aid of Equation 9.39, TR=1 2(ωxHx+ωyHy+ωzHz) =1 2⌊ωxωyωz⌋ IxIxy Ixz Ixy IyIyz Ixz Iyz Iz   ωx ωy ωz   Expanding, we obtain TR=1 2Ixω2 x+1 2Iyω2 y+1 2Izω2 z+Ixyωxωy+Ixzωxωz+Iyzωyωz (9.80) Obviously, if the xyzaxes are principal axes of inertia, then Equation 9.80 simplifies considerably, TR=1 2Aω2 x+1 2Bω2 y+1 2Cω2 z (9.81) Example 9.14A satellite in circular geocentric orbit of 300 km altitude has a mass of 1500 kg, and the moments of inertia relative to a body frame with origin at the center of mass Gare [I]= 2000 −1000 2500 −1500 3000 −1500 2500 −1500 4000 (kg·m2) If at a given instant the components of angular velocity in this frame of reference are ω=1ˆi−0.9ˆj+1.5ˆk(rad/s) calculate the total kinetic energy of the satellite. The speed of the satellite in its circular orbit is v=/radicalbiggµ r=/radicalbigg 398 600 6378+300=7.7258 km /s 9.8 The spinning top 443 The angular momentum of the satellite is {HG}=[I G]{ω}= 2000 −1000 2500 −1500 3000 −1500 2500 −1500 4000   1 −0.9 1.5  =  6650 −5950 9850  (kg·m2/s) Therefore, the total kinetic energy is T=1 2mv2 G+1 2ω·HG=1 2·1500·7725.82+1 2[1−0.91 .5]  6650 −5950 9850   =44.766 ×106+13 390 T=44.766 MJ Obviously, the kinetic energy is dominated by that due to the orbital motion. 9.8 The spinning top Let us analyze the motion of the simple axisymmetric top in Figure 9.18. It is constrained to rotate about point O. The moving coordinate system is chosen to have its origin at O.T h e zaxis is aligned with the spin axis of the top (the axis of rotational symmetry). The xaxis is the node line, which passes through Oand is perpendicular to the plane defined ˆ j Z Y mgdG φO xyz Xˆiˆ k ˆ I ˆJˆ K θωp /H11005 φωs⋅ ωn /H11005 θ⋅ Figure 9.18 Simple top rotating about the fixed point O. 444 Chapter 9 Rigid-body dynamics by the inertial Zaxis and the spin axis of the top. The yaxis is then perpendicular toxand z, such that ˆj=ˆk׈i. By symmetry, the moment of inertia matrix of the top relative to the xyzframe is diagonal, with Ix=Iy=Aand Iz=C. It is likely that A>C. From Equations 9.68 and 9.70, we have M0net=A˙ωxˆi+A˙ωyˆj+C˙ωzˆk+/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆi ˆj ˆk /Omega1 x/Omega1y/Omega1z AωxAωyCωz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle(9.82) The angular velocity ωof the top is the vector sum of the spin rate ω sand the rates of precession ωpand nutation ωn,w h e r e ωp=˙φω n=˙θ (9.83) Thus ω=ωnˆi+ωpˆK+ωsˆk From the geometry it follows that ˆK=sinθˆj+cosθˆk (9.84) Therefore, relative to the co-moving system, ω=ωnˆi+ωpsinθˆj+(ωs+ωpcosθ)ˆk (9.85) From Equation 9.85 we see that ωx=ωnωy=ωpsinθω z=ωs+ωpcosθ (9.86) Computing the time rates of these three expressions yields the components of angular acceleration relative to the xyzframe, ˙ωx=˙ωn˙ωy=˙ωpsinθ+ωpωncosθ˙ωz=˙ωs+˙ωpcosθ−ωpωnsinθ(9.87) The angular velocity /Omega1of the xyzsystem is /Omega1=ωpˆK+ωnˆi, so that, using Equation 9.84, /Omega1=ωnˆi+ωpsinθˆj+ωpcosθˆk (9.88) From Equation 9.88 we obtain /Omega1x=ωn/Omega1y=ωpsinθ/Omega1 z=ωpcosθ (9.89) The moment about Oin Figure 9.18 is that of the weight vector acting through the center of mass G: M0net=(dˆk)×(−mgˆK)=− mgdˆk×(sinθˆj+cosθˆk) or M0net=mgd sinθˆi (9.90) 9.8 The spinning top 445 Substituting Equations 9.86, 9.87, 9.89 and 9.90 into Equation 9.82, we get mgd sinθˆi=A˙ωnˆi+A(˙ωpsinθ+ωpωncosθ)ˆj+C(˙ωs+˙ωpcosθ−ωpωnsinθ)ˆk +/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆi ˆj ˆk ω nωpsinθω pcosθ AωnAωpsinθC(ωs+ωpcosθ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle(9.91) Let us consider the special case in which θ=constant, i.e., there is no nutation, so thatω n=˙ωn=0. Then Equation 9.91 reduces to mgd sinθˆi=A˙ωpsinθˆj+C(˙ωs+˙ωpcosθ)ˆk +/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆi ˆj ˆk 0ω psinθω pcosθ 0AωpsinθC(ωs+ωpcosθ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle(9.92) Expanding the determinant yields mgd sinθˆi=A˙ω psinθˆj+C(˙ωs+˙ωpcosθ)ˆk +[Cωpωssinθ+(C−A)ω2 pcosθsinθ]ˆi Equating the coefficients of ˆi,ˆjandˆkon each side of the equation leads to mgd sinθ=Cωpωssinθ+(C−A)ω2 pcosθsinθ (9.93a) 0=A˙ωpsinθ (9.93b) 0=C(˙ωs+˙ωpcosθ) (9.93c) Equation 9.93b implies ˙ωp=0, and from Equation 9.93c it follows that ˙ωs=0. There- fore, the rates of spin and precession are both constant. From Equation 9.93a we find (A−C)c o sθω2 p−Cωsωp+mgd=0( 0 <θ<180◦) (9.94) If the spin rate is zero, Equation 9.94 yields ωp/parenrightbig ωs=0=±/radicalBigg mgd (C−A)c o s θif (A−C)c o sθ< 0 (9.95) In this case, the top rotates about Oat this rate, without spinning. If A>C(prolate), its symmetry axis must make an angle between 90◦and 180◦to the vertical; otherwise ωpis imaginary. On the other hand, if A<C(oblate), the angle lies between 0◦and 90◦. Thus, in steady rotation without spin, the top’s axis sweeps out a cone which lies either below the horizontal plane ( A>C) or above the plane ( A<C). In the special case ( A−C)c o sθ=0, Equation 9.94 yields a steady precession rate which is inversely proportional to the spin rate, ωp=mgd Cωsif (A−C)c o sθ=0 (9.96) 446 Chapter 9 Rigid-body dynamics IfA=C, this precession apparently occurs irrespective of tilt angle θ.I f A/negationslash=C, this rate of precession occurs at θ=90◦, i.e., the spin axis is perpendicular to the precession axis. In general, Equation 9.94 is a quadratic equation in ωp, so we can use the quadratic formula to find ωp=C 2(A−C)c o sθ/parenleftBigg ωs±/radicalbigg ω2s−4mgd (A−C)c o sθ C2/parenrightBigg (9.97) Thus, for a given spin rate and tilt angle θ(θ/negationslash=90◦), there are two rates of precession ˙φ. Observe that if ( A−C)c o sθ>0, then ωpis imaginary when ω2 s<4mgd (A−C) cosθ/C2. Therefore, the minimum spin rate required for steady precession at a constant inclination θis ωs)min=2 C/radicalbig mgd (A−C)c o sθif (A−C)c o sθ> 0 (9.98) If (A−C)c o sθ<0, the radical in Equation 9.97 is real for all ωs. In this case, as ωs→0,ωpapproaches the value given above in Equation 9.95. Example 9.15For the top of Figure 9.18, let m=0.5k g , A(=Ix=Iy)=12×10−4kg·m2, C(=Iz)=4.5×10−4kg·m2and d=0.05 m. For an inclination of, say, 60◦, (A−C)c o sθ>0 so that Equation 9.98 requires ωs)min=407.01 rpm. Let us choose the spin rate to be ωs=1000 rpm =104.7r a d/s. Then, from Equation 9.97, the pre- cession rate as a function of the inclination θis given by either one of the following formulas ωp=31.421+√1−0.3312 cos θ cosθandωp=31.421−√1−0.3312 cos θ cosθ(a) These are plotted in Figure 9.19. 90 180/H110026000/H11002300006000 455055 180 90 0 03000 45 135 45 135 (a) (b) ω p (rpm) θ (degre es) θ (degre es) ω p (rpm) Figure 9.19 (a) High-energy precession rate (unlikely to be observed). (b) Low energy precession rate (the one most always seen). 9.8 The spinning top 447 G zy xRotor Rotating platformωp ωs Figure 9.20 A spinning rotor on a rotating platform. Figure 9.20 shows an axisymmetric rotor mounted so that its spin axis ( z) remains perpendicular to the precession axis (y ). In that case Equation 9.85 with θ=90◦ yields ω=ωpˆj+ωsˆk (9.99) Likewise, from Equation 9.88, the angular velocity of the co-moving xyz sys- tem is /Omega1=ωpˆj. If we assume that the spin rate and precession rate are constant (dωp/dt=dωs/dt=0), then Equation 9.68, written for the center of mass G, becomes MGnet=/Omega1×H=(ωpˆj)×(Aω pˆj+Cωsˆk) (9.100) where Aand Care the moments of inertia of the rotor about the xand zaxes, respectively. Setting Cωsˆk=Hs, the spin angular momentum, and ωpˆj=ωp,w e obtain MGnet=ωp×Hs (Hs=Cωsˆk) (9.101) This is the gyroscope equation, which is similar to Equation 9.96. Since the center of mass is the reference point, there is no restriction on the motion Gfor which Equation 9.101 is valid. Observe that the net gyroscopic moment MGnete x e r t e do nt h e rotor by its supports is perpendicular to the plane of the spin and precession vectors.If a spinning rotor is forced to precess, the gyroscopic moment M Gnetdevelops. Or, if a moment is applied normal to the spin axis of a rotor, it will precess so as to causethe spin axis to turn towards the moment axis. Example 9.16A uniform cylinder of radius r, length Land mass mspins at a constant angular velocity ωp. It rests on simple supports, mounted on a platform which rotates at an angular velocity of ωp. Find the reactions at AandB. Neglect the weight (i.e., calculate the reactions due just to gyroscopic effects). 448 Chapter 9 Rigid-body dynamics (Example 9.16 continued) GL/2 L/2 rωp ωsA Bzy R RL Figure 9.21 Illustration of the gyroscopic effect. The net vertical force on the cylinder is zero, so the reactions at each end are equal and opposite in direction, as shown on the free-body diagram insert in Figure 9.21. Noting that the moment of inertia of a uniform cylinder about its axis of rotational symmetry is1 2mr2, Equation 9.101 yields RLˆi=(ωpˆj)×/parenleftbigg1 2mr2ωsˆk/parenrightbigg =1 2mr2ωpωsˆi so that, R=mr2ωpωs 2L 9.9 Euler angles Three angles are required to specify the orientation of a rigid body relative to an inertial frame. The choice is not unique, but there are two sets in common use: theEuler angles and the yaw, pitch and roll angles. We will discuss each of them in turn. The three Euler angles give the orientation of a rigid, orthogonal xyzframe of reference relative to the XYZ inertial frame of reference. The orthogonal triad of unit vectors parallel to the inertial axes XYZ areˆI,ˆJandˆK, respectively. The orthogonal triad of unit vectors lying along the axes of the xyzframe are ˆi,ˆjandˆk, respectively. Figure 9.22 shows the ˆIˆJˆKtriad and the ˆiˆjˆktriad, along with the three successive rotations required to bring unit vectors initially aligned with ˆIˆJˆKinto alignment with ˆiˆjˆk. Since we are interested only in the relative orientation of the two frames, we can, for simplicity and without loss of generality, show the two frames sharing a commonorigin. The xyplane intersects the XYplane along a line (the node line) defined by the unit vector ˆi /primein the figure. The first rotation, ①, is around the ˆKaxis, through the Euler angle φ. It rotates the ˆI,ˆJdirections into the ˆi/prime,ˆj/primedirections. Viewed down the 9.9 Euler angles 449 φφ 1 θ φφθ2 133 ψψ 2 1 1 23 θθ 2ψψ 3ˆ K K /H11005 sin θj′′/H11001cosθk i′ /H11005 cos ψi/H11002sinψj j′′ /H11005 sin ψi/H11001cosψjˆj′′ˆj′′ ˆj′′ ˆj′ˆJ ˆ ˆ ˆˆj′ˆj′ ˆi′ˆjˆj ˆi ˆIˆ ˆ ˆ ˆˆ ˆˆiˆi′ ˆi′ˆkˆk ˆKˆIˆJ Figure 9.22 The Euler angles. Zaxis, this rotation appears as shown in the insert at the top of Figure 9.22, from which we see that ˆi/prime=cosφˆI+sinφˆJ ˆj/prime=− sinφˆI+cosφˆJ ˆk/prime=ˆK or  ˆi/prime ˆj/prime ˆk/prime  = cosφsinφ0 −sinφcosφ0 00 1   ˆI ˆJ ˆK   Therefore, the orthogonal transformation matrix associated with this rotation is [R3(φ)]= cosφsinφ0 −sinφcosφ0 00 1  (9.102) 450 Chapter 9 Rigid-body dynamics Recall from Section 4.5 that the subscript on Rdenotes that the rotation is around the ‘3’ direction, in this case the ˆKaxis. The second Euler rotation, ②, is around the node line ( ˆi/prime), through the angle θ required to bring the XYplane parallel to the xyplane. In other words, it rotates the Zaxis into alignment with the zaxis, and ˆj/primesimultaneously rotates into ˆj/prime/prime. The insert in the lower right of Figure 9.22 shows how this rotation appears when viewed from theˆi/primedirection. From that illustration we can deduce that ˆi/prime/prime=ˆi/prime ˆj/prime/prime=cosθˆj/prime+sinφˆK ˆk=− sinφˆj/prime+cosφˆK or  ˆi/prime/prime ˆj/prime/prime ˆk  = 10 0 0c o s θsinθ 0−sinθcosθ   ˆi/prime ˆj/prime ˆK  (9.103) Clearly, the orthogonal transformation matrix for this rotation is [R1(θ)]= 10 0 0c o s θsinθ 0−sinθcosθ  (9.104) Since the inverse of an orthogonal matrix is just its transpose, the inverse of Equation 9.103 is   ˆi/prime ˆj/prime ˆK  = 10 0 0c o s θ−sinθ 0 sin θ cosθ   ˆi/prime/prime ˆj/prime/prime ˆk   from which we get the particular result needed below, namely, ˆK=sinθˆj/prime/prime+cosθˆk (9.105) The third and final Euler rotation, ③,i si nt h e xyplane and rotates the unit vectors ˆi/prime andˆj/prime/primethrough the angle ψaround the zaxis so that they become aligned with ˆiand ˆj, respectively. This rotation appears from the zdirection as shown in the insert on the left of Figure 9.22. From that picture, we observe that ˆi=cosψˆi/prime+sinψˆj/prime/prime ˆj=− sinψˆi/prime+cosψˆj/prime/prime ˆk=ˆk or  ˆi ˆj ˆk  = cosψ sinψ 0 −sinψ cosψ 0 00 1   ˆi/prime ˆj/prime/prime ˆk  (9.106) 9.9 Euler angles 451 From this, the orthogonal transformation matrix is seen to be [R3(ψ)]= cosψ sinψ 0 −sinψ cosψ 0 00 1  (9.107) The inverse of Equations 9.106 is   ˆi/prime ˆj/prime/prime ˆk  = cosψ−sinψ 0 sinψ cosψ 0 00 1   ˆi ˆj ˆk   from which we obtain ˆi/prime=cosψˆi−sinψˆj (9.108a) ˆj/prime/prime=sinψˆi+cosψˆj (9.108b) Substituting Equation 9.108b into Equatio n 9.105 yields another result we will need below, ˆK=sinθsinψˆi+sinθcosψˆj+cosθˆk (9.109) The time rates of change of the Euler angles φ,θandψare, respectively, the precession ωp, the nutation ωnand the spin ωs. That is, ωp=˙φω n=˙θω s=˙ψ (9.110) If the absolute angular velocity ωof the rigid xyzframe is resolved into components ωx,ωyandωzalong the xyzaxes, we can express it analytically as ω=ωxˆi+ωyˆj+ωzˆk (9.111) On the other hand, in terms of the precession, nutation and spin, the absolute angular velocity can be written in terms of the non-orthogonal Euler angle rates ω=ωpˆK+ωnˆi/prime+ωsˆk (9.112) Substituting Equations 9.108a and 9.109 yields ω=(ωpsinθsinψ+ωncosψ)ˆi+(ωpsinθcosψ−ωnsinψ)ˆj +(ωs+ωpcosθ)ˆk (9.113) Comparing Equations 9.111 and 9.113, we see that ωx=ωpsinθsinψ+ωncosψ ωy=ωpsinθcosψ−ωnsinψ (9.114) ωz=ωs+ωpcosθ 452 Chapter 9 Rigid-body dynamics We can solve these three equations to obtain the Euler rates in terms of ωx,ωyandωz: ωp=˙φ=1 sinθ(ωxsinψ+ωycosψ) ωn=˙θ=ωxcosψ−ωysinψ (9.115) ωs=˙ψ=−1 tanθ(ωxsinψ+ωycosψ)+ωz Observe that if ωx,ωyandωzare given functions of time, found by solving Euler’s equations of motion (Equations 9.72), then E quations 9.115 are three coupled differ- ential equations which may be solved to o btain the three time-dependent Euler angles φ=φ(t)θ=θ(t)ψ=ψ(t) With this solution, the orientation of the xyzframe, and hence the body to which it is attached, is known for any given time t. Note, however, that Equations 9.115 ‘blow up’ when θ=0, i.e., when the xyplane is parallel to the XYplane. Finally, let us note that the transformation matrix [ Q]Xxfrom the inertial XYZ frame into the moving xyzframe is just the product of the three rotation matrices given by Equations 9.102, 9.104 and 9.107, i.e., [Q]Xx=[R3(ψ)][R1(θ)][R3(φ)] (9.116) Substituting the three matrices on the right and carrying out the matrix multiplica- tions yields [Q]Xx= cosφcosψ−sinφsinψcosθ sinφcosψ+cosφcosθsinψ sinθsinψ −cosφsinψ−sinφcosθcosψ−sinφsinψ+cosφcosθcosψ sinθcosψ sinφsinθ −cosφsinθ cosθ  (9.117) Remember, this is an orthogonal matrix, so that for the inverse transformation from xyztoXYZ we have [ Q]xX=([Q]Xx)T,o r [Q]xX= cosφcosψ−sinφsinψcosθ−cosφsinψ−sinφcosθcosψ sinφsinθ sinφcosψ+cosφcosθsinψ−sinφsinψ+cosφcosθcosψ−cosφsinθ sinθsinψ sinθcosψ cosθ  (9.118) Example 9.17At a given instant, the unit vectors of a body-fixed frame are ˆi=0.40825ˆI−0.40825ˆJ+0.8165ˆK ˆj=− 0.10102ˆI−0.90914ˆJ−0.40406 ˆK (a) ˆk=0.90726ˆI+0.082479 ˆJ−0.41239 ˆK and the angular velocity is ω=− 3.1ˆI+2.5ˆJ+1.7ˆK(rad/s) (b) Calculate ωp,ωnandωs(the precession, nutation and spin rates) at this instant. 9.9 Euler angles 453 We will ultimately use Equations 9.115 to find ωp,ωnandωs.T od os ow em u s t first obtain the Euler angles φ,θandψas well as the components of the angular velocity in the body frame. The procedure for determining φ,θandψis practically identical to that used to obtain the orbital elements i,/Omega1, and ωof a satellite orbit from its state vector (Algorithm 4.1). Referring to Figure 9.22, we first note that the angle between ˆkand ˆKis the inclination angle θ, so that θ=cos−1kZ=cos−1(−0.41239) =114.36◦(c) θlies between 0◦and 180◦. The ‘node’ vector Npoints in the direction of ˆi/primein Figure 9.22, and it is found by taking the cross product of ˆKintoˆk, N=ˆK׈k=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆI ˆJ ˆK 00 1 0.90726 0.082479 −0.41239/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =− 0.082479 ˆI+0.90726 ˆJ(/bardblN/bardbl=0.911) (d) The precession angle φ(analogous to RAof the ascending node /Omega1) is measured from theXaxis positive towards N. Therefore, taking care to place φin the proper quadrant, φ=  cos−1NX /bardblN/bardblifNY≥0 360◦−cos−1NX /bardblN/bardblifNY<0(e) Substituting (d), noting that in this case NY=0.90276 >0, we get φ=cos−1−0.082479 0.9110=95.194◦(f) The spin angle ψis measured positive from Ntoˆiin Figure 9.22. It plays the same role here as argument of perigee does for satellites. We can find the angle between Nandˆi by using the dot product operation, again being careful to put ψin the right quadrant, ψ=  cos−1N·ˆi /bardblN/bardblifiZ≥0 360◦−cos−1N·ˆi /bardblN/bardblifiZ<0(g) From (a) we note that iZ=0.8165 >0, so ψ=cos−1(−0.082479 ˆI+0.90726 ˆJ)·(0.40825 ˆI−0.40825 ˆJ+0.8165 ˆK) 0.9110 =cos−1/parenleftbigg−0.40406 0.9110/parenrightbigg =116.33◦(h) T o transform the components of the given angular velocity vector into components along the body frame, we need the matrix [ Q]Xxof the transformation from XYZ toxyz. The rows of [ Q]Xxare the direction cosines of ˆi,ˆjandˆk, which are given in (a). 454 Chapter 9 Rigid-body dynamics (Example 9.17 continued)Thus {ω}x=[Q]Xx{ω}X = 0.40825 −0.40825 0 .8165 −0.10102 −0.90914 −0.40406 0.90726 0 .082479 −0.41239   −3.1 2.5 1.7  =  −0.89815 −2.6466 −3.3074  (rad/s) That is, ωx=− 0.89815 rad /sωy=− 2.6466 rad /sωz=− 3.3074 rad /s(i) Finally, substituting (c), (f), (h) and (i) into Equations 9.115 yields ωp=1 sin 114 .36◦[−0.89815 ·sin 116 .33◦+(−2.6466) ·cos 116 .33◦] =0.40492 rad /s ωn=− 0.89815 ·cos 116 .33◦−(−2.6466) ·sin 116 .33◦=2.7704 rad /s ωs=−1 tan 114 .36◦[−0.89815 ·sin 116 .33◦+(−2.6466) ·cos 116 .33◦] +(−3.3074) =− 3.1404 rad /s Example 9.18The mass moments of inertia of a body about the principal body frame axes with origin at the center of mass Gare A=1000 kg ·m2B=2000 kg ·m2C=3000 kg ·m2(a) The Euler angles in radians are given as functions of time in seconds as follows: φ=2te−0.05t θ=0.02+0.3 sin 0.25t (b) ψ=0.6t Att=0, find (a) the net moment about Gand (b) the components αX,αYandαZ of the absolute angular acceleration in the inertial frame. (a) We must use Euler’s equations (Equations 9.72) to calculate the net moment, which means we must first obtain ωx,ωy,ωz,˙ωx,˙ωyand˙ωz.S i n c ew ea r eg i v e n the Euler angles as a function of time, we can compute their time derivatives and then use Equation 9.114 to find the body frame angular velocity components and their derivatives. Starting with (b) 1,w eg e t ωp=dφ dt=d dt(2te−0.05t)=2e−0.05t−0.1te−0.05t ˙ωp=dωp dt=d dt(2e−0.05t−0.1e−0.05t)=− 0.2e−0.05t+0.005te−0.05t 9.9 Euler angles 455 Proceeding to the remaining two Euler angles leads to ωn=dθ dt=d dt(0.02+0.3 sin 0 .25t)=0.075 cos 0 .25t ˙ωn=dωn dt=d dt(0.075 cos 0 .25t)=− 0.01875 sin 0 .25t ωs=dψ dt=d dt(0.6t )=0.6 ˙ωs=dωs dt=0 Evaluating all of these quantities, including those in (b), at t=0 yields φ=335.03◦ωp=0.60653 rad /s ˙ωp=− 0.09098 rad /s2 θ=11.433◦ωn=− 0.06009 rad /s ˙ωn=− 0.011221 rad /s2(c) ψ=343.77 ωs=0.6r a d /s ˙ωs=0 Equation 9.114 relates the Euler angle rates to the angular velocity components, ωx=ωpsinθsinψ+ωncosψ ωy=ωpsinθcosψ−ωnsinψ (d) ωz=ωs+ωpcosθ Taking the time derivative of each of these equations in turn leads to the following three equations, ˙ωx=ωpωncosθsinψ+ωpωssinθcosψ−ωnωssinψ +˙ωpsinθsinψ+˙ωncosψ ˙ωy=ωpωncosθcosψ−ωpωssinθsinψ−ωnωscosψ (e) +˙ωpsinθcosψ−˙ωnsinψ ˙ωz=−ωpωnsinθ+˙ωpcosθ+˙ωs Substituting the data in (c) into (d) and (e) yields ωx=− 0.091286 rad /sωy=0.098649 rad /sωz=1.1945 rad /s ˙ωx=0.063435 rad /s2˙ωy=2.2346 ×10−5rad/s2˙ωz=− 0.08195 rad /s2 (f) With (a) and (f) we have everything we need for Euler’s equations, Mxnet=A˙ωx+(C−B)ωyωz Mynet=B˙ωy+(A−C)ωzωx Mznet=C˙ωz+(B−A)ω xωy 456 Chapter 9 Rigid-body dynamics (Example 9.18 continued)from which we find Mxnet=181.27 N·m Mynet=218.12 N·m Mznet=− 254.86 N·m (b) Since the co-moving xyzframe is a body frame, rigidly attached to the solid, we know from Equation 9.73 that   αX αY αZ  =[Q]xX  ˙ωx ˙ωy ˙ωz  (g) In other words, the absolute angular acceleration and the relative angular accel- eration of the body are the same. All we have to do is project the components of relative acceleration in (f) onto the axes of the inertial frame. The required orthogonal transformation matrix is given in Equation 9.118, [Q]xX = cosφcosψ−sinφsinψcosθ−cosφsinψ−sinφcosθcosψ sinφsinθ sinφcosψ+cosφcosθsinψ−sinφsinψ+cosφcosθcosψ−cosφsinθ sinθsinψ sinθcosψ cosθ  Upon substituting the numerical values of the Euler angles from (c), this becomes [Q]xX= −0.90855 0 .20144 0 .3660 −0.29194 −0.93280 −0.21131 0.29884 −0.29884 0 .90631  Substituting this and the relative angular velocity rates from (c) into (g) yields   αX αY αZ  = −0.90855 0 .20144 0 .3660 −0.29194 −0.93280 −0.21131 0.29884 −0.29884 0 .90631   −0.027359 −0.32619 1.4532   =  0.4910 0.0051972 1.4063  (rad/s2) Example 9.19Figure 9.23 shows a rotating platform on which is mounted a rectangular paral- lelepiped shaft (with dimensions b,hand l) spinning about the inclined axis DE.I f the mass of the shaft is m, and the angular velocities ωpandωsare constant, calculate the bearing forces at Dand Eas a function of φandψ. Neglect gravity, since we are interested only in the gyroscopic forces. (The small extensions shown at each end of the parallelepiped are just for clarity; the distance between the bearings at Dand Eisl.) 9.9 Euler angles 457 D XYZ yzx OE φ (Measured in XY plane)u (Measured in Zz plane) Shaft Platformφ/H11005 ω p ψ /H11005 ω sψ x′ x′(Measured in x'x plane, perpendicular to shaft) xyz axes attached to the shaft l/2G l/2bh Figure 9.23 Spinning block mounted on rotating platform. The inertial XYZ frame is centered at Oon the platform, and it is right-handed (ˆI׈J=ˆK). The origin of the right-handed co-moving body frame xyzis at the shaft’s center of mass G, and it is aligned with the symmetry axes of the parallelepiped. The three Euler angles φ,θandψare shown in Figure 9.23. Since θis constant, the n u t a t i o nr a t ei sz e r o( ωn=0). Thus, Equations 9.114 reduce to ωx=ωpsinθsinψω y=ωpsinθcosψω z=ωpcosθ+ωs (a) Sinceωp,ωsandθare constant, it follows (recalling Equations 9.110) that ˙ωx=ωpωssinθcosψ˙ωy=−ωpωssinθsinψ˙ωz=0 (b) The principal moments of inertia of the parallelepiped are [see Figure 9.9(c)] A=Ix=1 12m(h2+l2) B=Iy=1 12m(b2+l2)( c) C=Iz=1 12m(b2+h2) Figure 9.24 is a free-body diagram of the shaft. Let us assume that the bearings at D and Eare such as to exert just the six body frame components of force shown. Thus, Dis a thrust bearing to which the axial torque TDis applied from, say, a motor of some kind. At Ethere is a simple journal bearing. 458 Chapter 9 Rigid-body dynamics (Example 9.19 continued) Gx yz b h TDDx Dz DyEyEx l/2l/2 Figure 9.24 Free-body diagram of the block in Figure 9.23. From Newton’s laws of motion we have Fnet=maG.B u t Gis fixed in inertial space, soaG=0. Thus, (Dxˆi+Dyˆj+Dzˆk)+(Exˆi+Eyˆj)=0 It follows that Ex=− Dx Ey=− Dy Dz=0( d ) Summing moments about Gwe get MGnet=l 2ˆk×(Exˆi+Eyˆj)+/parenleftbigg −l 2ˆk/parenrightbigg ×(Dxˆi+Dyˆj)+TDˆk =/parenleftbigg Dyl 2−Eyl 2/parenrightbigg ˆi+/parenleftbigg −Dxl 2+Exl 2/parenrightbigg ˆj+TDˆk =Dylˆi−Dxlˆj+TDˆk where we made use of Equation (d) 2. Thus, Mxnet=DylM ynet=− DxlM znet=TD (e) We substitute (a), (b), (c), and (e) into Euler’s equations (Equations 9.72): Mxnet=A˙ωx+(C−B)ωyωz Mynet=B˙ωy+(A−C)ωxωz (f) Mznet=C˙ωz+(B−A)ωxωy After making the substitutions and simplifying, the first Euler equation, Equation (f)1, becomes Dx=/braceleftbigg1 12m l[(l2−h2)ωpcosθ−2h2ωs]ωpsinθ/bracerightbigg cosψ (g) 9.10 Y aw, pitch and roll angles 459 Likewise, from Equation (f) 2we obtain Dy=/braceleftbigg1 12m l[(l2−b2)ωpcosθ−2b2ωs]ωpsinθ/bracerightbigg sinψ (h) Finally, Equation (f) 3yields TD=/bracketleftbigg1 24m(b2−h2)ω2 psin2θ/bracketrightbigg sin 2ψ (i) This completes the solution, since Ey=− Dyand Ez=− Dz. Note that the resultant transverse bearing load VatD(and E)i s V=/radicalBig D2x+D2y (j) As a numerical example, let l=1m h=0.1m b=0.025 m θ=30◦m=10 kg and ωp=100 rpm =10.47 rad /s ωs=2000 rpm =209.4r a d /s For these numbers, the variation of Vand TDwithψare as shown in Figure 9.25. 90 180 270102030 36040 0 0V, N ψ, degrees90 180 270/H110020.200.2 360 00.4 /H110020.4TD, N/H11080m ψ, degrees (a) (b) Figure 9.25 (a) Transverse bearing load. (b) Axial torque at D. 9.10 Yaw, pitch and roll angles The problem of the Euler angle relations, Equations 9.114, becoming singular when the nutation angle θis zero can be alleviated by using the yaw, pitch and roll angles illustrated in Figure 9.26. As in the Euler angles, the inertial ˆIˆJˆKtriad is rotated into the body ˆiˆjˆktriad by a sequence of three rotations, detailed in Figure 9.26. The first step is to rotate the ˆIandˆJdirections through a yaw angle φaround the ˆKaxis until they line up with the orthogonal unit vectors ˆi/prime,ˆj/prime.T h eˆi/primedirection is the projection of the body xaxis on the XYplane. This rotation appears as shown in insert ①at the 460 Chapter 9 Rigid-body dynamics φφ 1 θφθ 13ψ 11 θθ 2ψ 322 3 (yaw)φ(yaw) (pitch)(roll) ψ(roll) 2 3ˆJˆIˆJ ˆI/H9275 /H11005 ωyawK/H11001ωpitch j′/H11001ωrolli ωyaw/H11005 φ j′′ /H11005 cos ψj /H11002 sin ψk k′′ /H11005 sin ψj /H11001 cos ψk k′/H11005/H11002sin θi′′ /H11001 cos θk′′ˆj′′ˆjˆk′′ˆk ˆˆ ˆ ˆi′ ˆi′′ˆj ˆi′k′′ ˆkˆi′ˆj′ˆˆ ˆ ˆˆ ˆ ˆ ˆ ˆ k′′ˆ i′′ /H11005 iˆ ˆj′ /H11005 j′′ˆ ˆ k′ˆK /H11005 k′ˆ ˆˆ (pitch)ωpitch/H11005θω roll/H11005ψ ψ Figure 9.26 Y aw, pitch and roll angles. top of Figure 9.26, from which it can be seen that ˆi/prime=cosφˆI+sinφˆJ (9.119a) ˆj/prime=− sinφˆI+cosφˆJ (9.119b) ˆk/prime=ˆK (9.119c) or  ˆi/prime ˆj/prime ˆk/prime  = cosφsinφ0 −sinφcosφ0 00 1   ˆI ˆJ ˆK   Clearly, the yaw rotation matrix is [ R3(φ)], where [R3(φ)]= cosφsinφ0 −sinφcosφ0 00 1  The second rotation is a pitch around ˆj/primethrough the pitch angle θ. This carries ˆi/prime andˆk/prime(=ˆK) intoˆi/prime/prime(=ˆi) andˆk/prime/prime, while, of course, leaving ˆj/primeunchanged. We see this in 9.10 Y aw, pitch and roll angles 461 auxiliary view ②of Figure 9.26, from which we obtain ˆi/prime/prime=cosθˆi/prime−sinθˆk/prime ˆj/prime/prime=ˆj/prime ˆk/prime/prime=sinθˆi/prime+cosθˆk/prime or   ˆi/prime/prime ˆj/prime/prime ˆk/prime/prime  = cosθ0−sinθ 01 0 sinθ0c o s θ   ˆi/prime ˆj/prime ˆk/prime   This rotation about the intermediate y/primeaxis is therefore represented by [ R2(θ)], where [R2(θ)]= cosθ0−sinθ 01 0 sinθ0c o s θ  The inverse of this orthogonal matrix is its transpose, [R2(θ)]−1= cosθ0 sin θ 01 0 −sinθ0c o s θ  so that   ˆi/prime ˆj/prime ˆk/prime  = cosθ0 sin θ 01 0 −sinθ0c o s θ   ˆi/prime/prime ˆj/prime/prime ˆk/prime/prime   From this we see that ˆj/prime=ˆj/prime/prime ˆk/prime=− sinθˆi/prime/prime+cosθˆk/prime/prime(9.120) Finally, we roll around the body xaxis through the angle ψ, which brings ˆj/prime/primeandˆk/prime/prime into alignment with body unit vectors ˆjandˆk, respectively. Auxiliary view ③shows that ˆi=ˆi/prime/prime ˆj=cosψˆj/prime/prime+sinψˆk/prime/prime ˆk=− sinψˆj/prime/prime+cosψˆk/prime/prime or   ˆi ˆj ˆk  = 10 0 0c o s ψ sinψ 0−sinψ cosψ   ˆi/prime/prime ˆj/prime/prime ˆk/prime/prime  (9.121) 462 Chapter 9 Rigid-body dynamics Thus, the third and last rotation matrix is [R1(ψ)]= 10 0 0c o s ψ sinψ 0−sinψ cosψ  Taking the transpose of this array, we find the inverse of Equations 9.121   ˆi/prime/prime ˆj/prime/prime ˆk/prime/prime  = 10 0 0c o s ψ−sinψ 0 sin ψ cosψ   ˆi ˆj ˆk   which means ˆi/prime/prime=ˆi ˆj/prime/prime=cosψˆj−sinψˆk (9.122) ˆk/prime/prime=sinψˆj+cosψˆk The matrix [ Q]Xxof the transformation from ˆIˆJˆKintoˆiˆjˆkis the product of the three rotation matrices obtained above, [Q]Xx=[R1(ψ)][R2(θ)][R3(φ)] Carrying out the matrix multiplications yields [Q]Xx= cosφcosθ sinφcosθ −sinθ −sinφcosψ+cosφsinθsinψ cosφcosψ+sinφsinθsinψ cosθsinψ sinφsinψ+cosφsinθcosψ−cosφsinψ+sinφsinθcosψ cosθcosψ  (9.123) The inverse matrix which transforms xyzinto XYZ is just the transpose, [Q]xX= cosφcosθ−sinφcosψ+cosφsinθsinψ sinφsinψ+cosφsinθcosψ sinφcosθ cosφcosψ+sinφsinθsinψ−cosφsinψ+sinφsinθcosψ −sinθ cosθsinψ cosθcosψ  (9.124) The angular velocity ω, expressed in terms of the rates of yaw, pitch and roll, is ω=ωyawˆK+ωpitchˆj/prime+ωrollˆi in which ωyaw=˙φω pitch=˙θω roll=˙ψ Using Equation 9.119c, we can write ωas ω=ωyawˆk/prime+ωpitchˆj/prime+ωrollˆi Substituting Equation 9.120 into this expression yields ω=ωyaw(−sinθˆi/prime/prime+cosθˆk/prime/prime)+ωpitchˆj/prime/prime+ωrollˆi Finally, with Equations 9.122, we obtain from this ω=ωyaw[−sinθˆi+cosθ( sinψˆj+cosψˆk)]+ωpitch(cosψˆj−sinψˆk)+ωrollˆi Problems 463 After collecting terms, we see that ωx=ωroll−ωyawsinθpitch ωy=ωyawcosθpitch sinψroll+ωpitch cosψroll (9.125) ωz=ωyawcosθpitch cosψroll−ωpitch sinψroll wherein the subscript on each symbol helps us remember the rotation it describes. The inverse of these equations is ωyaw=ωysinψroll cosθpitch+ωzcosψroll cosθpitch ωpitch=ωycosψroll−ωzsinψroll (9.126) ωroll=ωx+ωytanθpitch sinψroll+ωztanθpitch cosψroll Notice that this system becomes singular (cos θpitch=0) when the pitch angle is ±90◦. Problems 9.1 Rigid, bent shaft 1 ( ABC ) rotates at a constant angular velocity of 2 ˆKrad/s around the positive Zaxis of the inertial frame. Bent shaft 2 ( CDE ) rotates around BCwith a constant angular velocity of 3 ˆjrad/s, relative to BC. Spinner 3 at Erotates around DE with a constant angular velocity of 4 ˆirad/s relative to DE. Calculate the magnitude of the absolute angular acceleration α3of the spinner at the instant shown. {Ans.: ||α3|| =/radicalbig 180+64 sin2θ−144 cos θ(rad/s2)} XYZ ABC DE 13 3 rad/s4 rad/s 2 rad/sAlways points in the direction BCDAlways points in the direction DE 2ˆi ˆjˆkAlways perpendicular to ˆi and j (k /H11005 i /H11003 j)ˆˆ ˆ ˆ θ φ Figure P .9.1 464 Chapter 9 Rigid-body dynamics 9.2 The body-fixed xyzframe is attached to the cylinder as shown. The cylinder rotates around the inertial Zaxis, which is collinear with the zaxis, with a constant absolute angular velocity ˙θˆk.R o d ABis attached to the cylinder and aligned with the y axis. Rod BCis perpendicular to ABand rotates around ABwith the constant angular velocity ˙φˆj relative to the cylinder. Rod CDis perpendicular to BCand rotates around BCwith the constant angular velocity ˙νˆmrelative to BC,w h e r e ˆmis the unit vector in the direction ofBC. The plate abcd rotates around CDwith a constant angular velocity ˙ψˆnrelative toCD, where the unit vector ˆnpoints in the direction of CD. Thus the absolute angular velocity of the plate is ωplate=˙θˆk+˙φˆj+˙νˆm+˙ψˆn. Show that (a)ωplate=(˙νsinφ−˙ψcosφsinν)ˆi+(˙φ+˙ψcosν)ˆj+(˙θ+˙νcosφ+˙ψsinφsinν)ˆk αplate=dωplate dt=[˙ν(˙φcosφ−˙ψcosφcosν)+˙ψ˙φsinφsinν−˙ψ˙θcosν−˙φ˙θ]ˆi (b) +[˙ν(˙θsinφ−˙ψsinν)−˙ψ˙θcosφsinν]ˆj +[˙ψ˙νcosνsinφ+˙ψ˙φcosφsinν−˙φ˙νsinφ]ˆk (c) aC=− l(˙φ2+˙θ2) sinφˆi+(2l˙φ˙θcosφ−5 4l˙θ2)ˆj−l˙φ2cosφˆk xyZ, z X Yθφψ ν BC abc d ˆiˆjˆI ˆJˆm ˆnˆpˆK, kˆ llll l/4 l/4r /H11005 l/4 O ADˆθk. ˆψn. ˆφj.ˆνm. Figure P .9.2 9.3 The mass center Gof a rigid body has a velocity v=t3ˆi+4ˆjm/s and an angular velocity ω=2t2ˆkrad/s, where tis time in seconds. The ˆi,ˆj,ˆkunit vectors are attached to and rotate with the rigid body. Calculate the magnitude of the acceleration aGof the center of mass at t=2 seconds. {Ans.: aG=− 20ˆi+64ˆj(m/s2)} 9.4 The inertial angular velocity of a rigid body is ω=ωxˆi+ωyˆj+ωzˆk,w h e r e ˆi,ˆj,ˆkare the unit vectors of a co-moving frame whose inertial angular velocity is /Omega1=ωxˆi+ωyˆj. Calculate the components of angular acceleration of the rigid body in the moving frame,assuming that ω x,ωyandωzare all constant. {Ans.: α=ωyωzˆi−ωxωzˆj} Problems 465 G XY ZOv ˆiˆj ˆk Figure P .9.3 9.5 Find the moments of inertia about the center of mass of the system of six point masses listed in the table. T able P .9.5 Point, i Mass mi(kg) xi(m) yi(m) zi(m) 11 0 1 1 1 21 0 −1 −1 −1 38 4 −44 48 −22 −2 51 2 3 −3 −3 61 2 −333 {Ans.: [I G]= 783.5 351.74 0 .27 351.7 783.5 −80.27 40.27 −80.27 783.5 (kg·m2)} 9.6 Find the mass moment of inertia of the configuration of Problem 9.5 about an axis through the origin and the point with coordinates (1 m, 2 m, 2 m).{Ans.: 621.3 kg ·m 2} 9.7 A uniform slender rod of mass mand length llies in the xyplane inclined to the xaxis by the angle θ. Use the results of Example 9.10 to find the mass moments of inertia about the xyzaxes passing through the center of mass G. {Ans.: [I G]=1 12ml2 sin2θ−1 2sin 2θ 0 −1 2sin 2θ cos2θ 0 00 1 } 466 Chapter 9 Rigid-body dynamics xy θ l/2l/2 G z Figure P .9.7 9.8 The uniform rectangular box has a mass of 1000 kg. The dimensions of its edges are shown. (a) Find the mass moments of inertia about the xyzaxes. {Ans.: [ IO]= 1666.7−1500 −750 −1500 3333 .3−500 −750 −500 4333 .3 (kg·m2)} (b) Find the principal moments of inertia and the principal directions about the xyz axes through O. {Partial ans.: I1=568.9k g·m2,ˆv1=0.8366ˆi+0.4960ˆj+0.2326ˆk} (c) Find the moment of inertia about the line through Oand the point with coordinates (3 m, 2 m, 1 m). {Ans.: 583.3 kg ·m2} xyz 1m 2m3m O Figure P .9.8 Problems 467 9.9 A taxiing airplane turns about its vertical axis with an angular velocity /Omega1while its propeller spins at an angular velocity ω=˙θ. Determine the components of the angular momentum of the propeller about the body-fixed xyzaxes centered at P. Treat the propeller as a uniform slender rod of mass mand length l. {Ans.: HP=1 12mωl2ˆi−1 24m/Omega1l2sin 2θˆj+/parenleftbig1 12ml2cos2θ+md2/parenrightbig /Omega1ˆk} z x yP θ ωd C/H9024 Figure P .9.9 9.10 R e l a t i v et oa n ˆi,ˆj,ˆkframe of reference the components of angular momentum Hare given by {H}= 1000 0 −300 0 1000 500 −300 500 1000   ωx ωy ωz  (kg·m2/s) where ωx,ωyandωzare the components of the angular velocity ω. Find the components ωsuch that {H}=1000{ ω}, where the magnitude of ωis 20 rad/s. {Ans.: ω=17.15ˆi+10.29ˆj(rad/s)} 9.11 Relative to a body-fixed xyzframe [I G]= 1 000 02 0 000 3 0 (kg·m 2) and ω=2t2ˆi+ 4ˆj+3tˆk(rad/s), where tis the time in seconds. Calculate the magnitude of the net moment about the center of mass Gatt=3s . {Ans.: 3374 N ·m} 9.12 In Example 9.11, the system is at rest when a 100 N force is applied to point Aas shown. Calculate the inertial components of angular acceleration at that instant.{Ans.: α X=143.9r a d /s2,αY=553.1r a d /s2,αZ=7.61 rad /s2} 468 Chapter 9 Rigid-body dynamics XZ 4321 0.4 m 0.5 m 0.3 m 0.2 mOG (0.075 m, 0.2536 m, 0.05714 m)100 NA Y xyz Inertial frameBody-fixed frame Figure P .9.12 9.13 The body-fixed xyzaxes pass through the center of mass Gof the airplane and are the principal axes of inertia. The moments of inertia about these axes are A,Band C, respectively. The airplane is in a level turn of radius Rw i t has p e e d v. (a) Calculate the bank angle θ. (b) Use Euler’s equations to calculate the rolling moment Mywhich must be applied by the aerodynamic surfaces. {Ans.: (a) θ=tan−1v2/Rg; (b) My=v2sin 2θ(C−A)/2R2} mgLθ Rxz y G Figure P .9.13 Problems 469 9.14 The airplane in Problem 9.11 is spinning with an angular velocity ωZabout the vertical Z axis. The nose is pitched down at the angle α. What external moments must accompany this maneuver?{Ans.: M y=Mz=0,Mx=ω2 Zsin 2α (C−B)/2} Z ωZz yα G Figure P .9.14 9.15 Two identical slender rods of mass mand length lare rigidly joined together at an angleθat point C, their 2/3 point. Determine the bearing reactions at Aand Bif the shaft rotates at a constant angular velocity ω. Neglect gravity and assume that the only bearing forces are normal to rod AB. {Ans.: ||FA|| =mω2lsinθ(1+2c o sθ)/18, ||FB|| =mω2lsinθ(1−cosθ)/9} 2l/3 l/3θ A By xC2l/3 l/3 Figure P .9.15 9.16 The flywheel ( A=B=5k g· m2,C=10 kg·m2) spins at a constant angular velocity ofωs=100ˆk(rad/s ) .I ti ss u p p o r t e db ya massless gimbal which is mounted on the platform as shown. The gimbal is initially stat ionary relative to the platform, which rotates with a constant angular velocity of ωp=0.5ˆj(rad/s). What will be the gimbal’s angular acceleration when the torquer applies a torque of 600 ˆi(N·m) to the flywheel? {Ans.: 70 ˆirad/s2} 470 Chapter 9 Rigid-body dynamics TorquerGimbaly z PlatformFlywheel xThe xyz axes are attached to the gimbal/H9275s /H9275p Figure P .9.16 9.17 A uniform slender rod of length Land mass mis attached by a smooth pin at Oto a vertical shaft which rotates at constant angular velocity ω. Use Euler’s equations and the body frame shown to calculate ωat the instant shown. {Ans.: ω=/radicalbig 3g/(2Lcosθ)} ω z ygO G θL/32L/3 Figure P .9.17 Problems 471 9.18 A uniform, thin, circular disk of mass 10 kg spins at a constant angular velocity of 630 rad/s about axis OG, which is normal to the disk, and pivots about the frictionless ball joint at O. Neglecting the mass of the shaft OG, determine the rate of precession ifOG remains horizontal as shown. Gravity acts down, as shown. Gis the center of mass, and the yaxis remains fixed in space. The moments of inertia about Gare IGz=0.02812 kg ·m2, and IGx=IGy=0.01406 kg ·m2. {Ans.: 1.38 rad/s} z xy 630 rad/s gGm/H1100510 kg 0.25 m90˚O Figure P .9.18 9.19 At the end of its take-off run, an airplane with retractable landing gear leaves the runway with a speed of 130 km/hr. The gear rotates into the wing with an angular velocity of0.8 rad/s with the wheels still spinning. Calculate the gyroscopic bending moment inthe wheel bearing B. The wheels have a diameter of 0.6 m, a mass of 25 kg and a radius of gyration of 0.2 m.{Ans.: 96.3 N ·m}. B Figure P .9.19 9.20 The gyro rotor, including shaft AB, has a mass of 4 kg and a radius of gyration 7 cm around AB. The rotor spins at 10 000 revolutions per minute while also being forced to 472 Chapter 9 Rigid-body dynamics rotate around the gimbal axis CCat 2 rad/s. What are the transverse forces exerted on the shaft at Aand B? Neglect gravity. {Ans.: 1.03 kN} ABC C10 000 rev/min 2 rad/sy 2c m 2c m z x Figure P .9.20 9.21 A jet aircraft is making a level, 2.5 km radius turn to the left at a speed of 650 km/hr. The rotor of the turbojet engine has a mass of 200 kg, a radius of gyration of 0.25 mand rotates at 15 000 revolutions per minute clockwise as viewed from the front of theairplane. Calculate the gyroscopic moment that the engine exerts on the airframe andspecify whether it tends to pitch the nose up or down.{Ans.: 1.418 kN ·m; pitch down} ki 1570.8 rad/s0.07222 rad/s jForwardG MGC /H11005 12.5 kg · m2ˆ ˆˆ Figure P .9.21 9.22 A cylindrical rotor of mass 10 kg, radius 0.05 m and length 0.60 m is simply-supported at each end in a cradle that rotates at a constant 20 rad/s counterclockwise as viewedfrom above. Relative to the cradle, the rotor spins at 200 rad/s counterclockwise as viewed from the right (from Btowards A). Assuming there is no gravity, calculate the bearing reactions R AandRB. Use the co-moving xyzframe shown, which is attached to the cradle but not to the rotor.{Ans.: R A=− RB=83.3N } Problems 473 200 rad/s 20 rad/s CA Gz yx RA RBGB Figure P .9.22 9.23 The Euler angles of a rigid body are φ=50◦,θ=25◦andψ=70◦. Calculate the angle (a positive number) between the body-fixed xaxis and the inertial Xaxis. {Ans.: 115.6◦} 9.24 Consider a rigid body experiencing rotational motion associated with angular velocity ω. The inertia tensor (relative to body-fixed axes through the center of mass G)i s  20−10 0 −10 30 0 00 4 0 (kg·m2) andω=10ˆi+20ˆj+30ˆk(rad/s). Calculate (a) the angular momentum HG, and (b) the rotational kinetic energy (about G). {Partial ans.: (b) TR=23 000 J} This page int entionally left blank 10Chapter Satellite attitude dynamics Chapter outline 10.1 Introduction 475 10.2 Torque-free motion 476 10.3 Stability of torque-free motion 486 10.4 Dual-spin spacecraft 491 10.5 Nutation damper 495 10.6 Coning maneuver 503 10.7 Attitude control thrusters 506 10.8 Yo-yo despin mechanism 509 10.9 Gyroscopic attitude control 516 10.10 Gravity-gradient stabilization 530 Problems 543 10.1 Introduction In this chapter we apply the equations of rigid body motion presented in Chapter 9 to the study of the attitude dynamics of satellites. We begin with spin-stabilized spacecraft. Spinning a satellite around its axis is a very simple way to keep the vehiclepointed in a desired direction. We investigate the stability of a spinning satellite toshow that only oblate spinners are stable over long times. Overcoming this restrictionon the shape of spin-stabilized spacecraft led to the development of dual-spin vehicles,which consist of two interconnected segments rotating at different rates about a com-mon axis. We consider the stability of that type of configuration as well. The nutationdamper and its effect on the stability of spin-stabilized spacecraft is covered next. 475 476 Chapter 10 Satellite attitude dynamics The rest of the chapter is devoted to some of the common means of changing the attitude or motion of a spacecraft by applying external or internal forces ortorques. The coning maneuver changes the attitude of a spinning spacecraft by usingthrusters to apply impulsive torque, which alters the angular momentum and hencethe orientation of the spacecraft. The much-used yo-yo despin maneuver reduces oreliminates the spin rate by releasing small masses attached to cords initially wrappedaround the cylindrical vehicle. An alternative to spin stabilization is three-axis stabilization by gyroscopic attitude control. In this case, the vehicle does not continuously rotate. Instead, the desiredattitude is maintained by the spin of small wheels within the spacecraft. These arecalled reaction wheels or momentum wheels. If allowed to pivot relative to the vehicle,they are known as control moment gyros. The attitude of the vehicle can be changed by varying the speed or orientation of these internal gyros. Small thrusters may alsobe used to supplement the gyroscopic attitude control and to hold the spacecraftorientation fixed when it is necessary to despin or reorient gyros that have becomesaturated (reached their maximum spin rate or deflection) over time. The chapter concludes with a discussion of how the earth’s gravitational field by itself can stabilize the attitude of large satellites such as the space shuttle or spacestation in low earth orbits. 10.2 Torque-free motion Gravity is the only force acting on a satellit e coasting in orbit (if we neglect secondary drag forces and the gravitational influence of bodies other than the planet beingorbited). Unless the satellite is unusually large, the gravitational force is concentratedat the center of mass G. Since the net moment about the center of mass is zero, the satellite is ‘torque-free’ , and according to Equation 9.30, ˙H G=0 (10.1) The angular momentum HGabout the center of mass does not depend on time. It is a vector fixed in inertial space. We will use HGto define the Zaxis of an inertial frame, as shown in Figure 10.1. The xyzaxes in the figure comprise the principal body frame, centered at G. The angle between the zaxis and HGis (by definition of the Euler angles) the nutation angle θ. Let us determine the conditions for which θis constant. From the dot product operation we know that cosθ=HG /bardblHG/bardbl·ˆk Differentiating this expression with resp ect to time, keeping in mind Equation 10.1, we get dcosθ dt=HG /bardblHG/bardbl·dˆk dt 10.2 T orque-free motion 477 x yz HG G ˆjˆk ˆiInertial frameˆIˆJˆK Body frame Figure 10.1 Rotationally symmetric satellite in torque-free motion. Butdˆk/dt=ω׈k, according to Equation 1.24, so dcosθ dt=HG·(ω׈k) /bardblHG/bardbl(10.2) Now ω׈k=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆiˆjˆk ω xωyωz 001/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=ω yˆi−ωxˆj Furthermore, we know from Equation 9.67 that the angular momentum is related to the angular velocity in the principal body frame by the expression HG=Aωxˆi+Bωyˆj+Cωzˆk Thus HG·(ω׈k)=(Aω xˆi+Bωyˆj+Cωzˆk)·(ωyˆi−ωxˆj)=(A−B)ωxωy so that Equation 10.2 can be written ˙θ=ωn=−(A−B)ωxωy /bardblHG/bardblsinθ(10.3) From this we see that the nutation rate vanishes only if A=B.I fA/negationslash=B, the nutation angleθwill not in general be constant. Relative to the body frame, Equation 10.1 is written (cf. Equation 1.28) ˙HGrel+ω×HG=0 478 Chapter 10 Satellite attitude dynamics This is Euler’s equation with MGnet=0, the components of which are given by Equations 9.72, A˙ωx+(C−B)ωzωy=0 B˙ωy+(A−C)ωxωz=0 (10.4) C˙ωz+(B−A)ωyωx=0 In the interest of simplicity, let us consider the special case illustrated in Figure 10.1, namely that in which the zaxis is an axis of rotational symmetry, so that A=B. Then Equations 10.4 become A˙ωx+(C−A)ωzωy=0 A˙ωy+(A−C)ωxωz=0 (10.5) C˙ωz=0 From Equation 10.5 3we see that ωz=ω0(constant) (10.6) The assumption of rotational symmetry therefore reduces the three differential equa- tions 10.4 to just two. Substituting Equation 10.6 into Equations 10.5 1and 10.5 2and introducing the notation λ=A−C Aω0 (10.7) they can be written ˙ωx−λωy=0 ˙ωy+λωx=0 (10.8) T o reduce these two equations in ωxandωydown to just one equation in ωx,w efi r s t differentiate Equation 10.8 1with respect to time to get ¨ωx−λ˙ωy=0 (10.9) We then solve Equation 10.8 2for˙ωyand substitute the result into Equation 10.9, which leads to ¨ωx+λ2ωx=0 (10.10) The solution of this well-known differential equation is ωx=/Omega1sinλt (10.11) where the constant amplitude /Omega1(/Omega1/negationslash=0) has yet to be determined. (Without loss of generality, we have set the phase angle, the other constant of integration, equal tozero.) Substituting Equation 10.11 back into Equation 10.8 1yields the solution for ωy, ωy=1 λdωx dt=1 λd dt(/Omega1sinλt) 10.2 T orque-free motion 479 or ωy=/Omega1cosλt (10.12) Equations 10.6, 10.11 and 10.12 give the components of the absolute angular velocity ωalong the three principal body axes, ω=/Omega1sinλtˆi+/Omega1cosλtˆj+ω0ˆk or ω=ω⊥+ω0ˆk (10.13) where ω⊥=/Omega1(sinλtˆi+cosλtˆj) (10.14) ω⊥(‘omega-perp’) is the component of ωnormal to the zaxis. It sweeps out a circle of radius /Omega1in the xyplane at an angular velocity λ. Thus, ωsweeps out a cone, as illustrated in Figure 10.2. From Equations 9.115, the three Euler orientation angles (and their rates) are related to the angular velocity components ωx,ωyandωzby ωp=˙φ=1 sinθ(ωxsinψ+ωycosψ) ωn=˙θ=ωxcosψ−ωysinψ ωs=˙ψ=−1 tanθ(ωxsinψ+ωycosψ)+ωz xyz λt/H9275⊥Cone swept out by /H9275 in the body frame/H9275Ω ω0k γˆ Figure 10.2 Components of the angular velocity * in the body frame. 480 Chapter 10 Satellite attitude dynamics Substituting Equations 10.6, 10.11 and 10.12 into these three equations yields ωp=/Omega1 sinθcos(λt−ψ) ωn=/Omega1sin(λt−ψ) (10.15) ωs=ω0−/Omega1 tanθcos(λt−ψ) Since A=B, we know from Equation 10.3 that ωn=0. It follows from Equation 10.15 2that ψ=λt (10.16) (Actually, λt−ψ=nπ,n=0, 1, 2, .... We can set n=0 without loss of generality.) Substituting Equation 10.16 into Equations 10.15 1and 10.15 3yields ωp=/Omega1 sinθ(10.17) and ωs=ω0−/Omega1 tanθ(10.18) We have thus obtained the Euler angle rates ωpandωsin terms of the components of the angular velocity ω. Differentiating Equation 10.16 with respect to time shows that λ=˙ψ=ωs (10.19) That is, the rate λat which ωrotates around the body zaxis equals the spin rate. Substituting the spin rate for λin Equation 10.7 shows that ωsis related to ω0alone, ωs=A−C Aω0 (10.20) Eliminating ωsfrom Equations 10.18 and 10.20 yields the relationship between the magnitudes of the orthogonal components of the angular velocity in Equation 10.13, /Omega1=C Aω0tanθ (10.21) A similar relationship exists between ωpandωs, which generally are notorthogonal. Substitute Equation 10.21 into Equation 10.17 to obtain ω0=A Cωpcosθ (10.22) Placing this in Equation 10.20 leaves an expression involving only ωpandωs,f r o m which we get a useful formula relating the precession of a torque-free body to its spin, ωp=C A−Cωs cosθ(10.23) 10.2 T orque-free motion 481 xyz /H9275 γA/H9024 Cω0kˆ /H9275⊥ H⊥ω0kˆθHG ωstθvs /H9275p/H9024 Figure 10.3 Angular velocity and angular momentum vectors in the body frame. Observe that if A>C(i.e., the body is prolate , like a soup can or an American football), thenωphas the same sign as ωs, which means the precession is prograde .F o ra n oblate body (like a tuna fish can or a frisbee), A<Cand the precession is retrograde . The components of angular momentum along the body frame axes are obtained from the body frame components of ω, HG=Aωxˆi+Aωyˆj+Cωzˆk or HG=H⊥+Cω0ˆk (10.24) where H⊥=A/Omega1(sin ωstˆi+cosωstˆj)=Aω⊥ (10.25) Sinceω0ˆkand Cω0ˆkare colinear, as are ω⊥and Aω⊥, it follows that ˆk,ωand HG all lie in the same plane. HGandωboth rotate around the zaxis at the same rate ωs. These details are illustrated in Figure 10.3. See how the precession and spin angular velocities, ωpandωs, add up vectorially to give ω. Note also that from the point of view of inertial space, where HGis fixed, ωandˆkrotate around HGwith angular velocity ωp. Letγbe the angle between ωand the spin axis z, as shown in Figures 10.2 and 10.3.γis sometimes referred to as the wobble angle. Then cosγ=ωz /bardblω/bardbl=ω0/radicalBig /Omega12+ω2 0=ω0/radicalBigg/parenleftbigg ω0C Atanθ/parenrightbigg2 +ω2 0=A√ A2+C2tan2θ γis constant, since A,Candθare fixed. Using trig identities, this expression can be recast as cosγ=cosθ/radicalBigg C2 A2+/parenleftbigg 1−C2 A2/parenrightbigg cos2θ(10.26) 482 Chapter 10 Satellite attitude dynamics zHG /H9275/H9275pHG /H9275vsθθ ωs Space cone Body cone Body coneSpace conezγ /H9275s (a) Prograde precession (b) Retrograde precessionA /H11022 CA /H11021 Cγ /H9275s /H9275p Figure 10.4 Space and body cones for a rotationally symmetric body in torque-free motion. (a) Prolate body. (b) Oblate body. From this we conclude that if A>C, then γ<θ , whereas C>Ameans γ>θ . That is, the angular velocity vector ωlies between the zaxis and the angular momentum vector HGwhen A>C(prolate body). On the other hand, when C>A(oblate body), HGlies between the zaxis and ω. These two situations are illustrated in Figure 10.4, which also shows the body cone andspace cone . The space cone is swept out in inertial space by the angular velocity vector as it rotates with angular velocity ωparound HG, whereas the body cone is the trace of ωin the body frame as it rotates with angular velocity ωsabout the zaxis. From inertial space, the motion may be visualized as the body cone rolling on the space cone, with the line of contact being the angular velocityvector. From the body frame it appears as though the space cone rolls on the bodycone. Figure 10.4 graphically confirms our deduction from Equation 10.23, namely,that precession and spin are in the same direction for prolate bodies and opposite in direction for oblate shapes. Finally, we know from Equations 10.24 and 10.25 that the magnitude /bardblH G/bardblof the angular momentum is /bardblHG/bardbl=/radicalBig A2/Omega12+C2ω2 0 Using Equation 10.21, we can write this as /bardblHG/bardbl=/radicalBigg A2/parenleftbigg ω0C Atanθ/parenrightbigg2 +C2ω2 0=Cω0/radicalbig 1+tan2θ=Cω0 cosθ Substituting Equation 10.22 into this expression yields a surprisingly simple formula for the magnitude of the angular momentum, /bardblHG/bardbl= Aωp (10.27) 10.2 T orque-free motion 483 Example 10.1A cylindrical shell is rotating in torque-free motion about its longitudinal axis. If the axis is wobbling slightly, determine the ratios of l/rfor which the precession will be prograde or retrograde. lr/H9275 z Figure 10.5 Cylindrical shell in torque-free motion. Figure 9.9(b) shows the moments of inertia of a thin-walled circular cylinder, C=mr2A=1 2mr2+1 12ml2 According to Equation 10.23 and Figure 10.4, direct or prograde precession exists if A>C, that is, if 1 2mr2+1 12ml2>mr2 or 1 12ml2>1 2mr2 Thus l>2.45r ⇒ Direct precession . l<2.45r ⇒ Retrograde precession . Example 10.2In the previous example, let r=1m , l=3m , m=100 kg and the nutation angle θis 20◦. How long does it take the cylinder to precess through 180◦if the spin rate is 2 π radians per minute? Since l>2.45r , the precession is direct. Furthermore, C=mr2=100·12=100 kg ·m2 A=1 2mr2+1 12ml2=1 2·100·12+1 12100·32=125 kg ·m2 484 Chapter 10 Satellite attitude dynamics (Example 10.2 continued)Thus, Equation 10.23 yields ωp=C A−Cωs cosθ=100 125−1002π cos 20◦=26.75 rad/min At this rate, the time for a precession angle of 180◦is t=π ωp=0.1175 min Example 10.3What is the torque-free motion of a satellite for which A=B=C? IfA=B=C, the satellite is spherically symmetric. Any orthogonal triad at Gis a principal body frame, so HGandωare collinear, HG=Cω Substituting this and MGnet=0, into Euler’s equations, Equation 10.72a, yields Cdω dt+ω×(Cω)=0 That is, ω=constant The angular velocity vector of a spherically symmetric satellite is fixed in magnitude and direction. Example 10.4The inertial components of the angular momentum of a torque-free rigid body are HG=320ˆI−375ˆJ+450ˆK(kg·m2/s) (a) The Euler angles are φ=20◦θ=50◦ψ=75◦(b) If the inertia tensor in the body-fixed principal frame is [IG]= 1000 0 0 0 2000 00 0 3000 (kg·m 2)( c ) calculate the inertial components of the (absolute) angular acceleration. Substituting the Euler angles from (b) into Equation 9.117, we obtain the matrix of the transformation from the inertial frame to the body frame, [Q]Xx= 0.03086 0 .6720 0 .7399 −0.9646 −0.1740 0 .1983 0.2620 −0.7198 0 .6428  (d) 10.2 T orque-free motion 485 We use this to obtain the components of HGin the body frame, {HG}x=[Q]Xx{HG}X= 0.03086 0.6720 0.7399 −0.9646 −0.1740 0.1983 0.2620 −0.7198 0.6428   320 −375 450   =  90.86 −154.2 643.0  (kg·m2/s) (e) In the body frame {HG}x=[IG]{ω} x,w h e r e {ω} xare the components of angular velocity in the body frame. Thus   90.86 −154.2 643.0  = 1000 0 0 0 2000 00 0 3000 {ω} x or, solving for {ω} x, {ω} x= 1000 0 0 0 2000 00 0 3000 −1  90.86 −154.2 643.0  =  0.09086 −0.07709 0.2144  (rad/s) (f) Euler’s equations of motion (Equation 9.72a) may be written for the case at hand as [IG]{α}x+{ω}x×([IG]{ω} x)={0} (g) where {α}xis the absolute acceleration in body frame components. Substituting (c) and (f) into this expression, we get  1000 0 0 0 2000 00 0 3000 {α} x+  0.09086 −0.07709 0.2144   ×  1000 0 0 0 2000 00 0 3000   0.09086 −0.07709 0.2144   =  0 00   1000 0 0 0 2000 00 0 3000 {α} x+  −16.52 −38.95−7.005  =  0 00   so that, finally, {α}x=− 1000 0 0 0 2000 00 0 3000 −1  −16.52 −38.95−7.005  =  0.01652 0.01948 0.002335  (rad/s 2)( h ) 486 Chapter 10 Satellite attitude dynamics (Example 10.4 continued)These are the components of the angular acceleration in the body frame. T o transformthem into the inertial frame we use {α} X=[Q]xX{α}x=([Q]Xx)T{α}x = 0.03086 −0.9646 0 .2620 0.6720 −0.1740 −0.7198 0.7399 0 .1983 0 .6428   0.01652 0.01948 0.002335  =  −0.01766 0.006033 0.01759  (rad/s2) That is, α=− 0.01766ˆI+0.006033 ˆJ+0.01759 ˆK(rad/s2) 10.3 Stability of torque-free motion Let a rigid body be in torque-free motion with its angular velocity vector directed along the principal body zaxis, so that ω=ω0ˆk,w h e r e ω0is constant. The nutation angle is zero and there is no precession. Let us perturb the motion slightly, as illustrated inFigure 10.6, so that ω x=δωxωy=δωyωz=ω0+δωz (10.28) As in Chapter 7,‘ δ’ means a very small quantity. In this case, δωx/lessmuchω0andδωy/lessmuchω0. Thus, the angular velocity vector has become slightly inclined to the zaxis. For torque- free motion, MGx=MGy=MGz=0, so that Euler’s equations (Equations 9.72b) Gω0/H9275δωz δωxδωy xyz { Figure 10.6 Principal body axes of a rigid body rotating primarily about the body zaxis. 10.3 Stability of torque-free motion 487 become A˙ωx+(C−B)ωyωz=0 B˙ωy+(A−C)ωxωz=0 (10.29) C˙ωz+(B−A)ω xωy=0 Observe that we have not assumed A=B, as we did in the previous section. Substi- tuting Equations 10.28 into Equations 10.29 and keeping in mind our assumptionthat˙ω 0=0, we get Aδ˙ωx+(C−B)ω0δωy+(C−B)δω yδωz=0 Bδ˙ωy+(A−C)ω0δωx+(C−B)δω xδωz=0 (10.30) Cδ˙ωz+(B−A)δω xδωy=0 Neglecting all products of the δωs (because they are arbitrarily small), Equations 10.30 become Aδ˙ωx+(C−B)ω0δωy=0 Bδ˙ωy+(A−C)ω0δωx=0 (10.31) Cδ˙ωz=0 Equation 10.31 3implies that δωzis constant. Differentiating Equation 10.31 1with respect to time, we get Aδ¨ωx+(C−B)ω0δ˙ωy=0 (10.32) Solving Equation 10.31 2forδ˙ωyyields δ˙ωy=− [(A−C)/B]ω0δωx, and substituting this into Equation 10.32 gives δ¨ωx−(A−C)(C−B) ABω2 0δωx=0 (10.33) Likewise, by differentiating Equation 10.31 2and then substituting δ˙ωxfrom Equation 10.31 1yields δ¨ωy−(A−C)(C−B) ABω2 0δωy=0 (10.34) If we define k=(A−C)(B−C) ABω2 0 (10.35) then both Equations 10.33 and 10.34 may be written in the form δ¨ω+kδω=0 (10.36) Ifk>0, then δω∝e±i√ kt, which means δωxandδωyvary sinusoidally with small amplitude. The motion is therefore bounded and neutrally stable. That means theamplitude does not die out with time, but it does not exceed the small amplitude ofthe perturbation. Observe from Equation 10.35 that k>0 if either C>Aand C>B 488 Chapter 10 Satellite attitude dynamics orC<Aand C<B. This means that the spin axis ( zaxis) is either the major axis of inertia or the minor axis of inertia. That is, if the spin axis is either the major orminor axis of inertia, the motion is stable. The stability is neutral for a rigid body,because there is no damping. On the other hand, if k<0, then δω ∝e±√ kt, which means that the initially small perturbations δωxandδωyincrease without bound. The motion is unstable. From Equation 10.35 we see that k<0 if either A>Cand C>BorA<Cand C<B. This means that the spin axis is the intermediate axis of inertia ( A>C>BorB>C>A). If the spin axis is the intermediate axis of inertia, the motion is unstable. If the angular velocity of a satellite lies in the direction of its major axis of inertia, the satellite is called a major axis spinner or oblate spinner. A minor axis spinneror prolate spinner has its minor axis of inertia aligned with the angular velocity. ‘Intermediate axis spinners’ are unstable and will presumably end up being major or minor axis spinners, if the satellite is a rigid body. However, the flexibility inherent inany real satellite leads to an additional instability, as we shall now see. Consider again the rotationally symmetr ic satellite in torque-free motion dis- cussed in Section 10.2. From Equations 10.24 and 10.25, we know that the angularmomentum H Gis given by HG=Aω⊥+Cωzˆk (10.37) Hence, H2 G=A2ω2 ⊥+C2ω2 z (10.38) Differentiating this equatio n with respect to time yields dH2 G dt=A2dω2 ⊥ dt+2C2ωz˙ωz (10.39) But, according to Equation 10.1, HGis constant, so that dH2 G/dt=0 and Equation 10.39 can be written dω2 ⊥ dt=− 2C2 A2ωz˙ωz (10.40) The rotary kinetic energy of a rotationally symmetric body ( A=B) is found using Equation 9.81, TR=1 2Aω2 x+1 2Aω2 y+1 2Cω2 z=1 2A(ω2 x+ω2 y)+1 2Cω2 z From Equation 10.13 we know that ω2 x+ω2 y=ω2 ⊥, which means TR=1 2Aω2 ⊥+1 2Cω2 z (10.41) The time derivative of TRis, therefore, ˙TR=1 2Adω2 ⊥ dt+Cωz˙ωz 10.3 Stability of torque-free motion 489 Solving this for ˙ωz,w eg e t ˙ωz=1 Cωz/parenleftbigg ˙TR−1 2Adω2 ⊥ dt/parenrightbigg Substituting this expression for ˙ωzinto Equation 10.40 and solving for dω2 ⊥/dtyields dω2 ⊥ dt=2C A˙TR C−A(10.42) Real bodies are not completely rigid, and their flexibility, however slight, gives rise to small dissipative effects which cause the kineti c energy to decrease over time. That is, ˙TR<0 For satellites with dissipation . (10.43) Substituting this inequality into Equation 10.42 leads us to conclude that dω2 ⊥ dt<0i f C>A(oblate spinner) (10.44) dω2 ⊥ dt>0i f C<A(prolate spinner) Ifdω2 ⊥/dtis negative, the spin is asymptotically stable. Should a non-zero value of ω⊥develop for some reason, it will drift back to zero over time so that once again the angular velocity lies completely in the spin direction. On the other hand, if dω2 ⊥/dtis positive, the spin is unstable. ω⊥does not damp out, and the angular velocity vector drifts away from the spin axis as ω⊥increases without bound. We pointed out above that spin about a minor axis of inertia is stable with respect to small disturbances. Nowwe see that only major axis spin is stable in the long run if dissipative mechanisms exist. For some additional insight into this phenomenon, solve Equation 10.38 for ω 2 ⊥, ω2 ⊥=H2 G−C2ω2 z A2 and substitute this result into the expression for kinetic energy, Equation 10.41, to obtain TR=1 2H2 G A+1 2(A−C)C Aω2 z (10.45) According to Equation 10.24, ωz=HGz C=HGcosθ C Substituting this into Equation 10.45 yields the kinetic energy as a function of just the inclination angle θ, TR=1 2H2 G A/parenleftbigg 1+A−C Ccos2θ/parenrightbigg (10.46) The extreme values of TRoccur at θ=0o rθ =π, TR=1 2H2 G C(major axis spinner) 490 Chapter 10 Satellite attitude dynamics andθ=π/2, TR=1 2H2 G A(minor axis spinner) Clearly, the kinetic energy of a torque-free satellite is smallest when the spin is around the major axis of inertia. We may think of a satellite with dissipation ( dTR/dt<0) as seeking the state of minimum kinetic energy that occurs when it spins about itsmajor axis. Example 10.5A rigid spacecraft is modeled by the solid cylinder Bwhich has a mass of 300 kg and the slender rod Rwhich passes through the cylinder and has a mass of 30 kg. Which of the principal axes x,y,zcan be an axis about which stable torque-free rotation can occur? 0.5 m xyz G 1.0 m 1.0 m0.5 m 0.5 m Figure 10.7 Built-up satellite structure. For the cylindrical shell A,w eh a v e rB=0.5m lB=1.0m mB=300 kg The principal moments of inertia about the center of mass are found in Figure 10.9(b), IBx=1 4mBr2 B+1 12mBl2 B=43.75 kg·m2 IBy=IBxx=43.75 kg·m2 IBz=1 2mBr2 B=37.5k g·m2 10.4 Dual-spin spacecraft 491 The properties of the transverse rod are lR=1.0m mR=30 kg Figure 10.9(a), with r=0, yields the moments of inertia, IRy=0 IRz=IRx=1 12mAr2 A=10.0k g ·m2 The moments of inertia of the assembly is the sum of the moments of inertia of the cylinder and the rod, Ix=IBx+IRx=53.75 kg ·m2 Iy=IBy+IRy=43.75 kg ·m2 Iz=IBz+IRz=47.50 kg ·m2 Since Izis the intermediate mass moment of inertia, rotation about the zaxis is unstable. With energy dissipation, rotation is stable in the long term only about themajor axis, which in this case is the xaxis. 10.4 Dual-spin spacecraft If a satellite is to be spin stabilized, it must be an oblate spinner. The diameter of the spacecraft is restricted by the cross-section of the launch vehicle’s upper stage, andits length is limited by stability requirements. Therefore, oblate spinners cannot takefull advantage of the payload volume available in a given launch vehicle, which afterall are slender, prolate shapes for aerodynamic reasons. The dual-spin design permitsspin stabilization of a prolate shape. The axisymmetric, dual-spin configuration, or gyrostat, consists of an axisymmet- ric rotor and a smaller axisymmetric platform joined together along a common lon-gitudinal spin axis at a bearing, as shown in Figure 10.8. The platform and rotor havetheir own components of angular velocity, ω pandωrrespectively, along the spin axis direction ˆk. The platform spins at a much slower rate than the rotor. The assembly acts like a rigid body as far as transverse rotations are concerned; i.e., the rotor and the plat-form have ω ⊥in common. An electric motor integrated into the axle bearing connect- ing the two components acts to overcome frictional torque which would otherwiseeventually cause the relative angular velocity between the rotor and platform to go tozero. If that should happen, the satellite would become a single spin unit, probably anunstable prolate spinner, since the rotor of a dual-spin spacecraft is likely to be prolate. The first dual-spin satellite was OSO-I (Orbiting Solar Observatory), which NASA launched in 1962. It was a major-axis spinner. The first prolate dual-spin spacecraftwas the two-storey tall TACSAT I (Tactical Communications Satellite). It was launchedinto geosynchronous orbit by the US Air Force in 1969. Typical of many of today’s 492 Chapter 10 Satellite attitude dynamics z G GrGp Bearing/H9275p /H9275r/H9275⊥Platform Rotor Figure 10.8 Axisymmetric, dual-spin satellite. communications satellites, TACSAT’s platform rotated at one revolution per day to keep its antennas pointing towards the earth. The rotor spun at about one revolutionper second. Of course, the axis of the spacecraft was normal to the plane of its orbit.The first dual-spin interplanetary spacecraft was Galileo, which we discussed brieflyin Section 8.9. Galileo’s platform was completely despun to provide a fixed orientation for cameras and other instruments. The rotor spun at three revolutions per minute. The equations of motion of a dual-spin spacecraft will be developed later in Section 10.8. Let us determine the stability of the motion by following the same ‘energy sink’ procedure employed in the previous section for a single-spin stabilized spacecraft. The angular momentum of the dual-spin configuration about the space-craft’s center of mass Gis the sum of the angular momenta of the rotor ( r) and the platform ( p)a b o u t G, H G=H(p) G+H(r) G(10.47) The angular momentum of the platform about the spacecraft center of mass is H(p) G=Cpωpˆk+Apω⊥ (10.48) 10.4 Dual-spin spacecraft 493 where Cpis the moment of inertia of the platform about the spacecraft spin axis, and Apis its transverse moment of inertia about G(not Gp). Likewise, for the rotor, H(r) G=Crωrˆk+Arω⊥ (10.49) where Crand Arare its longitudinal and transverse moments of inertia about axes through G. Substituting Equations 10.48 and 10.49 into 10.47 yields HG=(Crωr+Cpωp)ˆk+A⊥ω⊥ (10.50) where A⊥is the total transverse moment of inertia, A⊥=Ap+Ar From this it follows that H2 G=(Crωr+Cpωp)2+A2 ⊥ω2 ⊥ For torque-free motion, ˙HG=0, so that dH2 G/dt=0, or 2(C rωr+Cpωp)(C r˙ωr+Cp˙ωp)+A2 ⊥dω2 ⊥ dt=0 (10.51) Solving this for dω2 ⊥/dtyields dω2 ⊥ dt=−2 A2 ⊥(Crωr+Cpωp)(C r˙ωr+Cp˙ωp) (10.52) The total rotational kinetic energy of rotation of the dual spin spacecraft is the sum of that of the rotor and the platform, T=1 2Crω2 r+1 2Cpω2 p+1 2A⊥ω2 ⊥ Differentiating this expression w ith respect to time and solving for dω2 ⊥/dtyields dω2 ⊥ dt=2 A⊥(˙T−Crωr˙ωr−Cpωp˙ωp) (10.53) ˙Tis the sum of the power P(r)dissipated in the rotor and the power P(p)dissipated in the platform, ˙T=P(r)+P(p)(10.54) Substituting Equation 10.54 into 10.53 we find dω2 ⊥ dt=2 A⊥(P(r)−Crωr˙ωr+P(p)−Cpωp˙ωp) (10.55) Equating the two expressions for dω2 ⊥/dtin Equations 10.52 and 10.55 yields 2 A⊥(˙T−Crωr˙ωr−Cpωp˙ωp)=−2 A2 ⊥(Crωr+Cpωp)(C r˙ωr+Cp˙ωp) 494 Chapter 10 Satellite attitude dynamics Solve this for ˙Tto obtain ˙T=Cr A⊥[(A⊥−Cr)ωr−Cpωp]˙ωr+Cp A⊥[(A⊥−Cp)ωp−Crωr]˙ωp (10.56) Following Likins (1967), we identify the terms containing ˙ωrand˙ωpas the power dissipation in the rotor and platform, respectively. That is, comparing Equations 10.54and 10.56, P (r)=Cr A⊥[(A⊥−Cr)ωr−Cpωp]˙ωr (10.57a) P(p)=Cp A⊥[(A⊥−Cp)ωp−Crωr]˙ωp (10.57b) Solving these two expressions for ˙ωrand˙ωp, respectively, yields ˙ωr=A⊥ CrP(r) (A⊥−Cr)ωr−Cpωp(10.58a) ˙ωp=A⊥ CpP(p) (A⊥−Cp)ωp−Crωr(10.58b) Substituting these results into Equation 10.55 leads to dω2 ⊥ dt=2 A⊥/bracketleftBigg P(r) Cpωp ωr−(A⊥−Cr)+P(p) Cr−(A⊥−Cp)ωp ωr/bracketrightBigg/parenleftbigg Cr+Cpωp ωr/parenrightbigg (10.59) As pointed out above, for geosynchronous dual-spin communication satellites, ωp ωr≈2πrad/d 2πrad/s≈10−5 whereas for interplanetary dual-spin spacecraft, ωp=0. Therefore, there is an impor- tant class of spin stabilized spacecraft for which ωp/ωr≈0. For a despun platform wherein ωpis zero (or nearly so), Equation 10.59 yields dω2 ⊥ dt=2 A⊥/bracketleftbigg P(p)+Cr Cr−A⊥P(r)/bracketrightbigg (10.60) If the rotor is oblate ( Cr>A⊥), then, since P(r)and P(p)are both negative, it follows from Equation 10.60 that dω2 ⊥/dt<0. That is, the oblate dual spin configuration with a despun platform is unconditionally stable. In practice, however, the rotor islikely to be prolate ( C r<A⊥), so that Cr Cr−A⊥P(r)>0 In that case, dω2 ⊥/dt<0 only if the dissipation in the platform is significantly greater than that of the rotor. Specifically, for a prolate design it must be true that |P(p)|>/vextendsingle/vextendsingle/vextendsingle/vextendsingleC r Cr−A⊥P(r)/vextendsingle/vextendsingle/vextendsingle/vextendsingle 10.5 Nutation damper 495 The platform dissipation rate P(p)can be augmented by adding nutation dampers, which are discussed in the next section. For the despun prolate dual-spin configuration, Equations 10.58 imply ˙ωr=P(r) (A⊥−Cr)A⊥ Crωr ˙ωp=−P(p) CpA⊥ Crωr Clearly, the signs of ˙ωrand˙ωpare opposite. If ωr>0, then dissipation causes the spin rate of the rotor to decrease and that of the platform to increase. Were it not for theaction of the motor on the shaft connecting the two components of the spacecraft, eventually ω p=ωr. That is, the relative motion between the platform and rotor would cease and the dual-spinner would become an unstable single spin spacecraft. Settingω p=ωrin Equation 10.59 yields dω2 ⊥ dt=2Cr+Cp A⊥P(r)+P(p) (Cr+Cp)−A⊥ which is the same as Equation 10.42, the energy sink conclusion for a single spinner. 10.5 Nutation damper Nutation dampers are passive means of dissipating energy. A common type consists essentially of a tube filled with viscous fluid and containing a mass attached to springs,as illustrated in Figure 10.9. Dampers may contain just fluid, only partially filling the tube so it can slosh around. In either case, the purpose is to dissipate energy throughfluid friction. The wobbling of the spacecraft due to non-alignment of the angular z x yP m zmWz czm kzmNx Nym WxWy (b)/H9275 (a)RrG Figure 10.9 (a) Precessing oblate spacecraft with a nutation damper aligned with the zaxis. (b) Free-body diagram of the moving mass in the nutation damper. 496 Chapter 10 Satellite attitude dynamics velocity with the principal spin axis induces accelerations throughout the satellite, giving rise to the sloshing of fluids, stretching and flexing of non-rigid components,etc., all of which dissipate energy to one degree or another. Nutation dampers areadded to deliberately increase energy dissipation, which is desirable for stabilizingoblate single spinners and dual-spin spacecraft. Let us focus on the motion of the mass within the nutation damper of Figure 10.9 in order to gain some insight into how relative motion and deformation are inducedby the satellite’s precession. Note that point Pis the center of mass of the rigid satellite body itself. The center of mass Gof the satellite-damper mass combination lies between Pand m, as shown in Figure 10.9. We suppose that the tube is lined up with the zaxis of the body-fixed xyzframe, as shown. The mass min the tube is therefore constrained by the tube walls to move only in the zdirection. When the springs are undeformed, the mass lies in the xyplane. In general, the position vector ofmin the body frame is r=Rˆi+z mˆk (10.61) where zmis the zcoordinate of mand Ris the distance of the damper from the centerline of the spacecraft. The velocity and acceleration of mrelative to the satellite are, therefore, vrel=˙zmˆk (10.62) arel=¨zmˆk (10.63) The absolute angular velocity ωof the satellite (and, therefore, the body frame) is ω=ωxˆi+ωyˆj+ωzˆk (10.64) Recall Equation 9.73, which states that when ωis given in a body frame, we find the absolute angular acceleration by taking the time derivative of ω, holding the unit vectors fixed. Thus, ˙ω=˙ωxˆi+˙ωyˆj+˙ωzˆk (10.65) The absolute acceleration of mis found using Equation 1.42, which for the case at hand becomes a=aP+˙ω×r+ω×(ω×r)+2ω×vrel+arel (10.66) in which aPis the absolute acceleration of the reference point P. Substituting Equa- tions 10.61 through 10.65 into Equation 10.66, carrying out the vector operations,combining terms, and simplifying leads to the following expressions for the threecomponents of the inertial acceleration of m, a x=aPx−R(ω2 y+ω2 z)+zm˙ωy+zmωxωz+2˙zmωy ay=aPy+R˙ωz+Rωxωy−zm˙ωx+zmωyωz−2˙zmωx (10.67) az=aPz−zm(ω2 x+ω2 y)−R˙ωy+Rωxωz+¨zm Figure 10.9(b) shows the free-body diagram of the damper mass m. In the xand y directions the forces on mare the components of the force of gravity ( Wxand Wy) 10.5 Nutation damper 497 and the components Nxand Nyof the force of contact with the smooth walls of the damper tube. The directions assumed for these components are, of course, arbitrary.In the zdirection, we have the zcomponent W zof the weight, plus the force of the springs and the viscous drag of the fluid. The spring force ( −kz m) is directly proportional and opposite in direction to the displacement zm.kis the net spring constant. The viscous drag ( −c˙zm) is directly proportional and opposite in direction to the velocity ˙zmofmrelative to the tube. cis the damping constant. Thus, the three components of the net force on the damper mass mare Fnet x=Wx−Nx Fnet y=Wy−Ny (10.68) Fnet z=Wz−kzm−c˙zm Substituting Equations 10.67 and 10.68 into Newton’s second law, Fnet=ma, yields Nx=mR(ω2 y+ω2 z)−mzm˙ωy−mzmωxωy−2m˙zmωy+=0/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright (Wx−maPx) Ny=− mR˙ωz−mRωxωy+mzm˙ωx−mzmωyωz +2m˙zmωx+=0/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright (Wy−maPy) (10.69) m¨zm+c˙zm+[k−m(ω2 x+ω2 y)]zm=mR(˙ωy−ωxωz)+=0/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright (Wz−maPz) The last terms in parentheses in each of these expressions vanish if the acceleration of gravity is the same at mas at the reference point Pof the spacecraft. This will be true unless the satellite is of enormous size. If the damper mass mis vanishingly small compared to the mass Mof the rigid spacecraft body, then it will have little effect on the rotary motion. If the rotationalstate is that of an axisymmetric satellite in torque-free motion, then we know fromEquations 10.13, 10.14 and 10.19 that ω x=/Omega1sinωst ωy=/Omega1cosωst ωz=ω0 ˙ωx=/Omega1ω scosωst ˙ωy=−/Omega1ω ssinωst ˙ωz=0 in which case Equations 10.69 become Nx=mR(ω2 0+/Omega12cos2ωst)+m(ω s−ω0)/Omega1z msinωst−2m/Omega1˙zmcosωst Ny=− mR/Omega12cosωstsinωst+m(ω s−ω0)/Omega1z mcosωst+2m/Omega1˙zmsinωst(10.70) m¨zm+c˙zm+(k−m/Omega12)zm=− mR(ωs+ω0)/Omega1sinωst Equation 10.70 3is that of a single degree of freedom, damped oscillator with a sinu- soidal forcing function. The precession produces a force of amplitude m(ω 0+ωs)/Omega1R and frequency ωswhich causes the damper mass mto oscillate back and forth in the tube, such that zm=mR/Omega1(ω s+ω0) [k−m(ω2s+/Omega12)]2+(cωs)2{cωscosωst−[k−m(ω2 s+/Omega12) sinωst]} 498 Chapter 10 Satellite attitude dynamics Observe that the contact forces Nxand Nydepend exclusively on the amplitude and frequency of the precession. If the angular velocity lines up with the spin axis, so that/Omega1=0 (precession vanishes), then N x=mω2 0R Ny=0 No precession . zm=0 If precession is eliminated, so there is pure spin around the principal axis, the time- varying motions and forces vanish throughout the spacecraft, which thereafter rotatesas a rigid body with no energy dissipation. Now, the whole purpose of a nutation damper is to interact with the rotational motion of the satellite so as to damp out any tendencies to precess. Therefore, its massshould not be ignored in the equations of motion of the satellite. We will derive theequations of motion of the rigid satellite with nutation damper to show how rigid bodymechanics is brought to bear upon the problem and, simply, to discover precisely whatwe are up against in even this extremely simplified system. We will continue to use Pas the origin of our body frame. Since a moving mass has been added to the rigid satelliteand since we are not using the center of mass of the system as our reference point, wecannot use Euler’s equations. Applicable to the case at hand is Equation 9.33, accord-ing to which the equation of rotational motio n of the system of satellite plus damper is ˙H Prel+rG/P×(M+m)aP/G=MGnet (10.71) The angular momentum of the satellite body plus that of the damper mass, relative to point Pon the spacecraft, is HPrel=body of the spacecraft/bracehtipdownleft /bracehtipupright/bracehtipupleft /bracehtipdownright Aωxˆi+Bωyˆj+Cωzˆk+damper mass/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright r×m˙r (10.72) where the position vector ris given by Equation 10.61. According to Equation 1.28, ˙r=dr dt/parenrightbigg rel+ω×r=˙zmˆk+/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆiˆjˆk ω xωyωz R 0 z/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =ω yzmˆi+(ωzR−ωxzm)ˆj+(˙zm−ωyR)ˆk After substituting this into Equation 10.72 and collecting terms we obtain HPrel=[(A+mz2 m)ωx−mRz mωz]ˆi +[(B+mR2+mz2 m)ωy−mR˙zm]ˆj +[(C+mR2)ωz−mRz mωx]ˆk (10.73) T o calculate ˙HPrel, we again use Equation 1.28, ˙HPrel=dHPrel dt/parenrightbigg rel+ω×HPrel 10.5 Nutation damper 499 Carrying out the operations on the right leads eventually to ˙HPrel=[(A+mz2 m)˙ωx−mRz m˙ωz+(C−B−mz2 m)ωyωz −mRz mωxωy+2mz m˙zmωx]ˆi +{(B+mR2+mz2 m)˙ωy+mRz m(ω2 x−ω2 z) +[A+mz2 m−(C+mR2)]ω xωz+2mz m˙zmωy−mR¨zm}ˆj +[−mRz m˙ωx+(C+mR2)˙ωz+(B+mR2−A)ω xωy +mRz mωyωz−2mR˙zmωx]ˆk (10.74) T o calculate the second term on the left of Equation 10.71, we keep in mind that Pis the center of mass of the body of the satellite and first determine the position vectorof the center of mass Gof the vehicle plus damper relative to P, (M+m)r G/P=M(0)+mr (10.75) where r, the position of the damper mass mrelative to P, is given by Equation 10.61. Thus rG/P=m m+Mr=µr=µ(Rˆi+zmˆk) (10.76) in which µ=m m+M(10.77) Thus, rG/P×(M+m)a P/G=/parenleftbiggm M+m/parenrightbigg r×(M+m)aP/G=r×maP/G (10.78) The acceleration of Prelative to Gis found with the aid of Equation 1.32, aP/G=− ¨ rG/P=−µd2r dt2=−µ/bracketleftbiggd2r dt2/parenrightbigg rel+˙ω×r+ω×(ω×r)+2ω×dr dt/parenrightbigg rel/bracketrightbigg (10.79) where dr dt/parenrightbigg rel=dR dtˆi+dzm dtˆk=˙zmˆk (10.80) and d2r dt2/parenrightbigg rel=d2R dt2ˆi+d2zm dt2ˆk=¨zmˆk (10.81) Substituting Equations 10.61, 10.64, 10.65, 10.80 and 10.81 into Equation 10.79 yields aP/G=[−µz m˙ωy+µR(ω2 y+ω2 z)−µzmωxωz−2µ˙zmωy]ˆi +(µz m˙ωx−µR˙ωz−µRωxωy−µzmωyωz+2µ˙zmωx)ˆj +[µR˙ωy+µzm(ω2 x+ω2 y)−µRωxωz−µ¨zm]ˆk (10.82) 500 Chapter 10 Satellite attitude dynamics We move this expression into Equation 10.78 to get rG/P×(M+m)aP/G =[−µmz2 m˙ωx−2µm¨zmωx+µmR(ωxωy+˙ωz)+µmz2 mωyωz]ˆi +[−µm(R2+z2 m)˙ωy−2µmzm˙zmωy+µmRz m(ω2 z−ω2 x) +µm(R2−z2 m)ωxωz+µm¨Rzm]ˆj +(µmRz m˙ωx−µmR2˙ωz+2µmR˙zmωx −µmR2ωxωy−µmRz mωyωz)ˆk Placing this result and Equation 10.74 in Equation 10.71, and using the fact that MGnet=0, yields a vector equation whose three components are A˙ωx+(C−B)ωyωz+(1−µ)mz2 m˙ωx−(1−µ)mz2 mωyωz +2(1−µ)mzm˙zmωx−(1−µ)mRz mωxωy=0 [B+(1−µ)mR2]˙ωy+[A−C−(1−µ)mR2]ωxωz +(1−µ)mz2 m(ωxωz+˙ωy)+2(1−µ)mzm˙zmωy (10.83) −(1−µ)mR¨zm+(1−µ)mRz m(ω2 x−ω2 z)=0 [C+(1−µ)mR2]˙ωz+[B−A+(1−µ)mR2]ωxωy +(1−µ)mRz mωyωz−2(1−µ)mR˙zmωx−(1−µ)mRz m˙ωx=0 These are three equations in the four unknowns ωx,ωy,ωzand zm. The fourth equation is that of the motion of the damper mass min the zdirection, Wz−kzm−c˙zm=maz (10.84) where azis given by Equation 10.67 3, in which aPz=aPz−aGz+aGz=aP/Gz+aGz, so that az=aP/Gz+aGz−zm(ω2 x+ω2 y)−R˙ωy+Rωxωz+¨zm (10.85) Substituting the zcomponent of Equation 10.82 into this expression and that result into Equation 10.84 leads (with Wz=maGz)t o (1−µ)m¨zm+c˙zm+[k−(1−µ)m(ω2 x+ω2 y)]zm=(1−µ)mR[˙ωy−ωxωz] (10.86) Compare Equation 10.69 3with this expression, which is the fourth equation of motion we need. Equations 10.83 and 10.86 are a rather complicated set of non-linear, second order differential equations, which must be solved (numerically) to obtain a precise description of the motion of the semirigid spacecraft. That is beyond our scope.However, to study their stability we can linearize the equations in much the same wayas we did in Section 10.3. (Note that Equations 10.83 reduce to 10.29 when m=0.)W e assume the satellite is in pure spin with angular velocity ω 0about the zaxis and that the damper mass is at rest ( zm=0). This motion is slightly perturbed, in such a way that ωx=δωxωy=δωyωz=ω0+δωz zm=δzm (10.87) 10.5 Nutation damper 501 It will be convenient for this analysis to introduce operator notation for the time derivative, D=d/dt. Thus, given a function of time f(t), for any integer n, Dnf=dnf/dtn, and D0f(t)=f(t). Then the various time derivatives throughout the equations will, in accordance with Equation 10.87, be replaced as follows, ˙ωx=Dδωx˙ωy=Dδωy˙ωz=Dδωz˙zm=Dδzm¨zm=D2δzm(10.88) Substituting Equations 10.87 and 10.88 into Equations 10.83 and 10.86 and retaining only those terms which are at most linear in the small perturbations leads to ADδωx+(C−B)ω0δωy=0 [A−C−(1−µ)mR2]ω0δωx+[B+(1−µ)mR2]Dδωy −(1−µ)mR (D2+ω2 0)δzm=0 (10.89) [C+(1−µ)mR2]Dδωz=0 (1−µ)mRω0δωx−(1−µ)mRD δωy+[(1−µ)mD2+cD+k]δzm=0 δωzappears only in the third equation, which states that δωz=constant. The first, second and fourth equations may be combined in matrix notation,  AD (C−B)ω0 0 [A−C−(1−µ)mR2]ω0[B+(1−µ)mR2]D−(1−µ)mR (D2+ω2 0) (1−µ)mRω0 −(1−µ)mRD (1−µ)mD2+cD+k  ×  δwx δωy δzm  =  0 00  (10.90) This is a set of three linear differential equations in the perturbations δω x,δωyand δzm. We won’t try to solve them, since all we are really interested in is the stability of the satellite-damper system. It can be shown that the determinant /Delta1o ft h e3b y3 matrix in Equation 10.90 is /Delta1=a4D4+a3D3+a2D2+a1D+a0 (10.91) in which the coefficients of the characteristic equation /Delta1=0a r e a4=(1−µ)mAB a3=cA[B+(1−µ)mR2] a2=k[B+(1−µ)mR2]A+(1−µ)m[(A −C)(B−C) −(1−µ)AmR2]ω2 0 (10.92) a1=c{[A−C−(1−µ)mR2](B−C)}ω2 0 a0=k{[A−C−(1−µ)mR2](B−C)}ω2 0+[(B−C)(1−µ)2]m2R2ω4 0 According to the Routh–Hurwitz stability criteria (see any text on control systems, e.g., Palm, 1983), the motion represented by Equations 10.90 is asymptotically stableif and only if the signs of all of the following quantities, defined in terms of thecoefficients of the characteristic equation, are the same r 1=a4r2=a3r3=a2−a4a1 a3r4=a1−a3a0 a3a2−a4a1r5=a0(10.93) 502 Chapter 10 Satellite attitude dynamics Example 10.6A satellite is spinning about the zaxis of its principal body frame at 2 πradians per second. The principal moments of inertia about its center of mass are A=300 kg ·m2B=400 kg ·m2C=500 kg ·m2(a) For the nutation damper, the following properties are given R=1m µ=0.01 m=10 kg k=10 000 N /m c=150 N ·s/m (b) Use the Routh–Hurwitz stability criteria to assess the stability of the satellite as a major-axis spinner, a minor-axis spinner, and an intermediate-axis spinner. The data in (a) are for a major-axis spinner. Substituting into Equations 10.92 and 10.93, we find r1=+ 1.188×106kg3m4 r2=+ 18.44×106kg3m4/s r3=+ 1.228×109kg3m4/s2(c) r4=+ 92 820 kg3m4/s3 r5=+ 8.271×109kg3m4/s4 Since the rs are all positive, spin about the majo r axis is asymptotically stable. As we know from Section 10.3, without the damper the motion is neutrally stable. For spin about the minor axis, A=500 kg ·m2B=400 kg ·m2C=300 kg ·m2(d) For these moment of inertia values, we obtain r1=+ 1.980×106kg3m4 r2=+ 30.74×106kg3m4/s r3=+ 2.048×109kg3m4/s2(e) r4=− 304 490 kg3m4/s3 r5=+ 7.520×109kg3m4/s4 Since the rs are not all of the same sign, spin about the minor axis is not asymptotically stable. Recall that for the rigid satellite, such a motion was neutrally stable. Finally, for spin about the intermediate axis, A=300 kg ·m2B=500 kg ·m2C=400 kg ·m2(f) We know this motion is unstable, even without the nutation damper, but doing the Routh–Hurwitz stability check anyway, we get r1=+ 1.485×106kg3m4 r2=+ 22.94×106kg3m4/s 10.6 Coning maneuver 503 r3=+ 1.529×109kg3m4/s2 r4=− 192 800 kg3m4/s3 r5=− 4.323×109kg3m4/s4 The motion, as we expected, is not stable. 10.6 Coning maneuver Like the use of nutation dampers, the coning maneuver is an example of the attitude control of spinning spacecraft. In this case, the angular momentum is changed by theuse of on-board thrusters (small rockets) to apply pure torques. Consider a satellite in pure spin with angular momentum H G0. Suppose we wish to maintain the magnitude of the angular momentum but change its direction byrotating the spin axis through an angle θ, as illustrated in Figure 10.10. Recall from Section 9.4 that to change the angular momentum of the spacecraft requires applyingan external moment, /Delta1H G=/integraldisplay/Delta1t 0MGdt HG0HGf T TTTθ/2 θ/2 /H9004H G2 /H9004H G1 Figure 10.10 Impulsive coning maneuver. 504 Chapter 10 Satellite attitude dynamics ∆HG∆HG HG0 HGf θ/2 Figure 10.11 A sequence of small coning maneuvers. Thrusters may be used to provide the external impulsive torque required to produce an angular momentum increment /Delta1HG1normal to the spin axis. Since the spacecraft is spinning, this induces coning (precession) of the satellite about an axis at an angleθ/2t o H G0. The precession rate is given by Equation 10.23, ωp=C A−Cωs cos (θ 2)(10.94) After precessing 180◦, an angular momentum increment /Delta1HG2normal to the spin axis and in the same direction relative to the spacecraft as the initial torque impulse,with/bardbl/Delta1H G2/bardbl=/bardbl/Delta1HG1/bardbl, stabilizes the spin vector in the desired direction. The time required for an angular reorientation θusing a single coning maneuver is found by simply dividing the precession angle, πradians, by the precession rate ωp, t1=π ωp=πA−C Cωscosθ 2(10.95) Propellant expenditure is reflected in the magnitude of the individual angular momentum increments, in obvious analogy to delta-v calculations for orbital maneu-vers. The total delta-H required for the single coning maneuver is therefore given by /Delta1H total=/vextenddouble/vextenddouble/Delta1HG1/vextenddouble/vextenddouble+/vextenddouble/vextenddouble/Delta1HG2/vextenddouble/vextenddouble=2/parenleftbigg/vextenddouble/vextenddoubleHG0/vextenddouble/vextenddoubletanθ 2/parenrightbigg (10.96) Figure 10.11 illustrates the fact that /Delta1Htotal can be reduced by using a sequence of small coning maneuvers (small θs) rather than one big θ. The large number of small/Delta1Hs approximates a circular arc of radius /bardblHG0/bardbl, subtended by the angle θ. Therefore, approximately, /Delta1Htotal=2/parenleftbigg /bardblHG0/bardblθ 2/parenrightbigg =/bardbl HG0/bardblθ (10.97) This expression becomes more precise as the number of intermediate maneuvers increases. Figure 10.12 reveals the extent to which the multiple coning maneuver 10.6 Coning maneuver 505 10 50 700.80.9 30 900.71.0 HG0θ 2HG0tanθ 2      θ, degrees Figure 10.12 Ratio of delta-H for a sequence of small coning maneuvers to that for a single coning maneuver, as a function of the angle of swing of the spin axis. 2 4 6 810203040 θ /H11005 150° θ /H11005 120° θ /H11005 90° θ /H11005 60° θ /H11005 30° ntn t1 Figure 10.13 Time for a coning maneuver versus the number of intermediate steps. strategy reduces energy requirements. The difference is quite significant for large reorientation angles. One of the prices to be paid for the reduced energy of the multiple coning maneu- ver is time. (The other is the risk involved in repeating the maneuver over and overagain.) From Equation 10.95, the time required for nsmall-angle coning maneuvers through a total angle of θis t n=nπA−C Cωscosθ 2n(10.98) The ratio of this to the time t1required for a single coning maneuver is tn t1=ncosθ 2n cosθ 2(10.99) The time is directly proportional to the number of intermediate coning maneuvers, as illustrated in Figure 10.13. 506 Chapter 10 Satellite attitude dynamics 10.7 Attitude control thrusters As mentioned above, thrusters are small jets mounted in pairs on a spacecraft to control its rotational motion about the center of mass. These thruster pairs may bemounted in principal planes (planes normal to the principal axes) passing throughthe center of mass. Figure 10.14 illustrates a pair of thrusters for producing a torqueabout the positive yaxis. These would be accompanied by another pair of reaction motors pointing in the opposite directions to exert torque in the negative xdirection. If the position vectors of the thrusters relative to the center of mass are rand−r, and if Tis their thrust, then the impulsive moment they exert during a brief time interval /Delta1tis M=r×T/Delta1t+(−r)×(−T/Delta1t)=2r×T/Delta1t (10.100) If the angular velocity was initially zero, then after the firing, according to Equation 10.31, the angular momentum becomes H=2r×T/Delta1t (10.101) For Hin the principal xdirection, as in the figure, the corresponding angular velocity acquired by the vehicle is, from Equation 10.67, ω y=/bardblH/bardbl B(10.102) r /H11002rx yz GT /H11002T M Figure 10.14 Pair of attitude control thrusters mounted in the xzplane of the principal body frame. 10.7 Attitude control thrusters 507 Example 10.7A spacecraft of mass mand with the dimensions shown in Figure 10.15 is spinning without precession at the rate ω0about the zaxis of the principal body frame. At the instant shown in part (a) of the figure, the spacecraft initiates a coning maneuver to swing its spin axis through 90◦, so that at the end of the maneuver the vehicle is oriented as illustrated in Figure 10.15(b). Calculate the total delta-H required, and compare it with that required for the same reorientation without coning. Motion is to be controlled exclusively by the pairs of attitude thrusters shown, all of which have identical thrust T. x,Xy,Y z,Z z y T T TT x HG2 HG1 /H9004H G1/H9004H G2w w w/3w/3 (a)45°RCS-6RCS-5 RCS-2RCS-4RCS-3 RCS-1Precession of spin axisRCS-1 or RCS-2RCS-3 RCS-2 or RCS-1RCS-4 ww RCS-5 (b) Figure 10.15 (a) Initial orientation of spinning spacecraft. (b) Final configuration, with spin axis rotated 90◦. According to Figure 9.9(c), the moments of inertia about the principal body axes are A=B=1 12m/bracketleftbigg w2+/parenleftBigw 3/parenrightBig2/bracketrightbigg =5 54mw2C=1 12m(w2+w2)=1 6mw2 The initial angular momentum HG1points in the spin direction, along the positive z axis of the body frame, HG1=Cωzˆk=1 6mw2ω0ˆk We can presume that in the initial orientation, the body frame happens to coin- cide instantaneously with inertial frame XYZ . The coning motion is initiated by briefly firing the pair of thrusters RCS-1 and RCS-2, aligned with the body zaxis and lying in the yzplane. The impulsive torque will cause a change /Delta1H G1in angular momentum directed normal to the plane of the thrusters, in the positive body xdirection. The resultant angular momentum vector must lie at 45◦to the xand zaxes, bisecting the angle between the initial and final angular momenta. Thus, /bardbl/Delta1H G1/bardbl=/bardbl HG1/bardbltan 45◦=1 6mw2ω0 508 Chapter 10 Satellite attitude dynamics (Example 10.7 continued)After the coning is underway, the body axes of course move away from the XYZ frame. Since the spacecraft is oblate ( C>A), the precession of the spin axis will be opposite to the spin direction, as indicated in Figure 10.15. When the spin axis, after 180◦of precession, lines up with the xaxis the thrusters must fire again for the same duration as before so as to produce the angular momentum change /Delta1HG2, equal in magnitude but perpendicular to /Delta1HG1, so that HG1+/Delta1HG1+/Delta1HG2=HG2 where HG2=/bardbl HG1/bardblˆI=1 6mw2ω0ˆk For this to work, the plane of thrusters RCS-1 and RCS-2 – the yzplane – must be parallel to the XYplane when they fire, as illustrated in Figure 10.15(b). Since the thrusters can fire fore or aft, it does not matter which of them ends up on top or bottom. The vehicle must therefore spin through an integral number nof half rotations while it precesses to the desired orientation. That is, the total spin angle ψ between the initial and final configurations is ψ=nπ=ωst (a) where ωsis the spin rate and tis the time for the proper final configuration to be achieved. In the meantime, the precession angle φmust be πor 3πor 5π,o r ,i n general, φ=(2m−1)π=ωpt (b) where mis an integer and tis, of course, the same as that in (a). Eliminating tfrom both (a) and (b) yields nπ=(2m−1)πωs ωp Substituting Equation 10.94, with θ=π/2, gives n=(1−2m)4 91√ 2(c) Obviously, this equation cannot be valid if both mand nare integers. However, by tabulating nas a function of mwe find that when m=18,n=− 10.999. The minus sign simply reminds us that spin and precession are in opposite directions. Thus, the eighteenth time that the spin axis lines up with the xaxis the thrusters may be fired to almost perfectly align the angular momentum vector with the body zaxis. The slight misalignment due to the fact that /bardbln/bardblis not precisely 11 would probably occur in reality anyway. Passive or active nutation damping can drive this deviation to zero. Since/bardblHG1/bardbl=/bardbl HG2/bardbl, we conclude that /Delta1Htotal=2/parenleftbigg1 6mw2ω0/parenrightbigg =2 3mw2ω0 (d) 10.8 Y o-yo despin mechanism 509 An obvious alternative to the coning maneuver is to use thrusters RCS-3 and 4 to despin the craft completely, thrusters RCS-5 and 6 to initiate roll around the yaxis and stop it after 90◦, and then RCS-3 and 4 to respin the spacecraft to ω0around the zaxis. The combined delta-H for the first and last steps equals that of (d). Additional fuel expenditure is required to start and stop the roll around the yaxis. Hence, the coning maneuver is more fuel efficient. 10.8 Yo-yo despin mechanism A simple, inexpensive way to despin an axisymmetric satellite is to deploy small masses attached to cords wound around the girth of the satellite near the transverse plane through the center of mass. As the masses unwrap in the direction of the satellite’sangular velocity, they exert centrifugal force through the cords on the periphery of thesatellite, creating a moment opposite to the spin direction, thereby slowing down therotational motion. The cord forces are internal to the system of satellite plus weights,so as the strings unwind, the total angular momentum must remain constant. Sincethe total moment of inertia increases as the yo-yo masses spiral further away, theangular velocity must drop. Not only angular momentum but also rotational kineticenergy is conserved during this process. Y o-yo despin devices were introduced earlyin unmanned space flight (e.g., 1959 Transit 1-A) and continue to be used today (e.g.,1996 Mars Pathfinder, 1998 Mars Climate Orbiter, 1999 Mars Polar Lander, 2003Mars Exploration Rover). We will use the conservation of energy and momentum to determine the length of cord required to reduce the satellite’s angular velocity a specified amount. T o maintainthe position of the center of mass, two identical yo-yo masses are wound around thespacecraft in a symmetrical fashion, as illustrated in Figure 10.16. Both masses are rT AGφ ˆiˆj xy m/2R Hv ωH'A'm/2Pcord cordRφφ Figure 10.16 Two identical string and mass systems wrapped symmetrically around the periphery of an axisymmetric satellite. For simplicity, only one is shown being deployed. 510 Chapter 10 Satellite attitude dynamics released simultaneously by explosive bolts and unwrap in the manner shown (for only one of the weights) in the figure. In so doing, the point of tangency Tmoves around the circumference towards the split hinge device where the cord is attachedto the spacecraft. When Tand T /primereach the hinges Hand H/prime, the cords automatically separate from the spacecraft. Let each yo-yo weight have mass m/2. By symmetry, we need to track only one of the masses, to which we can ascribe the total mass m. Let the xyzsystem be a body frame rigidly attached to the satellite, as shown in Figure 10.16. As usual, the zaxis lies in the spin direction, pointing out of the page. The xaxis is directed from the center of mass of the system through the initial position of the yo-yo mass. The satellite and theyo-yo masses, prior to release, are rotating as a single rigid body with angular velocity ω 0=ω0ˆk. The moment of inertia of the satellite, excluding the yo-yo mass, is C,s o that the angular momentum of the satellite by itself is Cω0. The concentrated yo-yo masses are fastened a distance Rfrom the spin axis, so their total moment of inertia ismR2. Therefore, the initial angular momentum of the satellite plus yo-yo system is HG0=Cω0+mR2ω0 It will be convenient to write this as HG0=KmR2ω0 (10.103) where the non-dimensional factor Kis defined as K=1+C mR2(10.104) √ KR is the initial radius of gyration of the system. The initial rotational kinetic energy of the system, before the masses are released, is T0=1 2Cω2 0+1 2mR2ω2 0=1 2KmR2ω2 0 (10.105) At any state between the release of the weights and the release of the cords at the hinges, the velocity of the yo-yo mass must be found in order to compute the new angular momentum and kinetic energy. Observe that when the string has unwrappedan angle φ, the free length of string (between the point of tangency Tand the yo-yo mass P)i sRφ. From the geometry shown in Figure 10.16, the position vector of the mass relative to the body frame is seen to be r=rT/G/bracehtipdownleft /bracehtipupright/bracehtipupleft /bracehtipdownright (Rcosφˆi+Rsinφˆj)+rP/T/bracehtipdownleft /bracehtipupright/bracehtipupleft /bracehtipdownright (Rφsinφˆi−Rφcosφˆj) (10.106) =(Rcosφ+Rφsinφ)ˆi+(Rsinφ−Rφcosφ)ˆj Since ris measured in the moving reference, the absolute velocity vof the yo-yo mass is found using Equation 1.28, v=dr dt/parenrightbigg rel+/Omega1×r (10.107) 10.8 Y o-yo despin mechanism 511 where /Omega1is the angular velocity of the xyzaxes, which, of course, is the angular velocity ωof the satellite at that instant, /Omega1=ω (10.108) T o calculate dr/dt)rel, we hold ˆiandˆjconstant in Equation 10.106, obtaining dr dt/parenrightbigg rel=(−R˙φsinφ+R˙φsinφ+R˙φcosφ)ˆi+(R˙φcosφ−R˙φcosφ+R˙φsinφ)ˆj =R˙φcosφˆi+R˙φsinφˆj Thus v=R˙φcosφˆi+R˙φsinφˆj+/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆi ˆj ˆk 00 ω Rcosφ+RφsinφRsinφ−Rφcosφ0/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle or v=[Rφ(ω+˙φ)c o sφ−Rωsinφ]ˆi+[Rωcosφ+Rφ(ω+˙φ) sinφ]ˆj(10.109) From this we find the speed of the yo-yo weights, v=√ v·v=R/radicalBig ω2+(ω+˙φ)2φ2 (10.110) The angular momentum of the satellite plus the weights at an intermediate stage of the despin process is HG=Cωˆk+r×mv =Cωˆk+m/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆi ˆj ˆk Rcosφ+Rφsinφ Rsinφ−Rφcosφω Rφ(ω+˙φ)c o sφ−RωsinφRωcosφ+Rφ(ω+˙φ) sinφ0/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle Carrying out the cross product, combining terms and simplifying, leads to H G=Cω+mR2[ω+(ω+˙φ)φ2] which, using Equation 10.104, can be written HG=mR2[Kω+(ω+˙φ)φ2] (10.111) The kinetic energy of the satellite plus the yo-yo mass is T=1 2Cω2+1 2mv2 Substituting the speed from Equation 10.110 and making use again of Equation 10.104, we find T=1 2mR2[Kω2+(ω+˙φ)2φ2] (10.112) By the conservation of angular momentum, HG=HG0, we obtain from Equa- tions 10.103 and 10.111, mR2[Kω+(ω+˙φ)φ2]=KmR2ω0 512 Chapter 10 Satellite attitude dynamics which we can write as K(ω0−ω)=(ω+˙φ)φ2(10.113) Equations 10.105 and 10.112 and the conservation of kinetic energy, T=T0, combine to yield 1 2mR2[Kω2+(ω+˙φ)2φ2]=1 2KmR2ω2 0 or K/parenleftbig ω2 0−ω2/parenrightbig =(ω+˙φ)2φ2(10.114) Sinceω2 0−ω2=(ω0−ω)(ω0+ω), this can be written K(ω0−ω)(ω0+ω)=(ω+˙φ)2φ2 Replacing the factor K(ω0−ω) on the left using Equation 10.113 yields (ω+˙φ)φ2(ω0+ω)=(ω+˙φ)2φ2 After canceling terms, we find ω0+ω=ω+˙φ, or, simply ˙φ=ω0 (10.115) In other words, the cord unwinds at a constant rate (relative to the satellite), equal to the satellite’s initial angular velocity. Thus at any time tafter the release of the weights, φ=ω0t (10.116) By substituting Equation 10.115 into Equation 10.113, K(ω0−ω)=(ω+ω0)φ2 we find that φ=/radicalbigg Kω0−ω ω0+ωPartial despin . (10.117) Recall that the unwrapped length lof the cord is Rφ, which means l=R/radicalbigg Kω0−ω ω0+ωPartial despin . (10.118) We use Equation 10.118 to find the length of cord required to despin the spacecraft fromω0toω. T o remove all of the spin ( ω=0), φ=√ K⇒ l=R√ K Complete despin . (10.119) Surprisingly, the length of cord required to reduce the angular velocity to zero is independent of the initial angular velocity. We can solve Equation 10.117 for ωin terms of φ, ω=/parenleftbigg2K K+φ2−1/parenrightbigg ω0 (10.120) 10.8 Y o-yo despin mechanism 513 By means of Equation 10.116, this becomes an expression for the angular velocity as a function of time ω=/parenleftbigg2K K+ω2 0t2−1/parenrightbigg ω0 (10.121) Alternatively, since φ=l/R, Equation 10.120 yields the angular velocity as a function of cord length, ω=/parenleftbigg2KR2 KR2+l2−1/parenrightbigg ω0 (10.122) Differentiating ωwith respect to time in Equation 10.121 gives us an expression for the angular acceleration of the spacecraft, α=dω dt=−4Kω3 0t (K+ω2 0t2)2(10.123) whereas integrating ωwith respect to time yields the angle rotated by the satellite since release of the yo-yo mass, θ=2√ Ktan−1ω0t√ K−ω0t=2√ Ktan−1φ√ K−φ (10.124) For complete despin, this expression, together with Equation 10.119, yields θ=√ K/parenleftBigπ 2−1/parenrightBig (10.125) From the free-body diagram of the spacecraft shown in Figure 10.17, it is clear that the torque exerted by the yo-yo weights is MGz=− 2RN (10.126) T Gφ ˆiˆj xy R H H' T'N N R Figure 10.17 Free-body diagram of the satellite during the despin process. 514 Chapter 10 Satellite attitude dynamics where Nis the tension in the cord. From Euler’ s equations of motion, Equation 10.72, MGz=Cα (10.127) Combining Equations 10.123, 10.126 and 10.127 leads to a formula for the tension in the yo-yo cables, N=C R2Kω3 0t /parenleftbig K+ω2 0t2/parenrightbig2=Cω2 0 R2Kφ (K+φ2)2(10.128) Radial release Finally, we note that instead of releasing the yo-yo masses when the cables are tangent at the split hinges ( HandH/prime), they can be forced to pivot about the hinge and released when the string is directed radially outward, as illustrated in Figure 10.18. The aboveanalysis must be then extended to include the pivoting of the cord around the hinges.It turns out that in this case, the length of the cord as a function of the final angularvelocity is l=R/parenleftBigg/radicalBigg [(ω0−ω)K+ω]2 (ω2 0−ω2)K+ω2−1/parenrightBigg Partial despin, radial release . (10.129) so that for ω=0, l=R(√ K−1) Complete despin, radial release . (10.130) Gxy H H'Tangential release position Radial release position Figure 10.18 Radial versus tangential release of yo-yo masses. 10.8 Y o-yo despin mechanism 515 Example 10.8A satellite is to be completely despun using a two-mass yo-yo device with tangential release. Assume the spin axis of moment of inertia of the satellite is C=200 kg ·m2 and the initial spin rate is ω0=5 rad/s. The total yo-yo mass is 4 kg, and the radius of the spacecraft is 1 meter. Find (a) the required cord length l; (b) the time tto despin; (c) the maximum tension in the yo-yo cables; (d) the speed of the masses at release; (e) the angle rotated by the satellite during the despin; (f) the cord length required for radial release. (a) From Equation 10.104, K=1+C mR2=1+200 4·12=51 (a) From Equation 10.118 it follows that the cord length required for complete despin is l=R√ K=1·√ 51=7.1414 m (b) (b) The time for complete despin is obtained from Equations 10.116 and 10.118, ω0t=√ K⇒ t=√ K ω0=√ 51 5=1.4283 s (c) A graph of Equation 10.128 is shown in Figure 10.19. The maximum tension is 455 N , which occurs at 0.825 s. 0.2 0.4 0.6 0.8 1.0 1.2 1.4100200300400500 t (s)N (N) Figure 10.19 Variation of cable tension Nup to point of release. (d) From Equation 10.110, the speed of the yo-yo masses is v=R/radicalBig ω2+(ω+˙φ)2φ2 516 Chapter 10 Satellite attitude dynamics (Example 10.8 continued)According to Equation 10.115, ˙φ=ω0and at the time of release ( ω=0) Equation 10.118 states that φ=√ K.T h u s v=R/radicalBig ω2+(ω+ω0)2√ K2=1·/radicalBig 02+(0+5)2√ 512=35.71 m/s (e) The angle through which the satellite rotates before coming to rotational rest is given by Equation 10.124, θ=√ K/parenleftBigπ 2−1/parenrightBig =√ 51/parenleftBigπ 2−1/parenrightBig =4.076 rad (233 .5◦) (f) Allowing the cord to detach radially reduces the cord length required for complete despin from 7.141 m to l=R/parenleftBig√ K−1/parenrightBig =1·/parenleftBig√ 51−1/parenrightBig =6.141 m 10.9 Gyroscopic attitude control Momentum exchange systems (‘gyros’) are used to control the attitude of a spacecraft without throwing consumable mass overboard, as occurs with the use of thrusterjets. A momentum exchange system is illustrated schematically in Figure 10.20. n flywheels, labeled 1, 2, 3, etc., are attached to the body of the spacecraft at variouslocations. The mass of flywheel iism i. The mass of the body of the spacecraft is m0. xyz GG11 G2 G3 Gi/H9275 /H9275(2) /H9275(i)/H9275(1) /H9275(3)2 3 i Body-fixed frame Figure 10.20 Several attitude control flywheels, each with their own angular velocity, attached to the body of a spacecraft. 10.9 Gyroscopic attitude control 517 The total mass of the entire system – the ‘vehicle’ – is m, m=m0+n/summationdisplay i=1mi The vehicle’s center of mass is G, through which pass the three axes xyzof the vehicle’s body-fixed frame. The center of mass Giof each flywheel is connected rigidly to the spacecraft, but the wheel, driven by electric motors, rotates more or less independently,depending on the type of gyro. The body of the spacecraft has an angular velocity ω. The angular velocity of the ith flywheel is ω i, and differs from that of the body of the spacecraft unless the gyro is ‘caged’ . A caged gyro has no spin relative to the spacecraft,in which case ω i=ω. The angular momentum of the entire system about the vehicle’s center of mass G is the sum of the angular momenta of the individual components of the system, HG=H(v) G+H(w)(10.131) H(v) Gis the total angular momentum of the rigid body comprising the spacecraft and all of the flywheel masses concentrated at their centers of mass Gi. That system has the common vehicle angular velocity ω, which means, according to Equation 9.39, that /braceleftbig H(v) G/bracerightbig =/bracketleftbig I(v) G/bracketrightbig {ω} (10.132) where [ I(v) G] is the moment of inertia found by adding the moments of inertia of all the concentrated flywheel masses about Gto that of the body of the spacecraft. On the other hand, H(w)is the net angular momentum of the nflywheels about each of their individual centers of mass, H(w)=n/summationdisplay i=1H(i) Gi(10.133) H(i) Gi, the angular momentum of flywheel iabout its center of mass Gi, is obtained by once again using Equation 9.39, /braceleftbig H(i) Gi/bracerightbig =/bracketleftbig I(i) Gi/bracketrightbig/braceleftbig ω(i)/bracerightbig (10.134) [I(i) Gi] is the moment of inertia of flywheel iabout Gi, relative to axes which are parallel to the body-fixed xyzaxes. The mass distribution reflected in [ I(v) G]i sfi x e dr e l a t i v e to the body frame, which means this matrix does not vary with time. On the otherhand, since a momentum wheel might be one that pivots on gimbals relative to the body frame, the inertia tensor [ I(i) Gi] may be time dependent. Substituting Equation 10.131 into Equatio n 9.30 yields the equations of rotational motion of the gyro stabilized spacecraft, MGnet=˙H(v) G+˙H(w)(10.135) Since the angular momenta are computed in the non-inertial body-fixed frame, we must use Equation 1.28 to obtain the time derivatives on the right-hand side of 518 Chapter 10 Satellite attitude dynamics Equation 10.135. Therefore, MGnet=/bracketleftBigg dH(v) G dt/parenrightBigg rel+ω×H(v) G/bracketrightBigg +/bracketleftBigg dH(w) dt/parenrightBigg rel+ω×H(w)/bracketrightBigg (10.136) For torque-free motion, MGnet=0, in which case we have the conservation of angular momentum about the vehicle center of mass, H(v) G+H(w)=constant (10.137) Example 10.9Use Equation 10.136 to obtain the equations of motion of a torque-free, axisymmetric, dual-spin satellite, such as the one shown in Figure 10.21. xy GrG Rotorz Platform ω(r)ω(r) /H11001 ωpGpz z Figure 10.21 Dual-spin spacecraft. In the dual-spin satellite, we may arbitrarily choose the rotor as the body of the vehicle, to which the body frame is attached. The coaxial platform will play the role of the single reaction wheel. The center of mass Gof the satellite lies on the axis of rotational symmetry (the zaxis), between the center of mass of the rotor ( Gr) and that of the platform ( Gp). For this torque-free system, Equation 10.136 becomes dH(v) G dt/parenrightBigg rel+ω(r)×H(v) G+dH(p) Gp dt  rel+ω(r)×H(p) Gp=0 (a) in which rsignifies the rotor and pthe platform. The vehicle angular momentum about Gis that of the rotor plus that of the platform center of mass, /braceleftbig H(v) G/bracerightbig =/bracketleftbig I(r) G/bracketrightbig/braceleftbig ω(r)/bracerightbig +/bracketleftbig I(p) mG/bracketrightbig/braceleftbig ω(r)/bracerightbig =/parenleftbig/bracketleftbig I(r) G/bracketrightbig +/bracketleftbig I(p) mG/bracketrightbig/parenrightbig/braceleftbig ω(r)/bracerightbig (b) 10.9 Gyroscopic attitude control 519 /bracketleftbig I(p) mG/bracketrightbig is the moment of inertia tensor of the concentrated mass of the platform about the system center of mass, and it is calculated by means of Equation 9.44. The components of/bracketleftbig I(r) G/bracketrightbig and/bracketleftbig I(p) mG/bracketrightbig are constants, so from (b) we obtain d/braceleftbig H(v) G/bracerightbig dt/parenrightBigg rel=/parenleftbig/bracketleftbig I(r) G/bracketrightbig +/bracketleftbig I(p) mG/bracketrightbig/parenrightbig/braceleftbig ˙ω(r)/bracerightbig (c) The angular momentum of the platform about its own center of mass is /braceleftbig H(p) Gp/bracerightbig =/bracketleftbig I(p) Gp/bracketrightbig/braceleftbig ω(p)/bracerightbig (d) For both the platform and the rotor, the zaxis is an axis of rotational symmetry. Thus, even though the platform is not stationary in xyz, the moment of inertia matrix/bracketleftbig I(p) Gp/bracketrightbig is not time dependent. It follows that d/braceleftbig H(p) Gp/bracerightbig dt  rel=/bracketleftbig I(p) Gp/bracketrightbig/braceleftbig ˙ω(p)/bracerightbig (e) Using (b) through (e), we can write the equation of motion (a) as /parenleftbig/bracketleftbig I(r) G/bracketrightbig +/bracketleftbig I(p) mG/bracketrightbig/parenrightbig/braceleftbig ˙ω(r)/bracerightbig +/braceleftbig ω(r)/bracerightbig ×/parenleftbig/bracketleftbig I(r) G/bracketrightbig +/bracketleftbig I(p) mG/bracketrightbig/parenrightbig/braceleftbig ω(r)/bracerightbig +/bracketleftbig I(p) Gp/bracketrightbig/braceleftbig ˙ω(p)/bracerightbig +/braceleftbig ω(r)/bracerightbig ×/bracketleftbig I(p) Gp/bracketrightbig/braceleftbig ω(p)/bracerightbig =/braceleftbig 0/bracerightbig (f) The angular velocity ω(p)of the platform is that of the rotor, ω(r), plus the angular velocity of the platform relative to the rotor, ω(p) rel. Hence, we may replace/braceleftbig ω(p)/bracerightbig with {ω(r)}+/braceleftbig ω(p) rel/bracerightbig , so that, after a little rearrangement, (f) becomes /parenleftbig/bracketleftbig I(r) G/bracketrightbig +/bracketleftbig I(p) G/bracketrightbig/parenrightbig/braceleftbig ˙ω(r)/bracerightbig +/braceleftbig ω(r)/bracerightbig ×/parenleftbig/bracketleftbig I(r) G/bracketrightbig +/bracketleftbig I(p) G/bracketrightbig/parenrightbig/braceleftbig ω(r)/bracerightbig +/bracketleftbig I(p) Gp/bracketrightbig/braceleftbig ˙ω(p) rel/bracerightbig +/braceleftbig ω(r)/bracerightbig ×/bracketleftbig I(p) Gp/bracketrightbig/braceleftbig ω(p) rel/bracerightbig =/braceleftbig 0/bracerightbig (g) in which /bracketleftbig I(p) G/bracketrightbig =/bracketleftbig I(p) mG/bracketrightbig +/bracketleftbig I(p) Gp/bracketrightbig (Parallel axis formula.) (h) The components of the matrices and vectors in (g) relative to the principal xyzbody frame axes are /bracketleftbig I(r) G/bracketrightbig = Ar00 0 Ar0 00 Cr /bracketleftbig I(p) G/bracketrightbig = Ap00 0 Ap0 00 Cp /bracketleftbig I(p) Gp/bracketrightbig = Ap00 0 Ap 0 00 Cp  (i) and /braceleftbig ω(r)/bracerightbig =  ω(r) x ω(r) y ω(r) z  /braceleftbig ˙ω(r)/bracerightbig =  ˙ω(r) x ˙ω(r) y ˙ω(r) z  /braceleftbig ˙ω(p) rel/bracerightbig =  0 0 ωp  /braceleftbig ˙ω(p) rel/bracerightbig =  0 0 ˙ωp  (j) 520 Chapter 10 Satellite attitude dynamics (Example 10.9 continued)Ar,Cr,Apand Cpare the rotor and platform principal moments of inertia about the vehicle center of mass G, whereas Apis the moment of inertia of the platform about its own center of mass. We also used the fact that Cp=Cp, which of course is due to the fact that Gand Gpboth lie on the zaxis. This notation is nearly identical to that employed in our consideration of the stability of dual-spin satellites in Section 10.4 (wherein ωr=ω(r) zandω⊥=ω(r) xˆi+ω(r) yˆj). Substituting (i) and (j) into each of the four terms in (g), we get /parenleftbig/bracketleftbig I(r) G/bracketrightbig +/bracketleftbig I(p) G/bracketrightbig/parenrightbig/braceleftbig ˙ω(r)/bracerightbig = Ar+Ap 00 0 Ar+Ap 0 00 Cr+Cp   ˙ω(r) x ˙ω(r) y ˙ω(r) z   =  (A r+Ap)˙ω(r) x (Ar+Ap)˙ω(r) y (Cr+Cp)˙ω(r) z  (k) /braceleftbig ω(r)/bracerightbig ×/parenleftbig/bracketleftbig I(r) G/bracketrightbig +/bracketleftbig I(p) G/bracketrightbig/parenrightbig/braceleftbig ω(r)/bracerightbig =  ω(r) x ω(r) y ω(r) z  ×  (Ar+Ap)ω(r) x (Ar+Ap)ω(r) y (Cr+Cp)ω(r) z   =  [(Cp−Ap)+(Cr−Ar)]ω(r) yω(r) z [(Ap−Cp)+(Ar−Cr)]ω(r) xω(r) z 0  (l) /bracketleftbig I(p) Gp/bracketrightbig/braceleftbig ˙ω(p) rel/bracerightbig = Ap00 0 Ap0 00 Cp   0 0 ˙ωp  =  0 0 Cp˙ωp  (m) /braceleftbig ω(r)/bracerightbig ×/bracketleftbig I(p) Gp/bracketrightbig/braceleftbig ω(p) rel/bracerightbig =  ω(r) x ω(r) y ω(r) z  × Ap00 0 Ap0 00 Cp   0 0 ωp  =  Cpω(r) yωp −Cpω(r) xωp 0  (n) With these four expressions, (g) becomes   (Ar+Ap)˙ω(r) x (Ar+Ap)˙ω(r) y (Cr+Cp)˙ω(r) z  +  [(Cp−Ap)+(Cr−Ar)]ω(r) yω(r) z [(Ap−Cp)+(Ar−Cr)]ω(r) xω(r) z 0   +  0 0 Cp˙ωp  +  Cpω(r) yωp −Cpω(r) xωp 0  =  0 00  (o) Combining the four vectors on the left-hand side, and then extracting the three components of the vector equation finally yields the three equations of motion of the 10.9 Gyroscopic attitude control 521 dual-spin satellite in the body frame, A˙ω(r) x+(C−A)ω(r) yω(r) z+Cpω(r) yωp=0 A˙ω(r) y+(A−C)ω(r) xω(r) z−Cpω(r) xωp=0 (p) C˙ω(r) z+Cp˙ωp=0 where Aand Care the combined transverse and axial moments of inertia of the dual-spin vehicle about its center of mass, A=Ar+ApC=Cr+Cp (q) The three equations (p) involve four unknowns, ω(r) x,ω(r) y,ω(r) zandωp. A fourth equation is required to account for the means of providing the relative velocity ωp between the platform and the rotor. Friction in the axle bearing between the platform and the rotor would eventually cause ωpto go to zero, as pointed out in Section 10.4. We may assume that the electric motor in the bearing acts to keep ωpconstant at a specified value, so that ˙ωp=0. Then Equation (p) 3implies that ω(r) z=constant as well. Thus, ωpandω(r) zare removed from our list of unknowns, leaving ω(r) xandω(r) y to be governed by the first two equations in (p). Example 10.10A spacecraft in torque-free motion has three identical momentum wheels with their spin axes aligned with the vehicle’s principal body axes. The spin axes of momentum wheels 1, 2 and 3 are aligned with the x,yand zaxes, respectively. The inertia tensors of the rotationally symmetric momentum wheels about their centers of mass are, therefore, /bracketleftbig I(1) G1/bracketrightbig = I00 0J0 00 J /bracketleftbig I(2) G2/bracketrightbig = J00 0I0 00 J /bracketleftbig I(3) G3/bracketrightbig = J00 0J0 00 I  (a) The spacecraft moment of inertia tensor about the vehicle center of mass is /bracketleftbig I(v) G/bracketrightbig = A00 0B0 00 C  (b) Calculate the spin accelerations of the momentum wheels in the presence of external torque. The absolute angular velocity ωof the spacecraft and the angular velocities ω(1) rel,ω(2) rel,ω(3) relof the three flywheels relative to the spacecraft are {ω}=  ωx ωy ωz  {ω(1)}rel=  ω(1) 0 0  {ω(2)}rel=  0 ω(2) 0  {ω(3)}rel=  0 0 ω(3)   (c) 522 Chapter 10 Satellite attitude dynamics (Example 10.10 continued)Therefore, the angular momentum of the spacecraft and momentum wheels is /braceleftbig HG/bracerightbig =/bracketleftbig I(v) G/bracketrightbig/braceleftbig ω/bracerightbig +/bracketleftbig I(1) G1/bracketrightbig/parenleftbig/braceleftbig ω/bracerightbig +/braceleftbig ω(1)/bracerightbig rel/parenrightbig +/bracketleftbig I(2) G2/bracketrightbig/parenleftbig/braceleftbig ω/bracerightbig +/braceleftbig ω(2)/bracerightbig rel/parenrightbig +/bracketleftbig I(3) G3/bracketrightbig/parenleftbig/braceleftbig ω/bracerightbig +/braceleftbig ω(3)/bracerightbig rel/parenrightbig (d) Substituting Equations (a), (b) and (c) into this expression yields {HG}= I00 0I0 00 I   ω(1) ω(2) ω(3)  + A+I+2J 00 0 B+I+2J 0 00 C+I+2J   ωx ωy ωz   (e) In this case, Euler’s equations are {˙HG}rel+{ω}×{ HG}={ MG} (f) Substituting (e), we get  I00 0I0 00 I   ˙ω(1) ˙ω(2) ˙ω(3)  + A+I+2J 00 0 B+I+2J 0 00 C+I+2J   ˙ωx ˙ωy ˙ωz  +  ωx ωy ωz   ×  I00 0I0 00 I   ω(1) ω(2) ω(3)  + A+I+2J 00 0 B+I+2J 0 00 C+I+2J   ωx ωy ωz    =  MGx MGy MGz  (g) Expanding and collecting terms yields the time rates of change of the flywheel spins (relative to the spacecraft) in terms of those of the spacecraft’s absolute angular velocity components, ˙ω(1)=MGx I+B−C Iωyωz−/parenleftbigg 1+A I+2J I/parenrightbigg ˙ωx+ω(2)ωz−ω(3)ωy ˙ω(2)=MGy I+C−A Iωxωz−/parenleftbigg 1+B I+2J I/parenrightbigg ˙ωy+ω(3)ωx−ω(1)ωz (h) ˙ω(3)=MGz I+A−B Iωxωy−/parenleftbigg 1+C I+2J I/parenrightbigg ˙ωz+ω(1)ωy−ω(2)ωx Example 10.11The communications satellite is in a circular earth orbit of period T. The body zaxis always points towards the earth, so the angular velocity about the body yaxis is 2 π/T. The angular velocities about the body xand zaxes are zero. The attitude control system consists of three momentum wheels 1, 2 and 3 aligned with the principal x,y and zaxes of the satellite. Variable torque is applied to each wheel by its own electric motor. At time t=0 the angular velocities of the three wheels relative to the spacecraft are all zero. A small, constant environmental torque M0acts on the spacecraft. 10.9 Gyroscopic attitude control 523 Determine the axial torques C(1),C(2)and C(3)that the three motors must exert on their wheels so that the angular velocity ωof the satellite will remain constant. The moment of inertia of each reaction wheel about its spin axis is I. xyz xz y1 ω(1)ω(3) ω(2) 2 3 GG Figure 10.22 Three-axis stabilized satellite. The absolute angular velocity of the xyzframe is given by ω=ω0ˆj (a) where ω0=2π/T, a constant. At any instant, the absolute angular velocities of the three reaction wheels are, accordingly, ω(1)=ω(1)ˆi+ω0ˆj ω(2)=[ω(2)+ω0]ˆj (b) ω(3)=ω0ˆj+ω(3)ˆk From (a) it is clear that ωx=ωz=˙ωx=˙ωy=˙ωz=0. Therefore, Equations (h) of Example 10.10 become, for the case at hand, ˙ω(1)=MGx I+B−C Iω0(0)−/parenleftbigg 1+A I+2J I/parenrightbigg (0)+ω(2)(0)−ω(3)ω0 ˙ω(2)=MGy I+C−A I(0)(0) −/parenleftbigg 1+B I+2J I/parenrightbigg (0)+ω(3)(0)−ω(1)(0) ˙ω(3)=MGz I+A−B I(0)ω 0−/parenleftbigg 1+C I+2J I/parenrightbigg (0)+ω(1)ω0−ω(2)(0) which reduce to the following set of three first order differential equations, ˙ω(1)+ω0ω(3)=MGx I ˙ω(2)=MGy I(c) ˙ω(3)−ω0ω(1)=MGz I 524 Chapter 10 Satellite attitude dynamics (Example 10.11 continued)Equation (c) 2implies that ω(2)=MGyt/I+constant, and since ω(2)=0a tt=0, this means that for time thereafter, ω(2)=MGy It (d) Differentiating (c) 3with respect to tand solving for ˙ω(1)yields ˙ω(1)=¨ω(3)/ω0. Substituting this result into (c) 1we get ¨ω(3)+ω2 0ω(3)=ω0MGx I The well-known solution of this differential equation is ω(3)=acosω0t+bsinω0t+MGx Iω0 where aand bare constants of integration. According to the problem statement, ω(3)=0 when t=0. This initial condition requires a=− MGx/ω0I, so that ω(3)=bsinω0t+MGx Iω0(1−cosω0t)( e ) From this we obtain ˙ω(3)=bω0cosω0t+MGx Isinω0t, which, when substituted into (c)3, yields ω(1)=bcosω0t+MGx Iω0sinω0t−MGz Iω0(f) Sinceω(1)=0a tt=0, this implies b=MGz/ω0I. In summary, therefore, the angular velocities of wheels 1, 2 and 3 relative to the satellite are ω(1)=MGx Iω0sinω0t+MGz Iω0(cosω0t−1) (g1) ω(2)=MGy It (g2) ω(3)=MGz Iω0sinω0t+MGx Iω0(1−cosω0t) (g3) The angular momenta of the reaction wheels are H(1) G1=I(1) xω(1) xˆi+I(1) yω(1) yˆj+I(1) zω(1) zˆk H(2) G2=I(2) xω(2) xˆi+I(2) yω(2) yˆj+I(2) zω(2) zˆk (h) H(3) G3=I(3) xω(3) xˆi+I(3) yω(3) yˆj+I(3) zω(3) zˆk According to (b), the components of the flywheels’ angular velocities are ω(1) x=ω(1)ω(1) y=ω0 ω(1) z=0 ω(2) x=0 ω(2) y=ω(2)+ω0 ω(2) z=0 ω(3) x=0 ω(3) y=ω0 ω(3) z=ω(3) 10.9 Gyroscopic attitude control 525 Furthermore, I(1) x=I(2) y=I(3) z=I, so that (h) becomes H(1) G1=Iω(1)ˆi+I(1) yω0ˆj H(2) G2=I(ω(2)+ω0)ˆj (i) H(3) G3=I(3) yω0ˆj+Iω(3)ˆk Substituting (g) into these expressions yields the angular momenta of the wheels as a function of time, H(1) G1=/bracketleftbiggMGx ω0sinω0t+MGz ω0(cosω0t−1)/bracketrightbigg ˆi+I(1) yω0ˆj H(2) G2=(MGyt+Iω0)ˆj (j) H(3) G3=I(3) yω0ˆj+/bracketleftbiggMGz ω0sinω0t+MGx ω0(1−cosω0t)/bracketrightbigg ˆk The torque on the reaction wheels is found by applying Euler’s equation to each one. Thus, for wheel 1 MG1net=dH(1) G1 dt/parenrightBigg rel+ω×H(1) G1 =(MGxcosω0t−MGzsinω0t)ˆi+[MGz(1−cosω0t)−MGxsinω0t]ˆk Since the axis of wheel 1 is in the xdirection, the torque is the xcomponent of this moment (the zcomponent being a gyroscopic bending moment), C(1)=MGxcosω0t−MGzsinω0t For wheel 2, MG2net=dH(2) G2 dt/parenrightBigg rel+ω×H(2) G2=MGyˆj Thus C(2)=MGy Finally, for wheel 3 MG3net=dH(3) G3 dt/parenrightBigg rel+ω×H(3) G3 =[MGx(1−cosω0t)+MGzsinω0t]ˆi+(MGxsinω0t+MGzcosω0t)ˆk For this wheel, the torque direction is the zaxis, so C(3)=MGxsinω0t+MGzcosω0t 526 Chapter 10 Satellite attitude dynamics The external torques on the spacecraft of the previous example may be due to thruster misalignment or they may arise from environmental effects such as gravity gradientsor solar pressure. The example assumed that these torques were constant, which isthe simplest means of introducing their effects, but they actually vary with time.In any case, their magnitudes are extremely small, typically less than 10 −3N·m for ordinary-sized, unmanned spacecraft. Equation (g) 2of the example reveals that a small torque normal to the satellite’s orbital plane will cause the angular velocity ofmomentum wheel 2 to slowly but constantly increase. Over a long enough period oftime, the angular velocity of the gyro might approach its design limits, whereuponit is said to be saturated . At that point, attitude jets on the satellite would have to be fired to produce a torque around the yaxis while the wheel is ‘caged’ , i.e., its angular velocity is reduced to zero or to its non-zero bias value. Finally, note that if all of the external torques were zero, none of the momentum wheels in the example would be required. The constant angular velocity ω=(2π/T)ˆjof the vehicle, once initiated, would continue unabated. So far we have dealt with momentum wheels, which are characterized by the fact that their axes are rigidly aligned with the principal axes of the spacecraft, as shownin Figure 10.23. The speed of the electrically driven wheels is varied to produce therequired rotation rates of the vehicle in response to external torques. Depending onthe spacecraft, the nominal speed of a momentum wheel may be from zero to severalthousand rpm. Momentum wheels that are free to pivot on one or more gimbals are called control moment gyros. Figure 10.24 illustrates a double-gimbaled control moment gyro.These gyros spin at several thousand rpm. The motor-driven speed of the flywheel isconstant, and moments are exerted on the vehicle when torquers (electric motors) tiltthe wheel about a gimbal axis. The torque direction is normal to the gimbal axis. T osimplify the analysis of high-rpm gyros, we can assume that the angular momentumis directed totally along the spin axis. That is, in calculating the angular momentum x yz G /H9275s Figure 10.23 Momentum wheel aligned with a principal body axis. 10.9 Gyroscopic attitude control 527 H(w) Gwof a momentum wheel about its center of mass, we use the formula /braceleftbig H(w) Gw/bracerightbig =/bracketleftbig I(w) Gw/bracketrightbig/braceleftbig ω(w)/bracerightbig where ω(w)is the absolute angular velocity of the spinning flywheel, which may be written ω(w)=ω(v)+ω(w) p+ω(w) n+ω(w) s ω(v)is the angular velocity of the vehicle to which the gyro is attached, while ω(w) p,ω(w) n andω(w) sare the precession, nutation and spin rates of the gyro relative to the vehicle. The spin rate of the gyro is three or more orders of magnitude greater than any of theother rates. That is, under conditions in which a control moment gyro is designed to operate, /vextenddouble/vextenddoubleω (w) s/vextenddouble/vextenddouble/greatermuch/vextenddouble/vextenddoubleω(v)/vextenddouble/vextenddouble/vextenddouble/vextenddoubleω(w) s/vextenddouble/vextenddouble/greatermuch/vextenddouble/vextenddoubleω(w) p/vextenddouble/vextenddouble/vextenddouble/vextenddoubleω(w) s/vextenddouble/vextenddouble/greatermuch/vextenddouble/vextenddoubleω(w) n/vextenddouble/vextenddouble We may therefore accurately express the angular momentum of any high-rpm gyro as /braceleftbig H(w) Gw/bracerightbig =/bracketleftbig I(w) Gw/bracketrightbig/braceleftbig ω(w) s/bracerightbig (10.138) Since the spin axis of a gyro is an axis of symmetry, about which the moment of inertia is C(w), this can be written H(w) Gw=C(w)ω(w) sˆn(w) s Pivot bearing TorquerTorquerMounting fixture Spin axisInner gimbal axisOuter gimbal axis Drive motor Mounting fixturePivot bearing Flywheel Figure 10.24 Two-gimbal control moment gyro. 528 Chapter 10 Satellite attitude dynamics z xωs y φθSpin axisˆns Axes of vehicle body frame Figure 10.25 Inclination angles of the spin vector of a gyro. where ˆn(w) sis the unit vector along the spin axis, as illustrated in Figure 10.25. Relative to the body frame axes of the spacecraft, the components of ˆn(w) sappear as follows, ˆn(w) s=sinθcosφˆi+sinθsinφˆj+cosθˆk (10.139) If we let H(w)=C(w)ω(w) s then Equation 10.138 becomes, simply, H(w) Gw=H(w)ˆn(w) s (10.140) Let us consider the equation of motion of a spacecraft with a single gyro. From Equation 10.136, dH(v) G dt/parenrightBigg rel+ω×H(v) G+dH(w) Gw dt/parenrightBigg rel+ω×H(w) Gw=MGnet (10.141) Calculating each term on the left, we have, for the vehicle, H(v) G=Aωxˆi+Bωyˆj+Cωzˆk dH(v) G dt/parenrightBigg rel=A˙ωxˆi+B˙ωyˆj+C˙ωzˆk (10.142) ω×H(v) G=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆi ˆj ˆk ω xωyωz AωxBωyCωz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=(C−B)ω yωzˆi+(A−C)ωxωzˆj+(B−A)ˆk (10.143) 10.9 Gyroscopic attitude control 529 For the gyro, H(w) Gw=H(w)ˆn(w) s=H(w)sinθcosφˆi+H(w)sinθsinφˆj+H(w)cosθˆk dH(w) Gw dt/parenrightBigg rel=(˙H(w)sinθcosφ+H(w)˙θcosθcosφ−H(w)˙φsinθsinφ)ˆi +(˙H(w)sinθsinφ+H(w)˙θcosθsinφ+H(w)˙φsinθcosφ)ˆj +(˙H(w)cosθ−H(w)˙θsinθ)ˆk (10.144) ω×H(w) Gw=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆi ˆj ˆk ω x ωy ωz H(w)sinθcosφH(w)sinθsinφH(w)cosθ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =/parenleftbig H (w)ωycosθ−H(w)ωzsinφsinθ/parenrightbigˆi +/parenleftbig −H(w)ωxcosθ+H(w)ωzcosφsinθ/parenrightbigˆj +/parenleftbig −H(w)ωycosφsinθ+H(w)ωxsinφsinθ/parenrightbigˆk (10.145) Substituting Equations 10.142 through 10.145 into Equation 10.141 yields a vector equation with the following three components A˙ωx+H(w)˙θcosφcosθ−H(w)˙φsinφsinθ+˙H(w)cosφsinθ +/parenleftbig H(w)cosθ+Cωz/parenrightbig ωy−/parenleftbig H(w)sinφsinθ+Bωy/parenrightbig ωz=MGnetx(10.146a) B˙ωy+H(w)˙θsinφcosθ+H(w)˙φcosφsinθ+˙H(w)sinφsinθ −/parenleftbig H(w)cosθ+Cωz/parenrightbig ωx+/parenleftbig H(w)cosφsinθ+Aωx/parenrightbig ωz=MGnety(10.146b) C˙ωz−H(w)˙θsinθ+˙H(w)cosθ−/parenleftbig H(w)cosφsinθ+Aωx/parenrightbig ωy +/parenleftbig H(w)sinφsinθ+Bωy/parenrightbig ωx=MGnetz(10.146c) Additional gyros are accounted for by adding the components of Equations 10.144 and 10.145 for each additional unit. Example 10.12A satellite is in torque-free motion ( MGnet=0). A non-gimbaled gyro (momentum wheel) is aligned with the vehicle’s xaxis and is spinning at the rate ωs0. The spacecraft angular velocity is ω=ωxˆi. If the spin of the gyro is increased at the rate ˙ωs, find the angular acceleration of the spacecraft. Using Figure 10.25 as a guide, we set φ=0 andθ=90◦to align the spin axis with the xaxis. Since there is no gimbaling, ˙θ=˙φ=0. Equations 10.146 then yield A˙ωx+˙H(w)=0 B˙ωy=0 C˙ωz=0 530 Chapter 10 Satellite attitude dynamics (Example 10.12 continued)Clearly, the angular velocities around the yand zaxes remain zero, whereas, ˙ωx=−˙H(w) A=−C(w) A˙ωs Thus, a change in the vehicle’s roll rate around the xaxis can be initiated by accelerating the momentum wheel in the opposite direction. Example 10.13A satellite is in torque-free motion. A control moment gyro, spinning at the constant rateωs, is gimbaled about the spacecraft yand zaxes, with φ=0 and θ=90◦(cf. Figure 10.25). The spacecraft angular velocity is ω=ωzˆk. If the spin axis of the gyro, initially along the xdirection, is rotated around the yaxis at the rate ˙θ, what is the resulting angular acceleration of the spacecraft? Substituting ωx=ωy=˙H(w)=φ=0 andθ=90◦into Equations 10.146 gives A˙ωx=0 B˙ωy+H(w)(ωz+˙φ)=0 C˙ωz−H(w)˙θ=0 where H(w)=C(w)ωs. Thus, the components of vehicle angular acceleration are ˙ωx=0 ˙ωy=−C(w) Bωs(ωz+˙φ) ˙ωz=C(w) Cωs˙θ We see that pitching the gyro at the rate ˙θaround the vehicle yaxis alters only ωz, leaving ωxunchanged. However, to keep ωy=0 clearly requires ˙φ=−ωz. In other words, for the control moment gyro to control the angular velocity about only one vehicle axis, it must therefore be able to precess around that axis (the zaxis in this case). That is why the control moment gyro must have two gimbals. 10.10 Gravity-gradient stabilization Consider a satellite in circular orbit, as shown in Figure 10.26. Let rbe the position vector of a mass element dmrelative to the center of attraction, r0the position vector of the center of mass G, and/rho1the position of dmrelative to G. The force of gravity ondmis dFg=− GMdm r3r=−µr r3dm (10.147) where Mis the mass of the central body, and µ=GM. The net moment of the gravitational force around Gis MGnet=/integraldisplay m/rho1×dFgdm (10.148) 10.10 Gravity-gradient stabilization 531 x r0 zr/rho1 yGdm Circular orbit Figure 10.26 Rigid satellite in a circular orbit is the principal body frame. Since r=r0+/rho1, and r0=r0xˆi+r0yˆj+r0zˆk (10.149)/rho1=xˆi+yˆj+zˆk we have /rho1×dFg=−µdm r3/rho1×(r0+/rho1)=−µdm r3/rho1×r0=−µdm r3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleˆiˆjˆk xyzr 0xr0yr0z/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle Thus, /rho1×dF g=−µdm r3(r0zy−r0yz)ˆi−µdm r3(r0xz−r0zx)ˆj−µdm r3(r0yx−r0xy)ˆk Substituting this back into Equation 10.148 yields MGnet=/parenleftbigg −µr 0z/integraldisplay my r3dm+µr0y/integraldisplay mz r3dm/parenrightbigg ˆi+/parenleftbigg −µr 0x/integraldisplay mz r3dm+µr0z/integraldisplay mx r3dm/parenrightbigg ˆj +/parenleftbigg −µr 0y/integraldisplay mx r3dm+µr0x/integraldisplay my r3dm/parenrightbigg ˆk or MGnetx=−µr0z/integraldisplay my r3dm+µr0y/integraldisplay mz r3dm MGnety=−µr0x/integraldisplay mz r3dm+µr0z/integraldisplay mx r3dm (10.150) MGnetz=−µr0y/integraldisplay mx r3dm+µr0x/integraldisplay my r3dm 532 Chapter 10 Satellite attitude dynamics Now, since /bardbl/rho1/bardbl/lessmuch/bardbl r0/bardbl, it follows from Equation 7.12 that 1 r3=1 r3 0−3 r5 0r0·/rho1 or 1 r3=1 r3 0−3 r5 0(r0xx+r0yy+r0zz) Therefore, /integraldisplay mx r3dm=1 r3 0/integraldisplay mxd m−3r0x r5 0/integraldisplay mx2dm−3r0x r5 0/integraldisplay mxy dm −3r0x r5 0/integraldisplay mxz dm But the center of mass lies at the origin of the xyzaxes, which are principal moment of inertia directions. That means/integraldisplay mxd m=/integraldisplay mxy dm =/integraldisplay mxz dm =0 so that/integraldisplay mx r3dm=−3r0x r5 0/integraldisplay mx2dm (10.151) In a similar fashion, we can show that /integraldisplay my r3dm=−3r0y r5 0/integraldisplay my2dm (10.152) and/integraldisplay mz r3dm=−3r0y r5 0/integraldisplay mz2dm (10.153) Substituting these last three expressions into Equations 10.150 leads to MGnetx=3µr0yr0z r5 0/parenleftbigg/integraldisplay my2dm−/integraldisplay mz2dm/parenrightbigg MGnety=3µr0xr0z r5 0/parenleftbigg/integraldisplay mz2dm−/integraldisplay mx2dm/parenrightbigg (10.154) MGnetz=3µr0xr0y r5 0/parenleftbigg/integraldisplay mx2dm−/integraldisplay my2dm/parenrightbigg From Section 9.5 we recall that the moments of inertia are defined as A=/integraldisplay my2dm+/integraldisplay mz2dm B =/integraldisplay mx2dm+/integraldisplay mz2dm C =/integraldisplay mx2dm+/integraldisplay my2dm (10.155) from which we may write B−A=/integraldisplay mx2dm−/integraldisplay my2dm A −C=/integraldisplay mz2dm−/integraldisplay mx2dm C−B=/integraldisplay my2dm−/integraldisplay mz2dm 10.10 Gravity-gradient stabilization 533 It follows that Equations 10.154 reduce to MGnetx=3µr 0yr0z r5 0(C−B) MGnety=3µr 0xr0z r5 0(A−C) (10.156) MGnetz=3µr 0xr0y r5 0(B−A) These are the components, in the spacecraft body frame, of the gravitational torque produced by the variation of the earth’s gravitational field over the volume of thespacecraft. T o get an idea of these torque magnitudes, note first of all that r 0x/r0,r0y/r0 and r0z/r0are the direction cosines of the position vector of the center of mass, so that their magnitudes do not exceed 1. For a satellite in a low earth orbit of radius6700 km, 3µ/r 3 0∼=4×10−6s−2, which is therefore the maximum order of magnitude of the coefficients of the inertia terms in Equation 10.156. The moments of inertia ofthe space shuttle are on the order of 10 6kg·m2, so the gravitational torques on this large vehicle are on the order of 1 N ·m. Substituting Equations 10.156 into Euler’ s equations of motion (Equations 9.72), we get A˙ωx+(C−B)ωyωz=3µr 0yr0z r5 0(C−B) B˙ωy+(A−C)ωzωx=3µr 0xr0z r5 0(A−C) (10.157) C˙ωz+(B−A)ω xωy=3µr 0xr0y r5 0(B−A) Now consider the orbital reference frame shown in Figure 10.27. It is actually the Clohessy–Wiltshire frame of Chapter 7, with the axes relabeled. The z/primeaxis points radially outward from the center of the earth, the x/primeaxis is in the direction of the local horizon, and the y/primeaxis completes the right-handed triad by pointing in the direction of the orbit normal. This frame rotates around the y/primeaxis with an angular velocity equal to the mean motion nof the circular orbit. Suppose we align the satellite’s principal body frame axes xyzwith x/primey/primez/prime, respectively. When the body xaxis is aligned with the x/primedirection, it is called the rollaxis. The body yaxis, when aligned with the y/primedirection, is the pitch axis. The body zaxis, pointing outward from the earth in thez/primedirection, is the yaw axis. These directions are illustrated in Figure 10.28. With the spacecraft aligned in this way, the body frame components of the inertial angularvelocity ωareω x=ωz=0 andωy=n. The components of the position vector r0are r0x=r0y=0 and r0z=r0. Substituting this data into Equations 10.157 yields ˙ωx=˙ωy=˙ωz=0 That is, the spacecraft will orbit the planet with its principal axes remaining aligned with the orbital frame. If this motion is stable under the influence of gravityalone, without the use of thrusters, gyros or other devices, then the spacecraft isgravity gradient stabilized. We need to assess the stability of this motion so we can 534 Chapter 10 Satellite attitude dynamics x' r0z' y'Gv nOrbit Figure 10.27 Orbital reference frame x/primey/primez/primeattached to the center of mass of the satellite. x' r0z' y'Gx yz Orbit Roll PitchYaw Figure 10.28 Satellite body frame slightly misaligned with the orbital frame x/primey/primez/prime. determine how to orient a spacecraft to take advantage of this type of passive attitude stabilization. Let the body frame xyzbe slightly misaligned with the orbital reference frame, so that the yaw, pitch and roll angles between the xyzaxes and the x/primey/primez/primeaxes, respectively, 10.10 Gravity-gradient stabilization 535 are very small, as suggested in Figure 10.28. The absolute angular velocity ωof the spacecraft is the angular velocity ωrelrelative to the orbital reference frame plus the inertial angular velocity /Omega1of the x/primey/primez/primeframe, ω=ωrel+/Omega1 The components of ωrelin the body frame are found using the yaw, pitch and roll relations, Equations 9.125. In so doing, it must be kept in mind that all angles andrates are assumed to be so small that their squares and products may be neglected.Recalling that sin α=αand cos α=1 when α<<1, we therefore obtain ω xrel=ωroll−ωyaw=θpitch/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright sinθpitch=˙ψroll−neglect product/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright ˙φyawθpitch=˙ψroll (10.158) ωyrel=ωyaw=1/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright cosθpitch=ψ roll/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright sinψroll+ωpitch=1/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright cosψroll=neglect product/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright ˙φyawψroll+˙θpitch=˙θpitch (10.159) ωzrel=ωyaw=1/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright cosθpitch=1/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright cosψroll−ωpitch=ψ roll/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright sinψroll=˙φyaw−neglect product/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright ˙θpitchψroll=˙φyaw (10.160) The orbital frame’s angular velocity is the mean motion nof the circular orbit, so that /Omega1=nˆj/prime T o obtain the orbital frame’s angular velocity components along the body frame, we must use the transformation rule {/Omega1} x=[Q]x/primex{/Omega1} x/prime (10.161) where/bracketleftbig Q/bracketrightbig x/primexis given by Equation 9.123. (Keep in mind that x/primey/primez/primeare playing the role of XYZ in Figure 9.26.) Using the small angle approximations in Equation 9.123 leads to /bracketleftbig Q/bracketrightbig x/primex= 1 φyaw −θpitch −φ yaw 1 ψroll θpitch −ψ roll 1  With this, Equation 10.161 becomes   /Omega1x /Omega1y /Omega1z  = 1 φyaw −θpitch −φ yaw 1 ψroll θpitch −ψ roll 1   0 n 0  =  nφyaw n −nψ roll   Now we can calculate the components of the satellite’s inertial angular velocity along the body frame axes, ωx=ωxrel+/Omega1x=˙ψroll+nφyaw ωy=ωyrel+/Omega1y=˙θpitch+n (10.162) ωz=ωzrel+/Omega1z=˙φyaw−nψroll 536 Chapter 10 Satellite attitude dynamics Differentiating these with r espect to time, remembering that nis constant for a circular orbit, gives the components of inertial angular acceleration in the body frame, ˙ωx=¨ψroll+n˙φyaw ˙ωy=¨θpitch (10.163) ˙ωz=¨φyaw−n˙ψroll The position vector of the satellite’s center of mass lies along the z/primeaxis of the orbital frame, r0=r0ˆk/prime T o obtain the components of r0in the body frame we once again use the transformation matrix [ Q]x/primex   r0x r0y r0z  = 1 φyaw −θpitch −φyaw 1 ψroll θpitch −ψroll 1   0 0 r0  =  −r0θpitch r0ψroll r0  (10.164) Substituting Equations 10.162, 10.163 and 10.164, together with Equation 7.23, into Equations 10.157, and setting A=Iroll B=Ipitch C=Iyaw (10.165) yields Iroll(¨ψroll+n˙φyaw)+(Iyaw−Ipitch)(˙θpitch+n)(˙φyaw−nψroll) =3(Iyaw−Ipitch)n2ψroll Ipitch¨θpitch+(Iroll−Iyaw)(˙ψroll+nφyaw)(˙φyaw−nψroll) =− 3(Iroll−Iyaw)n2θpitch Iyaw(¨φyaw−n˙ψroll)+(Ipitch−Iroll)(˙θpitch+n)(˙ψroll+nφyaw) =− 3(Ipitch−Iroll)n2θpitchψroll Expanding terms and retaining terms at most linear in all angular quantities and their rates yields Iyaw¨φyaw+(Ipitch−Iroll)n2φyaw+(Ipitch−Iroll−Iyaw)n˙ψroll=0 (10.166) Iroll¨ψroll+(Iroll−Ipitch+Iyaw)n˙φyaw+4(Ipitch−Iyaw)n2ψroll=0 (10.167) Ipitch¨θpitch+3(Iroll−Iyaw)n2θpitch=0 (10.168) These are the differential equations governing the influence of gravity gradient torques on the small angles and rates of misalignment of the body frame with the orbital frame. Equation 10.168, governing the pitching motion around the y/primeaxis, is not coupled to the other two equations. We make the c lassical assumption that the solution is of the form θpitch=Pept(10.169) 10.10 Gravity-gradient stabilization 537 where and Parepconstants. Pis the amplitude of the small disturbance that initi- ates the pitching motion. Substituting Equation 10.169 into Equation 10.168 yields[I pitchp2+3(Iroll−Iyaw)n2]Pept=0 for all t, which implies that the bracketed term must vanish, and that means pmust have either of the two values p1,2=± i/radicalBigg 3(Iroll−Iyaw)n2 Ipitch(i=√ −1) Thus θpitch=P1ep1t+P2ep2t yields the stable, small-amplitude, steady-state harmonic oscillator solution only if p1 and p2are imaginary, that is, if Iroll>Iyaw For stability in pitch . (10.170) The stable pitch oscillation frequency is ωfpitch=n/radicalBigg 3(Iroll−Iyaw) Ipitch(10.171) (IfIyaw>Iroll, then p1and p2are both real, one positive, the other negative. The positive root causes θpitch→∞ , which is the undesirable, unstable case.) Let us now turn our attention to Equations 10.166 and 10.167, which govern yaw and roll motion under gravity gradient torque. Again, we assume the solution isexponential in form, φ yaw=Yeqtψroll=Reqt(10.172) Substituting these into Equations 10.166 and 10.167 yields [(Ipitch−Iroll)n2+Iyawq2]Y+(Ipitch−Iroll−Iyaw)nqR=0 (Iroll−Ipitch+Iyaw)nqY+[4(I pitch−Iyaw)n2+Irollq2]R=0 In the interest of simplification, we can factor Iyawout of the first equation and Iroll out of the second one to get /parenleftbiggIpitch−Iroll Iyawn2+q2/parenrightbigg Y+/parenleftbiggIpitch−Iroll Iyaw−1/parenrightbigg nqR=0 /parenleftbigg 1−Ipitch−Iyaw Iroll/parenrightbigg nqY+/parenleftbigg 4Ipitch−Iyaw Irolln2+q2/parenrightbigg R=0 (10.173) Let kY=Ipitch−Iroll IyawkR=Ipitch−Iyaw Iroll(10.174) It is easy to show from Equations 10.155, 10.165 and 10.174 that kY=/parenleftBig/integraltext mx2dm/slashBig/integraltext my2dm/parenrightBig −1 /parenleftBig/integraltext mx2dm/slashBig/integraltext my2dm/parenrightBig +1kR=/parenleftBig/integraltext mz2dm/slashBig/integraltext my2dm/parenrightBig −1 /parenleftBig/integraltext mz2dm/slashBig/integraltext my2dm/parenrightBig +1 538 Chapter 10 Satellite attitude dynamics which means |kY|<1|kR|<1 Using the definitions in Equation 10.174, we can write Equations 10.173 more compactly as (kYn2+q2)Y+(kY−1)nqR=0 (1−kR)nqY+(4kRn2+q2)R=0 or, using matrix notation, /bracketleftbiggkYn2+q2(kY−1)nq (1−kR)nq 4kRn2+q2/bracketrightbigg/braceleftbiggY R/bracerightbigg =/braceleftbigg0 0/bracerightbigg (10.175) In order to avoid the trivial solution ( Y=R=0), the determinant of the coefficient matrix must be zero. Expanding the determinant and collecting terms yields thecharacteristic equation for q, q 4+bn2q2+cn4=0 (10.176) where b=3kR+kYkR+1 c=4kYkR (10.177) This quadratic equation has four roots w hich, when substituted back into Equation 10.172, yield φyaw=Y1eq1t+Y2eq2t+Y3eq3t+Y4eq4t ψroll=R1eq1t+R2eq2t+R3eq3t+R4eq4t In order for these solutions to remain finite in time, the roots q1,...,q4must be negative (solution decays to zero) or imaginary (steady oscillation at initial small amplitude). T o reduce Equation 10.176 to a quadratic equation, let us introduce a new variable λand write, q=± n√ λ (10.178) Then Equation 10.176 becomes λ2+bλ+c=0 (10.179) the familiar solution of which is λ1=−1 2/parenleftBig b+/radicalbig b2−4c/parenrightBig λ2=−1 2/parenleftBig b−/radicalbig b2−4c/parenrightBig (10.180) T o guarantee that qin Equation 10.178 does not take a positive value, we must require thatλbe real and negative (so qwill be imaginary). For λto be real requires that b>2√c,o r 3kR+kYkR+1>4/radicalbig kYkR (10.181) Forλto be negative requires b2>b2−4c,w h i c hw i l lb et r u ei f c>0; i.e., kYkR>0 (10.182) 10.10 Gravity-gradient stabilization 539 Equations 10.181 and 10.182 are the conditions required for yaw and roll stability under gravity gradient torques, to which we must add Equation 10.170 for pitchstability. Observe that we can solve Equations 10.174 to obtain I yaw=1−kR 1−kYkRIpitch Iroll=1−kY 1−kYkRIpitch By means of these relationships, the pitch stability criterion, Iroll/Iyaw>1, becomes 1−kY 1−kR>1 In view of the fact that |kR|<1, this means kY<kR (10.183) Figure 10.29 shows those regions Iand IIon the kY−kRplane in which all three stability criteria (Equations 10.181, 10.182 and 10.183) are simultaneously satisfied,along with the requirement that the three moments of inertia I pitch,Irolland Iyaware positive. In the small sliver of region I, and kY<0 and kR<0; therefore, according to Equations 10.174, Iyaw>Ipitch and Iroll>Ipitch, which together with Equation 10.170 yield Iroll>Iyaw>Ipitch. Remember that the gravity gradient spacecraft is slowly‘spin- ning’ about the minor pitch axis (normal to the orbit plane) at an angular velocityequal to the mean motion of the orbit. So this criterion makes the spacecraft a ‘minoraxis spinner’ , the roll axis (flight direction) being the major axis of inertia. Withenergy dissipation, we know this orientation is not stable in the long run. On theother hand, in region II,k Yand kRare both positive, so that Equations 10.174 imply kRkY 11 –1 0Stable regions: I : Iroll /H11022 I yaw /H11022 I pitchII : Ipitch /H11022 I roll /H11022 I yaw –1 Figure 10.29 Regions in which the values of kYandkRyield neutral stability in yaw, pitch and roll of a gravity gradient satellite. 540 Chapter 10 Satellite attitude dynamics Ipitch>Iyawand Ipitch>Iroll. Thus, along with the pitch criterion ( Iroll>Iyaw), we have Ipitch>Iroll>Iyaw. In this, the preferred, configuration, the gravity gradient spacecraft is a ‘major axis spinner’ about the pitch axis, and the minor yaw axis is theminor axis of inertia. It turns out that all of the known gravity-gradient stabilizedmoons of the solar system, like the earth’s, whose ‘captured’ rate of rotation equalsthe orbital period, are major axis spinners. In Equation 10.171 we presented the frequency of the gravity gradient pitch oscillation. For completeness we should also point out that the coupled yaw and rollmotions have two oscillation frequencies, which are obtained from Equations 10.178 and 10.180, ω fyaw/roll )1,2=n/radicalbigg 1 2(b±/radicalbig b2−4c) (10.184) Recall that band care found in Equation 10.177. We have assumed throughout this discussion that the orbit of the gravity gradient satellite is circular. Kaplan (1976) shows that the effect of a small eccentricity turnsup only in the pitching motion. In particular, the natural oscillation expressed byEquation 10.170 is augmented by a forced oscillation term, θ pitch=P1ep1t+P2ep2t+2esinnt 3/parenleftbiggIroll−Iyaw Ipitch/parenrightbigg −1(10.185) where eis the (small) eccentricity of the orbit. From this we see that there is a pitch resonance. When ( Iroll−Iyaw)/Ipitch approaches 1/3, the amplitude of the last term grows without bound. Example 10.14The uniform, monolithic 10 000 kg slab, having the dimensions shown in Figure 10.30, is in a circular LEO. Determine the orientation of the satellite in its orbit for gravity gradient stabilization, and compute the periods of the pitch and yaw/roll oscillationsin terms of the orbital period T. xy z 3m1m 9mGa bcd Figure 10.30 Parallelepiped satellite. 10.10 Gravity-gradient stabilization 541 According to Figure 9.9(c), the principal moments of inertia around the xyzaxes through the center of mass are A=10 000 12(12+92)=68 333 kg ·m2 B=10 000 12(32+92)=75 000 kg ·m2 C=10 000 12(32+12)=8333.3k g ·m2 Let us first determine whether we can stabilize this object as a minor axis spinner. In that case, Ipitch=C=8333.3k g ·m2Iyaw=A=68 333 kg ·m2Iroll=B=75 000 kg ·m2 Since Iroll>Iyaw, the satellite would be stable in pitch. T o check yaw/roll stability, we first compute kY=Ipitch−Iroll Iyaw=− 0.97561 kR=Ipitch−Iyaw Iroll=− 0.8000 We see that kYkR>0, which is one of the two requirements. The other one is found in Equation 10.181, but in this case 1+3kR+kYkR−4/radicalbig kYkR=− 4.1533 <0 so that condition is not met. Hence, the object cannot be gravity-gradient stabilized as a minor axis spinner. As a major axis spinner, we must have Ipitch=B=75 000 kg ·m2Iyaw=C=8333.3k g ·m2Iroll=A=68 333 kg ·m2 Then Iroll>Iyaw, so the pitch stability condition is satisfied. Furthermore, since kY=Ipitch−Iroll Iyaw=0.8000 kR=Ipitch−Iyaw Iroll=0.97561 we have kYkR=0.7805 >0 1+3kR+kYkR−4√kYkR=1.1735 >0 which means the two criteria for stability in the yaw and roll modes are met. The satellite should therefore be orbited as shown in Figure 10.31, with its minor axis aligned with the radial from the earth’s center, the plane abcd lying in the orbital plane, and the body xaxis aligned with the local horizon. According to Equation 10.171, the frequency of the pitch oscillation is ωfpitch=n/radicalBigg 3Iroll−Iyaw Ipitch=n/radicalbigg 368 333 −8333.3 75 000=1.5492n 542 Chapter 10 Satellite attitude dynamics (Example 10.14 continued)where nis the mean motion. Hence, the period of this oscillation, in terms of that of the orbit, is Tpitch=2π ωfpitch=0.64552π n=0.6455 T a bcdz x Figure 10.31 Orientation of the parallelepiped for gravity-gradient stabilization. For the yaw/roll frequencies, we use Equation 10.184, ωfyaw/roll/parenrightbig 1=n/radicalbigg 1 2/parenleftBig b+/radicalbig b2−4c/parenrightBig where b=1+3kR+kYkR=4.7073 and c=4kYkR=3.122 Thus, ωfyaw/roll/parenrightbig 1=2.3015 n Likewise, ωfyaw/roll/parenrightbig 2=/radicalbigg 1 2/parenleftBig b−/radicalbig b2−4c/parenrightBig =1.977n From these we obtain Tyaw/roll1=0.5058 T Tyaw/roll2=0.4345 T Finally, observe that Iroll−Iyaw Ipitch=0.8 so that we are far from the pitch resonance condition that exists if the orbit has a small eccentricity. Problems 543 Problems 10.1 The axisymmetric satellite has axial and transverse mass moments of inertia about axes through the mass center GofC=1200 kg ·m2andA=2600 kg ·m2, respectively. If it is spinning at ωs=6 rad/s when it is launched, determine its angular momentum. Precession occurs about the inertial Zaxis. {Ans.: /bardblHG/bardbl=13 450 kg ·m2/s} z Z 6° Gωs Figure P .10.1 10.2 A spacecraft is symmetrical about its body-fixed zaxis. Its principal mass moments of inertia are A=B=300 kg ·m2andC=500 kg ·m2.T h e zaxis sweeps out a cone with a total vertex angle of 10◦as it precesses around the angular momentum vector. If the spin velocity is 6 rad/s, compute the period of precession.{Ans.: 0.417 s} xyz 10° G Figure P .10.2 544 Chapter 10 Satellite attitude dynamics 10.3 A thin ring tossed into the air with a spin velocity of ωshas a very small nutation angle θ(in radians). What is the precession rate ωp? {Ans.: ωp=2ωs(1+θ2/2), retrograde} Figure P .10.3 10.4 For an axisymmetric rigid satellite, [IG]= Ixx 00 0 Iyy 0 00 Izz = 1000 0 0 0 1000 00 0 5000 kg·m 2 It is spinning about the body zaxis in torque-free motion, precessing around the angular momentum vector Hat the rate of 2 rad/s. Calculate the magnitude of H. {Ans.: 2000 N ·m·s} 10.5 At a given instant the box-shaped 500 kg satellite (in torque-free motion) has an absolute angular velocity ω=0.01ˆi−0.03ˆj+0.02ˆk(rad/s). Its moments of iner- tia about the principal body axes xyzareA=385.4k g·m2,B=416.7k g·m2and C=52.08 kg·m2, respectively. Calculate the magnitude of its absolute angular acceleration.{Ans.: 6.167 ×10 −4m/s2} y zxG0.5 m1.0 m 3m Figure P .10.5 Problems 545 10.6 An 8 kg thin ring in torque-free motion is spinning with an angular velocity of 30 rad/s and a constant nutation angle of 15◦. Calculate the rotational kinetic energy ifA=B=0.36 kg ·m2,C=0.72 kg ·m2. {Ans.: 370.5 J} 15° xyz Figure P .10.6 10.7 The rectangular block has an angular velocity ω=1.5ω 0ˆi+0.8ω 0ˆj+0.6ω 0ˆk,w h e r e ω0has units of rad/s. (a) Determine the angular velocity ωof the block if it spins around the body zaxis with the same rotational kinetic energy. (b) Determine the angular velocity ωof the block if it spins around the body zaxis with the same angular momentum. {Ans.: (a) ω=1.31ω 0, (b)ω=1.04ω 0} 2ll3l xyz G Figure P .10.7 10.8 For a rigid axisymmetric satellite, the mass moment of inertia about its long axis is 1000 kg ·m2, and the moment of inertia about transverse axes through the centroid is 5000 kg ·m2. It is spinning about the minor principal body axis in torque-free motion at 6 rad/s with the angular velocity lined up with the angular momentum vector H. Over time, the energy degrades due to inter nal effects and the satellite is eventually spinning about a major principal body axis with the angular velocity lined up with 546 Chapter 10 Satellite attitude dynamics the angular momentum vector H. Calculate the change in rotational kinetic energy between the two states.{Ans.: −14.4 kJ} 10.9 Let the object in Example 9.11 be a highly dissipative torque-free satellite, whose angular velocity at the instant shown is ω=10ˆirad/s. Calculate the decrease in kinetic energy after it becomes, as eventually it must, a major axis spinner.{Ans.: −0.487 J} 4321 0.4 m 0.5 m 0.3 m 0.2 mO xyz Body-fixed frame G Figure P .10.9 10.10 For a non-precessing, dual-spin satellite, Cr=1000 kg ·m2and Cp=500 kg ·m2.T h e angular velocity of the rotor is 3 ˆkrad/s and the angular velocity of the platform relative to the rotor is 1 ˆkrad/s. If the relative angular velocity of the platform is reduced to 0.5ˆkrad/s, what is the new angular velocity of the rotor? {Ans.: 3.17 rad/s} Rotorx yz Gp Platform G Grωz(r)ωz /H11001 ωp(r) Figure P .10.10 Problems 547 10.11 For a rigid axisymmetric satellite, the mass moment of inertia about its long axis is 1000 kg ·m2, and the moment of inertia about transverse axes through the center of mass is 5000 kg ·m2. It is initially spinning about the minor principal body axis in torque-free motion at ωs=0.1 rad/s, with the angular velocity lined up with the angular momentum vector H0. A pair of thrusters exert an external impulsive torque on the satellite, causing an instantaneous change /Delta1H of angular momentum in the direction normal to H0(no change in spin rate), so that the new angular momentum is H1, at an angle of 20◦toH0, as shown in the figure. How long does it take the satellite to precess (‘cone’) through an angle of 180◦around H1? {Ans.: 118 s} 20° GH0/H9004H H1Position just after the impulsive torque Position after 180° precession at the rate ωp ωs ωp Figure P .10.11 10.12 The solid right-circular cylinder of mass 500 kg is set into torque-free motion with its symmetry axis initially aligned with the fixed spatial line a–a. Due to an injection error,the vehicle’s angular velocity vector ωis misaligned 5◦(the wobble angle) from the symmetry axis. Calculate to three significant figures the maximum angle φbetween fixed line a–a and the axis of the cylinder.{Ans.: 31 ◦} 10.13 A satellite is spinning at 0.01 rev/s. The moment of inertia of the satellite about the spin axis is 2000 kg ·m2. Paired thrusters are located at a distance of 1.5 m from the spin axis. They deliver their thrust in pulses, each thruster producing an impulse of15 N·s per pulse. At what rate will the satellite be spinning after 30 pulses? {Ans.: 0.0637 rev/s} 10.14 A satellite has moments of inertia A=2000 kg ·m 2,B=4000 kg ·m2and C= 6000 kg ·m2about its principal body axes xyz. Its angular velocity is ω=0.1ˆi+0.3ˆj+ 0.5ˆk(rad/s). If thrusters cause the angular momentum vector to undergo the change /Delta1H G=50ˆi−100ˆj+3000ˆk(kg·m2/s), what is the magnitude of the new angular velocity?{Ans.: 0.628 rad/s} 10.15 The body-fixed xyzaxes are principal axes of inertia passing through the center of mass of the 300 kg cylindrical satellite, which is spinning at 1 revolution per second about thezaxis. What impulsive torque about the yaxis must the thrusters impart to cause the satellite to precess at 0.1 revolution per second?{Ans.: 137 N ·m·s} 548 Chapter 10 Satellite attitude dynamics 5° G/H9275 aφa 2m0.5 m Figure P .10.12 z x yG1.5 m1.5 m1 rev/s Figure P .10.15 10.16 A satellite is to be despun by means of a tangential-release yo-yo mechanism consisting of two masses, 3 kg each, wound around the mid plane of the satellite. The satellite isspinning around its axis of symmetry with an angular velocity ω s=5 rad/s. The radius of the cylindrical satellite is 1.5 m and the moment of inertia about the spin axis isC=300 kg ·m 2. Problems 549 (a) Find the cord length and the deployment time to reduce the spin rate to 1 rad/s. (b) Find the cord length and time to reduce the spin rate to zero. {Ans.: (a) l=5.902 m, t=0.787 s; (b) l=7.228 m, t=0.964 s} 10.17 A cylindrical satellite of radius 1 m is init ially spinning about the axis of symmetry at the rate of 2 revolutions per second with a nutation angle of 15◦. The principal moments of inertia are Ix=Iy=30 kg·m2,Iz=60 kg·m2. An energy dissipation device is built into the satellite, so that it eventually ends up in pure spin around the zaxis. (a) Calculate the final spin rate about the zaxis. (b) Calculate the loss of kinetic energy. (c) A tangential release yo-yo despin device is also included in the satellite. If the two yo-yo masses are each 7 kg, what cord length is required to completely despin thesatellite? Is it wrapped in the proper direction in the figure? {Ans.: (a) 2.071 rad/s; (b) 8.62 J; (c) 2.3 m} Hz xy15° Yo-yo cord and massωs Figure P .10.17 10.18 A communications satellite is in a GEO (geostationary equatorial orbit) with a period of 24 hours. The spin rate ωsabout its axis of symmetry is 1 revolution per minute, and the moment of inertia about the spin axis is 550 kg ·m2. The moment of inertia about transverse axes through the mass center Gis 225 kg ·m2. If the spin axis is initially pointed towards the earth, calculate the magnitude and direction of the applied torqueM Grequired to keep the spin axis pointed always towards the earth. {Ans.: 0.00420 N ·m, about the negative xaxis} 10.19 The moments of inertia of a satellite about its principal body axes xyz are A=1000 kg ·m2,B=600 kg ·m2and C=500 kg ·m2, respectively. The moments of inertia of a momentum wheel at the center of mass of the satellite and aligned withthexaxis are I x=20 kg·m2and Iy=Iz=6k g·m2. The absolute angular velocity of the satellite with the momentum wheel locked is ω0=0.1ˆi+0.05ˆj(rad/s). Calcu- late the angular velocity ωfof the momentum wheel (relative to the satellite) required to reduce the xcomponent of the absolute angular velocity of the satellite to 0.003 rad/s. {Ans.: 4.95 rad/s} 550 Chapter 10 Satellite attitude dynamics zz z xx x Earth ωsωs ωs Figure P .10.18 xyz G Figure P .10.19 10.20 A satellite has principal moments of inertia I1=300 kg ·m2,I2=400 kg ·m2, I3=500 kg ·m2. Determine the permissible orientations in a circular orbit for gravity- gradient stabilization. Specify which axes m ay be aligned in the pitch, roll and yaw directions. (Recall that, relative to a Clohessy–Wiltshire frame at the center of mass ofthe satellite, yaw is about the xaxis (outward radial from earth’s center); roll is about theyaxis (velocity vector); pitch is about the zaxis (normal to orbital plane). 11Chapter Rocket vehicle dynamics Chapter outline 11.1 Introduction 551 11.2 Equations of motion 552 11.3 The thrust equation 555 11.4 Rocket performance 557 11.5 Restricted staging in field-free space 560 11.6 Optimal staging 570 11.6.1 Lagrange multiplier 570 Problems 578 11.1 Introduction In previous chapters we have made frequent reference to delta-v maneuvers of spacecraft. These require a propulsion system of some sort whose job it is to throw vehicle mass (in the form of propellants) overboard. Newton’s balance of momentumprinciple dictates that when mass is ejected from a system in one direction, the massleft behind must acquire a velocity in the opposite direction. The familiar and oft-quoted example is the rapid release of air from an inflated toy balloon. Another is thatof a diver leaping off a small boat at rest in the water, causing the boat to acquire amotion of its own. The unfortunate astronaut who becomes separated from his shipin the vacuum of space cannot with any amount of flailing of arms and legs ‘swim’back to safety. If he has tools or other expe ndable objects of equipment, accurately 551 552 Chapter 11 Rocket vehicle dynamics throwing them in the direction opposite to his spacecraft may do the trick. Spewing compressed gas from a tank attached to his back through to a nozzle pointed awayfrom the spacecraft would be a better solution. The purpose of a rocket motor is to use the chemical energy of solid or liquid pro- pellants to steadily and rapidly produce a larg e quantity of hot, high pressure gas which is then expanded and accelerated through a nozzle. This large mass of combustionproducts flowing out of the nozzle at supersonic speed possesses a lot of momentumand, leaving the vehicle behind, causes the vehicle itself to acquire a momentum in theopposite direction. This is represented as the action of the force we know as thrust.The design and analysis of rocket propulsion systems is well beyond our scope. This chapter contains a necessarily brief introduction to some of the fundamentals of rocket vehicle dynamics. The equations of motion of a launch vehicle in a gravity turn trajectory are presented first. This is followed by a simple development of thethrust equation, which brings in the concept of specific impulse. The thrust equationand the equations of motion are then combine d to produce the rocket equation, which relates delta-v to propellant expenditure and specific impulse. The sounding rocketprovides an important but relatively simple application of the concepts introducedto this point. The chapter concludes with an elementary consideration of multi-stagelaunch vehicles. Those seeking a more detailed introduction to the subject of rockets and rocket performance will find the texts by Wiesel (1997) and Hale (1994), as well as referencescited therein, useful. 11.2 Equations of motion Figure 11.1 illustrates the trajectory of a satellite launch vehicle and the forces actingon it during the powered ascent. Rockets at the base of the booster produce the TD mgv /H9253 /H9253 Cunut /H9267 xhy Trajectory's center of curvatureTrajectory Local horizon To earth's centerˆˆ Figure 11.1 Launch vehicle boost trajectory. γis the flight path angle. 11.2 Equations of motion 553 thrust Twhich acts along the vehicle’s axis in the direction of the velocity vector v. The aerodynamic drag force Dis directed opposite to the velocity, as shown. Its magnitude is given by D=qAC D (11.1) where q=1 2/rho1v2is the dynamic pressure, in which /rho1is the density of the atmosphere andvis the speed, i.e., the magnitude of v.Ais the frontal area of the vehicle and CDis the coefficient of drag. CDdepends on the speed and the external geometry of the rocket. The force of gravity on the booster is mg,w h e r e mis its mass and gis the local gravitational acceleration, pointing towards the center of the earth. As discussedin Section 1.2, at any point of the trajectory, the velocity vdefines the direction of the unit tangent ˆu tto the path. The unit normal ˆunis perpendicular to vand points towards the center of curvature C. The distance of point Cfrom the path is /rho1(not to be confused with density). /rho1is the radius of curvature. In Figure 11.1 the vehicle and its flight path are shown relative to the earth. In the interest of simplicity we will ignore the earth’s spin and write the equations of motionrelative to a non-rotating earth. The small acceleration terms required to account forthe earth’s rotation can be added for a more refined analysis. Let us resolve Newton’ssecond law, F net=ma, into components along the path directions ˆutandˆun. Recall from Section 1.2 that the acceleration along the path is at=dv dt(11.2) and the normal acceleration is an=v2//rho1(where /rho1is the radius of curvature). It was shown in Example 1.4 (Equation 1.9) that for flight over a flat surface, v//rho1=− dγ/dt, in which case the normal acceleration can be expressed in terms of the flight pathangle as a n=−vdγ dt T o account for the curvature of the earth, as was done in Section 1.6, one can use polar coordinates with origin at the earth’s center to show that a term must be addedto this expression, so that it becomes a n=−vdγ dt+v2 RE+hcosγ (11.3) where REis the radius of the earth and his the altitude of the rocket. Thus, in the direction of ˆutNewton’s second law requires T−D−mgsinγ=mat (11.4) whereas in the ˆundirection mgcosγ=man (11.5) After substituting Equations 11.2 and 11.3, these latter two expressions may be written dv dt=T m−D m−gsinγ (11.6) 554 Chapter 11 Rocket vehicle dynamics vdγ dt=−/parenleftbigg g−v2 RE+h/parenrightbigg cosγ (11.7) T o these we must add the equations for downrange distance xand altitude h, dx dt=RE RE+hvcosγdh dt=vsinγ (11.8) Recall that the variation of gwith altitude is given by Equation 1.8. Numerical meth- ods must be used to solve Equations 11.6, 11.7 and 11.8. T o do so, one must accountfor the variation of the thrust, booster mass, atmospheric density, the drag coefficient,and the acceleration of gravity. Of course, the vehicle mass continuously decreases as propellants are consumed to produ ce the thrust, which we shall discuss in the following section. The free-body diagram in Figure 11.1 does not include a lifting force, which, if the vehicle were an airplane, would act normal to the velocity vector. Launch vehicles aredesigned to be strong in lengthwise compression, like a column. T o save weight theyare, unlike an airplane, made relatively weak in bending, shear and torsion, which arethe kinds of loads induced by lifting surfaces. Transverse lifting loads are held closelyto zero during powered ascent through the atmosphere by maintaining zero angleof attack, i.e., by keeping the axis of the booster aligned with its velocity vector (therelative wind). Pitching maneuvers are done early in the launch, soon after the rocketclears the launch tower, when its speed is still low. At the high speeds acquired withina minute or so after launch, the slightest angle of attack can produce destructivetransverse loads in the vehicle. The space shuttle orbiter has wings so it can act as aglider after re-entry into the atmosphere. However, the launch configuration of theorbiter is such that its wings are at the zero-lift angle of attack throughout the ascent. Satellite launch vehicles take off vertically and, at injection into orbit, must be flying parallel to the earth’s surface. During the initial phase of the ascent, the rocketbuilds up speed on a nearly vertical trajectory taking it above the dense lower layersof the atmosphere. While it transitions the thinner upper atmosphere, the trajectorybends over, trading vertical speed for horizontal speed so the rocket can achieve orbitalperigee velocity at burnout. The gradual transition from vertical to horizontal flight,illustrated in Figure 11.1, is caused by the force of gravity, and it is called a gravity turn trajectory. At lift off, the rocket is vertical and the flight path angle γis 90 ◦. After clearing the tower and gaining speed, vernier thrusters or gimbaling of the main enginesproduce a small, programmed pitchover, establishing an initial flight path angle γ 0, slightly less than 90◦. Thereafter, γwill continue to decrease at a rate dictated by Equation 11.7. (For example, if γ=85◦,v=110 m/s (250 mph), and h=2 km, then dγ/dt=− 0.44◦/s.) As the speed vof the vehicle increases, the coefficient of cos γ in Equation 11.7 decreases, which means the rate of change of the flight path anglebecomes increasingly smaller, tending towards zero as the booster approaches orbitalspeed, v circular orbit =/radicalbig g(R+h). Ideally, the vehicle is flying horizontally ( γ=0) at that point. The gravity turn trajectory is just one example of a practical trajectory, tailored for satellite boosters. On the other hand, sounding rockets fly straight up from launchthrough burnout. Rocket-powered guided missiles must execute high-speed pitch and 11.3 The thrust equation 555 yaw maneuvers as they careen towards moving targets, and require a rugged structure to withstand the accompanying side loads. 11.3 The thrust equation T o discuss rocket performance requires an expression for the thrust Tin Equa- tion 11.6. It can be obtained by a simple one-dimensional momentum analysis.Figure 11.2(a) shows a system consisting of a rocket and its propellants. The exteriorof the rocket is surrounded by the static pressure p aof the atmosphere everywhere except at the rocket nozzle exit where the pressure is pe.peacts over the nozzle exit area Ae. The value of pedepends on the design of the nozzle. For simplicity, we assume no other forces act on the system. At time tthe mass of the system is mand the absolute velocity in its axial direction is v. The propellants combine chemically in the rocket’s combustion chamber, and during the small time interval /Delta1ta small mass /Delta1m of combustion products is forced out of the nozzle, to the left. As a result of this expulsion, the velocity of the rocket changes by the small amount /Delta1v, to the right. The absolute velocity of /Delta1m isve, assumed to be to the left. According to Newton’s second law of motion, (momentum of the system at t+/Delta1t)−(momentum of the system at t) =net external impulse or /bracketleftBig (m−/Delta1m)(v +/Delta1v)ˆi+/Delta1m/parenleftBig −veˆi/parenrightBig/bracketrightBig −mvˆi=(pe−pa)Ae/Delta1tˆi (11.9) Let˙me(a positive quantity) be the rate at which exhaust mass flows across the nozzle exit plane. The mass mof the rocket decreases at the rate dm/dt , and conservation of mass requires the decrease of mass to equal the mass flow rate out of the nozzle.Thus, dm dt=−˙ me (11.10) υ/H11001∆υυepam /H11002 ∆m υpa pem Time t (a) (b)Time t /H11001 ∆t x∆m Figure 11.2 (a) System of rocket and propellant at time t. (b) The system an instant later, after ejection of a small element /Delta1m of combustion products. 556 Chapter 11 Rocket vehicle dynamics Assuming ˙meis constant, the vehicle mass as a function of time (from t=0) may therefore be written m(t)=m0−˙met (11.11) where m0is the initial mass of the vehicle. Since /Delta1mis the mass which flows out in the time interval /Delta1t,w eh a v e /Delta1m=˙me/Delta1t (11.12) Let us substitute this expression into Equation 11.9 to obtain /bracketleftBig (m−˙me/Delta1t)(v+/Delta1v)ˆi+˙me/Delta1t/parenleftBig −veˆi/parenrightBig/bracketrightBig −mvˆi=(pe−pa)Ae/Delta1tˆi Collecting terms, we get m/Delta1vˆi−˙me/Delta1t(ve+v)ˆi−˙me/Delta1t/Delta1vˆi=(pe−pa)Ae/Delta1tˆi Dividing through by /Delta1t, taking the limit as /Delta1t→0, and canceling the common unit vector leads to mdv dt−˙meca=(pe−pa)Ae (11.13) where cais the speed of the exhaust relative to the rocket, ca=ve+v (11.14) Rearranging terms, Equation 11.13 may be written ˙meca+(pe−pa)Ae=mdv dt(11.15) The left-hand side of this equation is the unbalanced force responsible for the acceleration dv/dtof the system in Figure 11.2. This unbalanced force is the thrust T, T=˙meca+(pe−pa)Ae (11.16) where ˙mecais the jet thrust and ( pe−pa)Aeis the pressure thrust. We can write Equation 11.16 as T=˙me/bracketleftbigg ca+(pe−pa)Ae ˙me/bracketrightbigg (11.17) The term in brackets is called the effective exhaust velocity c, c=ca+(pe−pa)Ae ˙me(11.18) In terms of the effective exhaust velocity, the thrust may be expressed simply as T=˙mec (11.19) 11.4 Rocket performance 557 The specific impulse Ispis defined as the thrust per sea-level weight rate (per second) of propellant consumption. That is, Isp=T ˙meg0(11.20) where g0is the standard sea-level acceleration of gravity. The unit of specific impulse is force ÷(force/second) or seconds. T ogether, Equations 11.19 and 11.20 imply that c=Ispg0 (11.21) Obviously, one can infer the jet velocit y directly from the specific impulse. Specific impulse is an important performance parameter for a given rocket engine and propel- lant combination. However, large specific impulse equates to large thrust only if themass flow rate is large, which is true of chemical rocket engines. The specific impulsesof chemical rockets typically lie in the range 200–300 s for solid fuels and 250–450 sfor liquid fuels. Ion propulsion systems have very high specific impulse ( >10 4s), but their very low mass flow rates produce much smaller thrust than chemical rockets. 11.4 Rocket performance From Equations 11.10 and 11.20 we have T=− Ispg0dm dt(11.22) or dm dt=−T Ispg0 If the thrust and specific impulse are constant, then the integral of this expression over the burn time /Delta1tis /Delta1m=−T Ispg0/Delta1t from which we obtain /Delta1t=Ispg0 T(m0−mf)=Ispg0 Tm0/parenleftbigg 1−mf m0/parenrightbigg (11.23) where m0and mfare the mass of the vehicle at the beginning and end of the burn, respectively. The mass ratio is defined as the ratio of the initial mass to final mass, n=m0 mf(11.24) Clearly, the mass ratio is always greater than unity. In terms of the initial mass ratio, Equation 11.23 may be written /Delta1t=n−1 nIsp T/m0g0(11.25) 558 Chapter 11 Rocket vehicle dynamics T/mg0is the thrust-to-weight ratio. The thrust-to-weight ratio for a launch vehicle at lift-off is typically in the range 1.3 to 2. Substituting Equation 11.22 into Equation 11.6, we get dv dt=− Ispg0dm/dt m−D m−gsinγ Integrating with respect to time, from t0totf, yields /Delta1v=Ispg0lnm0 mf−/Delta1vD−/Delta1vG (11.26) where the drag loss /Delta1vDand the gravity loss /Delta1vgare given by the integrals /Delta1vD=/integraldisplaytf t0D mdt /Delta1vG=/integraldisplaytf t0gsinγdt (11.27) Since the drag D, acceleration of gravity g, and flight path angle γare unknown functions of time, these integrals cannot be computed. (Equations 11.6 through 11.8,together with 11.3, must be solved numerically to obtain v(t) andγ(t); but then /Delta1v would follow from those results.) Equation 11.26 can be used for rough estimateswhere previous data and experience provide a basis for choosing conservative valuesof/Delta1v Dand/Delta1vG. Obviously, if drag can be neglected, then /Delta1vD=0. This would be a good approximation for the last stage of a satellite booster, for which it can also besaid that /Delta1v G=0, since γ∼=0◦when the satellite is injected into orbit. Sounding rockets are launched vertically and fly straight up to their maximum altitude before falling back to earth, usually by parachute. Their purpose is to measureremote portions of the earth’s atmosphere. (‘Sound’ in this context means to measureor investigate.) If for a sounding rocket γ=90 ◦, then /Delta1vG≈g0(tf−t0), since gis within 90 percent of g0out to 300 km altitude. Example 11.1A sounding rocket of initial mass m0and mass mfafter all propellant is consumed is launched vertically ( γ=90◦). The propellant mass flow rate ˙meis constant. Neglect- ing drag and the variation of gravity with altitude, calculate the maximum height h attained by the rocket. For what flow rate is the greatest altitude reached? The vehicle mass as a function of time, up to burnout, is m=m0−˙met (a) At burnout, m=mf, so the burnout time tbois tbo=m0−mf ˙me(b) The drag loss is assumed to be zero, and the gravity loss is /Delta1vG=/integraldisplaytbo 0g0sin(90◦)dt=g0tbo 11.4 Rocket performance 559 Recalling that Ispg0=cand using (a), it follows from Equation 11.26 that, up to burnout, the velocity as a function of time is v=clnm0 m0−˙met−g0t (c) Since dh/dt=v, the altitude as a function of time is h=/integraldisplayt 0vdt=/integraldisplayt 0/parenleftbigg clnm0 m0−˙met−g0t/parenrightbigg dt =c ˙me/bracketleftbigg (m0−˙met)lnm0−bt m0+˙met/bracketrightbigg −1 2g0t2(d) T h eh e i g h ta tb u r n o u th bois found by substituting (b) into this expression, hbo=c ˙me/parenleftbigg mflnmf m0+m0−mf/parenrightbigg −1 2/parenleftbiggm0−mf ˙me/parenrightbigg2 g (e) Likewise, the burnout velocity is obtained by substituting (b) into (c), vbo=clnm0 mf−g0 ˙me(m0−mf)( f) After burnout, the rocket coasts upward with the constant downward acceleration of gravity, v=vbo−g0(t−tbo) h=hbo+vbo(t−tbo)−1 2g0(t−tbo)2 Substituting (b), (e) and (f) into these expressions yields, for t>tbo, v=clnm0 mf−g0t h=c ˙me/parenleftbigg m0lnmf m0+m0−mf/parenrightbigg +ctlnm0 mf−1 2g0t2(g) The maximum height hmaxis reached when v=0, clnm0 mf−g0tmax=0⇒ tmax=c g0lnm0 mf Substituting tmaxinto (g) leads to our result, hmax=cm0 ˙me(1+lnn−n)+1 2c2 g0ln2n where nis the mass ratio ( n>1). Since n>(1+lnn), it follows that (1 +lnn−n) is negative. Hence, hmaxcan be increased by increasing the mass flow rate ˙me.I nf a c t , the greatest height is achieved when ˙me→∞ , i.e., all of the propellant is expended at once, like a mortar shell. 560 Chapter 11 Rocket vehicle dynamics 11.5 Restricted staging in field-free space In field-free space we neglect drag and gravitational attraction. In that case, Equation 11.26 becomes /Delta1v=Ispg0lnm0 mf(11.28) This is at best a poor approximation for high-thrust rockets, but it will suffice to shed some light on the rocket staging problem. Observe that we can solve this equation forthe mass ratio to obtain m 0 mf=e/Delta1v Ispg0 (11.29) The amount of propellant expended to produce the velocity increment /Delta1vism0−mf. If we let /Delta1m=m0−mf, then Equation 11.29 can be written as /Delta1m m0=1−e−/Delta1v Ispg0 (11.30) This relation is used to compute the prop ellant required to produce a given delta-v. The gross mass m0of a launch vehicle consists of the empty mass mE, the propellant mass mpand the payload mass mPL, m0=mE+mp+mPL (11.31) The empty mass comprises the mass of the structure, the engines, fuel tanks, control systems, etc. mEis also called the structural mass, although it embodies much more than just structure. Dividing Equation 11.31 through by m0, we obtain πE+πp+πPL=1 (11.32) where πE=mE/m0,πp=mp/m0andπPL=mPL/m0are the structural fraction, pro- pellant fraction and payload fraction, r espectively. It is convenient to define the payload ratio λ=mPL mE+mp=mPL m0−mPL(11.33) and the structural ratio ε=mE mE+mp=mE m0−mPL(11.34) The mass ratio nwas introduced in Equation 11.24. Assuming all of the propellant is consumed, that may now be written n=mE+mp+mPL mE+mPL(11.35) λ,εand nare not independent. From Equation 11.34 we have mE=ε 1−εmp (11.36) 11.5 Restricted staging in field-free space 561 0.1246 0.001 0.01 1.0 0.00011357 00.01/H9255 /H11005 0.001 0.05 0.1 0.2 0.5υbo Ispg0 λ Figure 11.3 Dimensionless burnout speed versus payload ratio. whereas Equation 11.33 gives mPL=λ(m E+mp)=λ/parenleftbiggε 1−εmp+mp/parenrightbigg =λ 1−εmp (11.37) Substituting Equations 11.36 and 11.37 into Equation 11.35 leads to n=1+λ ε+λ(11.38) Thus, given any two of the ratios λ,εandn, we obtain the third from Equation 11.38. Using this relation in Equation 11.28 and setting /Delta1vequal to the burnout speed vbo, when the propellants have been used up, yields vbo=Ispg0lnn=Ispg0ln1+λ ε+λ(11.39) This equation is plotted in Figure 11.3 for a range of structural ratios. Clearly, for a given empty mass, the greatest possible /Delta1voccurs when the payload is zero. However, what we want to do is maximize the amount of payload while keeping the structuralweight to a minimum. Of course, the mass of load-bearing structure, rocket motors,pumps, piping, etc., cannot be made arbitrarily small. Current materials technologyplaces a lower limit on εof about 0.1. For this value of the structural ratio and λ=0.05, Equation 11.39 yields v bo=1.94I spg0=0.019I sp(km/s) 562 Chapter 11 Rocket vehicle dynamics Stage 1Stage 2PayloadmPL mf2mE2 mp2 mp1mE1mf1m02 m01 Figure 11.4 Tandem two-stage booster. The specific impulse of a typical chemical rocket is about 300 s, which in this case would provide /Delta1v=5.7 km/s. However, the circular orbital velocity at the earth’s surface is 7.905 km/s. So this booster by itself could not orbit the payload. Theminimum specific impulse required for a single stage to orbit would be 416 s. Onlytoday’s most advanced liquid hydrogen/liquid oxygen engines, e.g., the space shuttlemain engines, have this kind of performance. Practicality and economics would likelydictate going the route of a multi-stage booster. Figure 11.4 shows a series or tandem two-stage rocket configuration, with one stage sitting on top of the other. Each stage has its own engines and propellant tanks.The dividing line between the stages is where they separate during flight. The firststage drops off first, the second stage next, etc. The payload of an Nstage rocket is actually stage N+1. Indeed, satellites commonly carry their own propulsion systems into orbit. The payload of a given stage is everything above it. Therefore, as illustratedin Figure 11.4, the initial mass m 01of stage 1 is that of the entire vehicle. After stage 1 expels all of its fuel, the mass mf1which remains is stage 1’s empty mass mE1plus the mass of stage 2 and the payload. After separation of stage 1, the process continueslikewise for stage 2, with m 02being its initial mass. Titan II, the launch vehicle for the US Gemini program, had the two-stage, tandem configuration. So did the Saturn 1B, used to launch earth orbital flights early in theUS Apollo program, as well as to send crews to Skylab and an Apollo spacecraft todock with a Russian Soyuz spacecraft in 1975. 11.5 Restricted staging in field-free space 563 Figure 11.5 Parallel staging. Figure 11.5 illustrates the concept of parallel staging. Two or more solid or liquid rockets are attached (‘strapped on’) to a core vehicle carrying the payload. Whereas inthe tandem arrangement, the motors in a given stage cannot ignite until separationof the previous stage, all of the rockets ignite at once in the parallel-staged vehicle.The strap-on boosters fall away after they burn out early in the ascent. The spaceshuttle is the most obvious example of parallel staging. Its two solid rocket boostersare mounted on the external tank, which fuels the three ‘main’ engines built into theorbiter. The solid rocket boosters and the external tank are cast off after they are depleted. In more common use is the combination of parallel and tandem staging,in which boosters are strapped to the first stage of a multi-stage stack. Examplesinclude the United States’ Titan III and IV and Delta launchers, Europe’s Ariane 4 and5, Russia’s Proton and Soyuz variants, Japan’s H-2, and China’s Long March launchvehicles. The original Atlas, used in many variants, for among other things, to launch the orbital flights of the US Mercury program, had three main liquid-fuel engines at itsbase. They all fired simultaneously at launch, but several minutes into the flight, theouter two ‘boosters’ dropped away, leaving the central sustainer engine to burn therest of the way to orbit. Since the booster engines shared the sustainer’s propellanttanks, the Atlas exhibited partial staging, and is sometimes referred to as a one and ahalf stage rocket, the discarded boosters comprising the half stage. We will for simplicity focus on tandem staging, although parallel-staged systems are handled in a similar way (Wiesel, 1997). Restricted staging involves the simple 564 Chapter 11 Rocket vehicle dynamics but unrealistic assumption that all stages are similar. That is, each stage has the same specific impulse Isp, the same structural ratio ε, and the same payload ratio λ.F r o m Equation 11.38 it follows that the mass ratios nare identical, too. Let us investigate the effect of restricted staging on the final burnout speed vbofor a given payload mass mPLand overall payload fraction πPL=mPL m0(11.40) where m0is the total mass of the tandem-stacked vehicle. For a single-stage vehicle, the payload ratio is λ=mPL m0−mPL=1 m0 mPL−1=πPL 1−πPL(11.41) so that, from Equation 11.38, the mass ratio is n=1 πPL(1−ε)+ε(11.42) According to Equation 11.39, the burnout speed is vbo=Ispg0ln1 πPL(1−ε)+ε(11.43) Letm0be the total mass of the two-stage rocket of Figure 11.4, i.e., m0=m01 (11.44) The payload of stage 1 is the entire mass m02of stage 2. Thus, for stage 1 the payload ratio is λ1=m02 m01−m02=m02 m0−m02(11.45) The payload ratio of stage 2 is λ2=mPL m02−mPL(11.46) By virtue of the two stages’ being similar, λ1=λ2,o r m02 m0−m02=mPL m02−mPL Solving this equation for m02yields m02=√m0√mPL Butm0=mPL/πPL, so the gross mass of the second stage is m02=/radicalBigg 1 πPLmPL (11.47) 11.5 Restricted staging in field-free space 565 Putting this back into Equation 11.45 (or 11.46), we obtain the common two-stage payload ratio λ=λ1=λ 2, λ2-stage=πPL1 2 1−πPL1 2(11.48) This together with Equation 11.38 and the assumption that ε1=ε2=εleads to the common mass ratio for each stage, n2-stage=1 πPL1 2(1−ε)+ε(11.49) Assuming that stage 2 ignites immediately after burnout of stage 1, the final velocity of the two-stage vehicle is the sum of the burnout velocities of the individual stages, vbo=vbo1+vbo2 or vbo2-stage=Ispg0lnn2-stage+Ispg0lnn2-stage=2Ispg0lnn2-stage so that, with Equation 11.49, we get vbo2-stage=Ispg0ln/bracketleftBigg 1 πPL1 2(1−ε)+ε/bracketrightBigg2 (11.50) The empty mass of each stage can be found in terms of the payload mass using the common structural ratio ε, mE1 m01−m02=εmE2 m02−mPL=ε Substituting Equations 11.40 and 11.44 together with 11.47 yields mE1=/parenleftBig 1−πPL1 2/parenrightBig ε πPLmPL mE2=/parenleftBig 1−πPL1 2/parenrightBig ε πPL1 2mPL (11.51) Likewise, we can find the propellant mass for each stage from the expressions mp1=m01−(mE1+m02) mp2=m02−(mE2+mPL) (11.52) Substituting Equations 11.40 and 11.44, together with 11.47, 11.51 and 11.52, we get mp1=/parenleftBig 1−πPL1 2/parenrightBig (1−ε) πPLmPL mp2=/parenleftBig 1−πPL1 2/parenrightBig (1−ε) πPL1 2mPL (11.53) 566 Chapter 11 Rocket vehicle dynamics Example 11.2The following data is given mPL=10 000 kg πPL=0.05 ε=0.15 (a) Isp=350 s g0=0.00981 km /s2 Calculate the payload velocity vboat burnout, the empty mass of the launch vehicle and the propellant mass for (a) a single stage and (b) a restricted, two-stage vehicle. (a) From Equation 11.43 we find vbo=350·0.00981 ln1 0.05(1+0.15)+0.15=5.657 km /s Equation 11.40 yields the gross mass m0=10 000 0.05=200 000 kg from which we obtain the empty mass using Equation 11.34, mE=ε(m0−mPL)=0.15(200 000 −10 000) =28 500 kg The mass of propellant is mp=m0−mE−mPL=200 000 −28 500 −10 000 =161 500 kg (b) For a restricted two-stage vehicle, the burnout speed is given by Equation 11.50, vbo2-stage=350·0.00981 ln/bracketleftBigg 1 0.051 2(1−0.15)+0.15/bracketrightBigg2 =7.407 km /s The empty mass of each stage is found using Equations 11.51, mE1=/parenleftBig 1−0.051 2/parenrightBig ·0.15 0.05·10 000 =23 292 kg mE2=/parenleftBig 1−0.051 2/parenrightBig ·0.15 0.051 2·10 000 =5208 kg For the propellant masses, we turn to Equations 11.53 mp1=/parenleftBig 1−0.051 2/parenrightBig ·(1−0.15) 0.05·10 000 =131 990 kg 11.5 Restricted staging in field-free space 567 mp2=/parenleftBig 1−0.051 2/parenrightBig ·(1−0.15) 0.051 2·10 000 =29 513 kg The total empty mass, mE=mE1+mE2, and the total propellant mass, mp=mp1+ mp2, are the same as for the single stage rocket. The mass of the second stage, including the payload, is 22.4 percent of the total vehicle mass. Observe in the previous example that, although the total vehicle mass was unchanged, the burnout velocity increased 31 percent for the two-stage arrangement. The reasonis that the second stage is lighter and can therefore be accelerated to a higher speed.Let us determine the velocity gain associated with adding another stage, as illustrated in Figure 11.6. The payload ratios of the three stages are λ 1=m02 m01−m02λ2=m03 m02−m03λ3=mPL m03−mPL Since the stages are similar, these payload ratios are all the same. Setting λ1=λ2and recalling that m01=m0, we find m2 02−m03m0=0 Payload Stage 2 Stage 1Stage 3mf3mPL mE3 mE2 mp2 mp1mE1mp3 mf2 mf1m02 m01m03 Figure 11.6 Tandem three-stage launch vehicle. 568 Chapter 11 Rocket vehicle dynamics Similarly, λ1=λ3yields m02m03−m0mPL=0 These two equations imply that m02=mPL π2 3 PLm03=mPL π1 3 PL(11.54) Substituting these results back into any one of the above expressions for λ1,λ2orλ3 yields the common payload ratio for the restricted three-stage rocket, λ3-stage=π1 3 PL 1−π1 3 PL With this result and Equation 11.38 we find the common mass ratio, n3-stage=1 πPL1 3(1−ε)+ε(11.55) Since the payload burnout velocity is vbo=vbo1+vbo2+vbo3,w eh a v e vbo3-stage=3Ispg0lnn3-stage=Ispg0ln 1 π1 3 PL(1−ε)+ε 3 (11.56) Because of the common structural ratio across each stage, mE1 m01−m02=εmE2 m02−m03=εmE3 m03−mPL=ε Substituting Equations 11.40 and 11.54 and solving the resultant expressions for the empty stage masses yields mE1=/parenleftbigg 1−π1 3 PL/parenrightbigg ε πPLmPL mE2=/parenleftbigg 1−π1 3 PL/parenrightbigg ε π2 3 PLmPL mE3=/parenleftbigg 1−π1 3 PL/parenrightbigg ε π1 3 PLmPL (11.57) The stage propellant masses are mp1=m01−(mE1+m02)mp2=m02−(mE2+m03)mp3=m03−(mE3+mPL) Substituting Equations 11.40, 11.54 and 11.57 leads to mp1=/parenleftbigg 1−π1 3 PL/parenrightbigg (1−ε) πPLmPL mp2=/parenleftbigg 1−π1 3 PL/parenrightbigg (1−ε) π2 3 PLmPL (11.58) mp3=/parenleftbigg 1−π1 3 PL/parenrightbigg (1−ε) π1 3 PLmPL 11.5 Restricted staging in field-free space 569 Example 11.3Repeat Example 11.2 for the restricted three-stage launch vehicle. Equation 11.56 gives the burnout velocity for three stages, vbo=350·0.00981 ·ln/parenleftBigg 1 0.051 3(1−0.15)+0.15/parenrightBigg3 =7.928 km /s Substituting mPL=10 000 kg, πPL=0.05 and ε=0.15 into Equations 11.57 and 11.58 yields mE1=18 948 kg mE2=6980 kg mE3=2572 kg mp1=107 370 kg mp2=39 556 kg mp3=14 573 kg Again, the total empty mass and total propellant mass are the same as for the single and two-stage vehicles. Notice that the velocity increase over the two-stage rocket is just 7 percent, which is much less than the advantage the two-stage had over the singlestage vehicle. Looking back over the velocity formulas for one, two and three stage vehicles (Equations 11.43, 11.50 and 11.56), we can induce that for an N-stage rocket, v boN-stage=Ispg0ln/parenleftBigg 1 πPL1 N(1−ε)+ε/parenrightBiggN =Ispg0Nln/parenleftBigg 1 πPL1 N(1−ε)+ε/parenrightBigg (11.59) What happens as we let Nbecome very large? First of all, it can be shown using Taylor series expansion that, for large N, πPL1 N≈1+1 NlnπPL (11.60) Substituting this into Equation 11.59, we find that vb0N-stage≈Ispg0Nln/bracketleftBigg 1 1+1 N(1−ε)l nπPL/bracketrightBigg Since the term1 N(1−ε)l nπPLis arbitrarily small, we can use the fact that 1/(1+x)=1−x+x2−x3+ ··· to write 1 1+1 N(1−ε)l nπPL≈1−1 N(1−ε)l nπPL which means vb0N-stage≈Ispg0Nln/bracketleftbigg 1−1 N(1−ε)l nπPL/bracketrightbigg Finally, since ln(1 −x)=− x−x2/2−x3/3−x4/4− ··· , we can write this as vb0N-stage≈Ispg0N/bracketleftbigg −1 N(1−ε)l nπPL/bracketrightbigg 570 Chapter 11 Rocket vehicle dynamics 23 4 67 8 N6 59 15 9 1 08.74 78νbo, km/sIsp = 350 s ε = 0.15 πPL = 0.05 Figure 11.7 Burnout velocity versus number of stages (Equation 11.59). Therefore, as N, the number of stages, tends towards infinity, the burnout velocity approaches vbo∞=Ispg0(1−ε)l n1 πPL(11.61) Thus, no matter how many similar stages we use, for a given specific impulse, payload fraction and structural ratio, we cannot exceed this burnout speed. For example,using I sp=350 s, πPL=0.05 and ε=0.15 from the previous two examples yields vb0∞=8.743 km/s, which is only 10 percent greater than vboof a three-stage vehicle. The trend of vbotowards this limiting value is illustrated by Figure 11.7. Our simplified analysis does not take into account the added weight and com- plexity accompanying additional stages. Practical reality has limited the number ofstages of actual launch vehicles to rarely more than three. 11.6 Optimal staging Let us now abandon the restrictive assumption that all stages of a tandem-stacked vehicle are similar. Instead, we will specify the specific impulse Ispiand structural ratio εiof each stage, and then seek the minimum-mass N-stage vehicle that will carry a given payload mPLto a specified burnout velocity vbo. T o optimize the mass requires using the Lagrange multiplier method, which we shall briefly review. 11.6.1 Lagrange multiplier Consider a bivariate function fon the xyplane. Then z=f(x,y)i sas u r f a c el y i n g above or below the plane, or both. f(x,y) is stationary at a given point if it takes on a local maximum or a local minimum, i.e., an extremum, at that point. For fto be 11.6 Optimal staging 571 stationary means df=0; i.e., ∂f ∂xdx+∂f ∂ydy=0 (11.62) where dxand dyare independent and not necessarily zero. It follows that for an extremum to exist, ∂f ∂x=∂f ∂y=0 (11.63) Now let g(x,y)=0 be a curve in the xyplane. Let us find the points on the curve g=0a tw h i c h fis stationary. That is, rather than searching the entire xyplane for extreme values of f, we confine our attention to the curve g=0, which is therefore a constraint. Since g=0, it follows that dg=0, or ∂g ∂xdx+∂g ∂ydy=0 (11.64) If Equations 11.62 and 11.64 are both valid at a given point, then dy dx=−∂f/∂x ∂f/∂y=−∂g/∂x ∂g/∂y That is, ∂f/∂x ∂g/∂x=∂f/∂y ∂g/∂y=−η From this we obtain ∂f ∂x+η∂g ∂x=0∂f ∂y+η∂g ∂y=0 But these, together with the constraint g(x,y)=0, are the very conditions required for the function h(x,y,η)=f(x,y)+ηg(x,y) (11.65) to have an extremum, namely, ∂h ∂x=∂f ∂x+η∂g ∂x=0 ∂h ∂y=∂f ∂y+η∂g ∂y=0 ∂h ∂η=g=0(11.66) ηis the Lagrange multiplier. The procedure generalizes to functions of any number of variables. One can determine mathematically whether the extremum is a maximum or a minimum by checking the sign of the second differential d2hof the function hin Equation 11.65, d2h=∂2h ∂x2dx2+2∂2h ∂x∂ydxdy+∂2h ∂y2dy2(11.67) 572 Chapter 11 Rocket vehicle dynamics Ifd2h<0 at the extremum for all dxand dysatisfying the constraint condition, Equation 11.64, then the extremum is a local maximum. Likewise, if d2h>0, then the extremum is a local minimum. Example 11.4(a) Find the extrema of the function z=− x2−y2. (b) Find the extrema of the same function under the constraint y=2x+3. (a) T o find the extrema we must use Equations 11.63. Since ∂z/∂x=− 2xand ∂z/∂y=− 2y, it follows that ∂z/∂x=∂z/∂y=0a tx=y=0, at which point z=0. Since zis negative everywhere else (see Figure 11.8), it is clear that the extreme value is the maximum value. (b) The constraint may be written g=y−2x−3. Clearly, g=0. Multiply the con- straint by the Lagrange multiplier ηand add the result (zero!) to the function −(x2+y2) to obtain h=− (x2+y2)+η(y−2x−3) This is a function of the three variables x,yandη. For it to be stationary, the partial derivatives with respect to all three of these variables must vanish. First we have ∂h ∂x=− 2x−2η Setting this equal to zero yields x=−η (a) Next, ∂h ∂y=− 2y+η For this to be zero means y=η 2(b) Finally ∂h ∂η=y−2x−3 Setting this equal to zero gives us back the constraint condition, y−2x−3=0( c ) Substituting (a) and (b) into (c) yields η=1.2, from which (a) and (b) imply, x=− 1.2 y=0.6 (d) These are the coordinates of the point on the line y=2x+3a tw h i c h z=− x2−y2is stationary. Using (d), we find that z=− 1.8 at this point. Figure 11.8 is an illustration of this problem, and it shows that the com- puted extremum (a maximum, in the sense that small negative numbers exceed 11.6 Optimal staging 573 }/H110021.8y /H11005 2x /H11001 3 xy z (/H110021.2, 0.6, 0)z /H11005 /H11002x2 /H11002 y2 Figure 11.8 Location of the point on the line y=2x+3 at which the surface z=− x2−y2is closest to the xyplane. large negative numbers) is where the surface z=− x2−y2is closest to the line y=2x+3, as measured in the zdirection. Note that in this case, Equation 11.67 yields d2h=− 2dx2−2dy2, which is negative, confirming our conclusion that the extremum is a maximum. Now let us return to the optimal staging problem. It is convenient to introduce the step mass miof the ith stage. The step mass is the empty mass plus the propellant mass of the stage, exclusive of all the other stages, mi=mEi+mpi (11.68) The empty mass of stage ican be expressed in terms of its step mass and its structural ratioεias follows, mEi=εi(mEi+mpi)=εimi (11.69) The total mass of the rocket excluding the payload is M, which is the sum of all of the step masses, M=N/summationdisplay i=1mi (11.70) Thus, recalling that m0is the total mass of the vehicle, we have m0=M+mPL (11.71) Our goal is to minimize m0. 574 Chapter 11 Rocket vehicle dynamics For simplicity, we will deal first with a two-stage rocket, and then generalize our results to Nstages. For a two-stage vehicle, m0=m1+m2+mPL, so we can write, m0 mPL=m1+m2+mPL m2+mPLm2+mPL mPL(11.72) The mass ratio of stage 1 is n1=m01 mE1+m2+mPL=m1+m2+mPL ε1m1+m2+mPL(11.73) where Equation 11.69 was used. Likewise, the mass ratio of stage 2 is n2=m02 ε2m2+mPL=m2+mPL ε2m2+mPL(11.74) We can solve Equations 11.73 and 11.74 to obtain the step masses from the mass ratios, m2=n2−1 1−n2ε2mPL m1=n1−1 1−n1ε1(m2+mPL)(11.75) Now, m1+m2+mPL m2+mPL=1−ε1 1−ε1m1+m2+mPL m2+mPL+(ε1m1−ε1m1)1 ε1m1+m2+mPL 1 ε1m1+m2+mPL These manipulations leave the right-hand side unchanged. Carrying out the multiplications proceed as follows, m1+m2+mPL m2+mPL=(1−ε1)(m1+m2+mPL) ε1m1+m2+mPL−ε1(m1+m2+mPL)1 ε1m1+m2+mPL 1 ε1m1+m2+mPL =(1−ε1)m1+m2+mPL ε1m1+m2+mPL ε1m1+m2+mPL ε1m1+m2+mPL−ε1m1+m2+mPL ε1m1+m2+mPL Finally, with the aid of Equation 11.73, this algebraic trickery reduces to m1+m2+mPL m2+mPL=(1−ε1)n1 1−ε1n1(11.76) Likewise, m2+mPL mPL=(1−ε2)n2 1−ε2n2(11.77) 11.6 Optimal staging 575 so that Equation 11.72 may be written in terms of the stage mass ratios instead of the step masses, m0 mPL=(1−ε1)n1 1−ε1n1(1−ε2)n2 1−ε2n2(11.78) Taking the natural logarithm of both sides, we get lnm0 mPL=ln(1−ε1)n1 1−ε1n1+ln(1−ε2)n2 1−ε2n2 Expanding the logarithms on the right side leads to lnm0 mPL=[ln(1−ε1)+lnn1−ln(1−ε1n1)] +[ln(1−ε2)+lnn2−ln(1−ε2n2)] (11.79) Observe that for mPLfixed, ln( m0/mPL) is a monotonically increasing function of m0, d dm0/parenleftbigg lnm0 mPL/parenrightbigg =1 m0>0 Therefore, ln ( m0/mPL) is stationary when m0is stationary. From Equations 11.21 and 11.39, the burnout velocity of the two-stage rocket is vbo=vbo1+vbo2=c1lnn1+c2lnn2 (11.80) which means that, given vbo, our constraint equation is vbo−c1lnn1−c2lnn2=0 (11.81) Introducing the Lagrange multiplier η, we combine Equations 11.79 and 11.81 to obtain h=[ln(1−ε1)+lnn1−ln(1−ε1n1)]+[ln(1−ε2)+lnn2−ln(1−ε2n2)] +η(v bo−c1lnn1−c2lnn2) (11.82) Finding the values of n1and n2for which his stationary will extremize ln( m0/mPL) (and, hence, m0) for the prescribed burnout velocity vbo.his stationary when ∂h/∂n1=∂h/∂n2=∂h/∂η=0. Thus, ∂h ∂n1=1 n1+ε1 1−ε1n1−ηc1 n1=0 ∂h ∂n2=1 n2+ε2 1−ε2n2−ηc2 n2=0 ∂h ∂η=vbo−c1lnn1−c2lnn2=0 These three equations yield, respectively, n1=c1η−1 c1ε1ηn2=c2η−1 c2ε2ηvbo=c1lnn1+c2lnn2 (11.83) 576 Chapter 11 Rocket vehicle dynamics Substituting n1and n2into the expression for vbo,w eg e t c1ln/parenleftbiggc1η−1 c1ε1η/parenrightbigg +c2ln/parenleftbiggc2η−1 c2ε2η/parenrightbigg =vbo (11.84) This equation must be solved iteratively for η, after which ηis substituted into Equa- tions 11.83 1,2to obtain the stage mass ratios n1and n2. These mass ratios are used in Equations 11.75 together with the assumed s tructural ratios, exhaust velocities, and payload mass to obtain the step masses of each stage. We can now generalize the optimization procedure to an N-stage vehicle, for which Equation 11.82 becomes h=N/summationdisplay i=1/bracketleftbig ln(1−εi)+lnni−ln(1−εini)/bracketrightbig −η/parenleftBigg vbo−N/summationdisplay i=1cilnni/parenrightBigg (11.85) At the outset, we know the required burnout velocity vbo, the payload mass mPL, and for every stage we have the structural ratio εiand the exhaust velocity ci(i.e., the specific impulse). The first step is to solve for the Lagrange parameter ηusing Equation 11.84, which, for Nstages, is written N/summationdisplay i=1cilnciη−1 ciεiη=vbo Expanding the logarithm, this can be written N/summationdisplay i=1ciln(ciη−1)−lnηN/summationdisplay i=1ci−N/summationdisplay i=1cilnciεi=vbo (11.86) After solving this equation iteratively for η, we use that result to calculate the optimum mass ratio for each stage (cf. Equation 11.83), ni=ciη−1 ciεiη,i=1, 2,...,N (11.87) Of course, each nimust be greater than 1. Referring to Equations 11.75, we next obtain the step masses of each stage, beginning with stage Nand working our way down the stack to stage 1, mN=nN−1 1−nNεNmPL mN−1=nN−1−1 1−nN−1εN−1(mN+mPL) mN−2=nN−2−1 1−nN−2εN−2(mN−1+mN+mPL) (11.88) ... m1=n1−1 1−n1ε1(m2+m3+··· mPL) 11.6 Optimal staging 577 Having found each step mass, each empty stage mass is mEi=εimi (11.89) and each stage propellant mass is mpi=mi−mEi (11.90) For the function hin Equation 11.85 it is easily shown that ∂2h ∂ni∂nj=0, i,j=1,..., N(i/negationslash=j) It follows that the second differential of his d2h=N/summationdisplay i=1N/summationdisplay j=1∂2h ∂ni∂njdnidnj=N/summationdisplay i=1∂2h ∂n2 i(dn i)2(11.91) where it can be shown, again using Equation 11.85, that ∂2h ∂n2 i=ηci(εini−1)2+2εini−1 (εini−1)2n2 i(11.92) Forhto be minimum at the mass ratios nigiven by Equation 11.87, it must be true that d2h>0. Equations 11.91 and 11.92 indicate that this will be the case if ηci(εini−1)2+2εini−1>0, i=1,..., N (11.93) Example 11.5Find the optimal mass for a three-stage launch vehicle which is required to lift a 5000 kg payload to a speed of 10 km/s. For each stage, we are given that Stage 1 Isp1=400 s (c 1=3.924 km /s) ε1=0.10 Stage 2 Isp2=350 s (c 2=3.434 km /s) ε2=0.15 Stage 3 Isp3=300 s (c 3=2.943 km /s) ε3=0.20 Substituting this data into Equation 11.86, we get 3.924 ln(3.924η −1)+3.434 ln(3.434η −1)+2.943 ln(2.943η −1) −10.30 ln η+7.5089 =10 As can be checked by substitution, the it erative solution of this equation is η=0.4668 Substituting ηinto Equations 11.87 yields the optimum mass ratios, n1=4.541 n2=2.507 n3=1.361 For the step masses, we appeal to Equations 11.88 to obtain m1=165 700 kg m2=18 070 kg m3=2477 kg 578 Chapter 11 Rocket vehicle dynamics (Example 11.5 continued)Using Equations 11.89 and 11.90, the empty masses and propellant masses are foundto be m E1=16 570 kg mE2=2710 kg mE3=495.4k g mp1=149 100 kg mp2=15 360 kg mp3=1982 kg The payload ratios for each stage are λ1=m2+m3+mPL m1=0.1542 λ2=m3+mPL m2=0.4139 λ3=mPL m3=2.018 The total mass of the vehicle is m0=m1+m2+m3+mPL=191 200 kg and the overall payload fraction is πPL=mPL m0=5000 191 200=0.0262 Finally, let us check Equation 11.93, ηc1(ε1n1−1)2+2ε1n1−1=0.4541 ηc2(ε2n2−1)2+2ε2n2−1=0.3761 ηc3(ε3n3−1)2+2ε3n3−1=0.2721 A positive number in every instance means we have indeed found a local minimum of the function in Equation 11.85. Problems 11.1 Suppose a spacecraft in permanent orbit around the earth is to be used for delivering payloads from low earth orbit (LEO) to geostationary equatorial orbit (GEO). Beforeeach flight from LEO, the spacecraft is re fueled with propellant which it uses up in its round trip to GEO. The outbound leg requires four times as much propellant asthe inbound return leg. The delta-v for transfer from LEO to GEO is 4.22 km/s (seeExample 6.12). The specific impulse of the propulsion system is 430 s. If the payloadmass is 3500 kg, calculate the empty mass of the vehicle.{Ans.: 2733 kg} 11.2 A two stage, solid-propellant sounding rocket has the following properties: First stage: m 0=249.5k g mf=170.1k g ˙me=10.61 kg/sIsp=235 s Second stage: m0=113.4k g mf=58.97 kg ˙me=4.053 kg /sIsp=235 s Delay time between burnout of first stage and ignition of second stage: 3 seconds. Problems 579 As a preliminary estimate, neglect drag and the variation of earth’s gravity with altitude to calculate the maximum height reached by the second stage after burnout.{Ans.: 322 km} 11.3 A two-stage launch vehicle has the following properties: First stage: 2 solid propellant rockets. Each one has a total mass of 525 000 kg, 450 000 kg of which is propellant. I sp=290 s. Second stage: 2 liquid rockets with Isp=450 s. Dry mass =30 000 kg, propellant mass=600 000 kg. Calculate the payload mass to a 300 km orbit if launched due east from KSC. Let the total gravity and drag loss be 2 km/s.{Ans.: 114 000 kg} 11.4 Consider a rocket comprising three similar stages (i.e., each stage has the same specific impulse, structural ratio and payload ratio). The common specific impulse is 310 s. Thetotal mass of the vehicle is 150 000 kg, the total structural mass (empty mass) is 20 000 kgand the payload mass is 10 000 kg. Calculate (a) The mass ratio nand the total /Delta1vfor the three-stage rocket. {Ans.: n=2.04,/Delta1v=6.50 km/s} (b) m p1,mp2, and mp3. (c) mE1,mE2and mE3. (d) m01,m02and m03. 11.5 A small two-stage vehicle is to propel a 1 0 kg payload to a speed of 6.2 km/s. The properties of the stages are: for the first stage, Isp=300 s and ε=0.2; for the second stage, Isp=235 s and ε=0.3. Estimate the optimum mass of the vehicle. {Ans.: 1125 kg} 11.6 Find the extrema of the function z=x2+y2+2xy subject to the constraint x2−2x+y2=0. {Ans.: zmin=0.1716 at ( x,y)=(0.2929, −0.7071) and zmax=5.828 at (x ,y)=(1.707, 0.7071)} This page int entionally left blank References Bate, R. R., Mueller, D., and White, J. E. (1971). Fundamentals of Astrodynamics , Dover Publications. Battin, R. H. (1999). An Introduction to the Mathematics and Methods of Astrody- namics, Revised Edition, AIAA Education Series. Beyer, W. H., ed. (1991). Standard Mathematical Tables and Formulae , 29th Edition, CRC Press. Bond, V . R. and Allman, M. C. (1996). Modern Astrodynamics: Fundamentals and Perturbation Methods , Princeton University Press. Boulet, D. L. (1991). Methods of Orbit Determination for the Microcomputer , Willmann-Bell. Chobotov, V . A., ed. (2002). Orbital Mechanics, Third Edition, AIAA Education Series, AIAA. Coriolis, G. (1835). ‘On the Equations of Relative Motion of a System of Bodies’ , J. École Polytechnique , Vol. 15, No. 24, 142–54. Hahn, B. D. (2002). Essential MATLAB ®for Scientists and Engineers, Second Edition, Butterworth-Heinemann. Hale, F. J. (1994). Introduction to Space Flight , Prentice-Hall, Englewood Cliffs, New Jersey. Hohmann, W. (1925). The Attainability of Celestial Bodies (in German), R. Oldenbourg. Kaplan, M. H. (1976). Modern Spacecraft Dynamics and Control , Wiley. Kermit, S. and Davis, T. A. (2002). MA TLAB Primer , Sixth Edition, Chapman & Hall/CRC. Likins, P . W. (1967). ‘Attitude Stability Criteria for Dual Spin Spacecraft’ , Journal of Spacecraft and Rockets , Vol. 4, No. 12, 1638–43. Magrab, E. B., ed. (2000). An Engineer’s Guide to MATLAB® , Prentice-Hall. NASA Goddard Space Flight Center (2003). National Space Science Data Center , http://nssdc.gsfc.nasa.gov. Nise, N. S. (2003). Control Systems Engineering , Fourth Edition, Wiley. Ogata, K. (2001). Modern Control Engineering , Fourth Edition, Prentice-Hall. Palm, W. J. (1983). Modeling, Analysis and Control of Dynamic Systems , Wiley. Prussing, J. E. and Conway, B. A. (1993). Orbital Mechanics , Oxford University Press. 581 582 References Seidelmann, P . K., ed. (1992). Explanatory Supplement to the Astronomical Almanac , University Science Books. Standish, E. M., Newhall, X.X., Williams, J. G., and Y eomans, D. K. (1992). ‘Orbital Ephemerides of the Sun, Moon and Planets’ . In Explanatory Supplement to the Astronomical Almanac (P . K. Seidelmann, ed.), p. 316, University Science Books. US Naval Observatory (2004). The Astronomical Almanac , GPO. Wiesel, W. E. (1997). Spacecraft Dynamics , Second Edition, McGraw-Hill. Further reading Brown, C. D. (1992). Spacecraft Mission Design , AIAA Education Series. Chobotov, V . A. (1991). Spacecraft Attitude Dynamics and Control ,K r i e g e r . Chobotov, V . A., ed. (2002). Orbital Mechanics, Third Edition, AIAA Education Series. Danby, J. M. A. (1988). Fundamentals of Celestial Mechanics , Second Edition, Willmann-Bell. Escobal, P . R. (1976). Methods of Orbit Determination , Second Edition, Krieger. Griffin, M. D. and French, J. R. (1991). Space Vehicle Design , AIAA Education Series. Hill, P . P . and Peterson, C. R. (1992). Mechanics and Thermodynamics of Propulsion , Addison-Wesley. Kane, T. R., Likins, P . W., and Levinson, D. A. (1983). Spacecraft Dynamics , McGraw-Hill. Larson, W. J. and Wertz, J. R., ed. (1992). Space Mission Analysis and Design , Second Edition, Microcosm Press and Kluwer Academic Publishers. Logsdon, T. (1998). Orbital Mechanics: Theory and Applications , Wiley. McCuskey, S. W. (1963). Introduction to Celestial Mechanics , Addison-Wesley. Meeus, J. (1998). Astronomical Algorithms , Second Edition, Willmann-Bell. Moulton, F. R. (1970). An Introduction to Celestial Mechanics , Second Edition, Dover Publications. Schaub, S. and Junkins, J. L. (2003). Analytical Mechanics of Space Systems , AIAA Education Series. Sellers, J. J. (1994). Understanding Space: An Introduction to Astronautics , McGraw-Hill. Sutton, G. P . and Biblarz, O. (2001). Rocket Propulsion Elements , Seventh Edition, Wiley. Thomson, W. T. (1986). Introduction to Space Dynamics , Dover Publications. Vallado, D. A. (2001). Fundamentals of Astrodynamics and Applications , Second Edition, Microcosm Press and Kluwer Academic Publishers. Wertz, J. R. (1978). Spacecraft Attitude Determination and Control ,K l u w e r Academic Publishers. AAppendix Physical data The following tables contain information that is commonly available and may be found in the literature and on the world wide web. See, for example, the Astronomical Almanac (US Naval Observatory, 2004) and National Space Science Data Center (NASA Goddard Space Flight Center, 2003). T able A.1 Astronomical data for the sun, the planets and the moon Inclination Inclination of orbit Sidereal of equator Semimajor to the Orbit Radius rotation to orbit axis of Orbit ecliptic sidereal Object (km) Mass (kg) period plane orbit (km) eccentricity plane period Sun 696 000 1.989 ×103025.38d 7.25◦–– –– Mercury 2440 330.2 ×102158.65d 0.01◦57.91×1060.2056 7.00◦87.97d Venus 6052 4.869 ×1024243d* 177.4◦108.2×1060.0067 3.39◦224.7d Earth 6378 5.974 ×102423.9345h 23.45◦149.6×1060.0167 0.00◦365.256d (Moon) 1737 73.48 ×102127.32d 6.68◦384.4×1030.0549 5.145◦27.322d Mars 3396 641.9 ×102124.62h 25.19◦227.9×1060.0935 1.850◦1.881y Jupiter 71 490 1.899 ×10279.925h 3.13◦778.6×1060.0489 1.304◦11.86y Saturn 60 270 568.5 ×102410.66h 26.73◦1.433×1090.0565 2.485◦29.46y Uranus 25 560 86.83 ×102417.24h* 97.77◦2.872×1090.0457 0.772◦84.01y Neptune 24 760 102.4 ×102416.11h 28.32◦4.495×1090.0113 1.769◦164.8y Pluto 1195 12.5 ×10216.387d* 122.5◦5.870×1090.2444 17.16◦247.7y *R e t r o g r a d e 583 584 Appendix A Physical data T able A.2 Gravitational parameter ( µ) and sphere of influence (SOI) radius for the sun, the planets and the moon Celestial body µ(km3/s2) SOI radius (km) Sun 132 712 000 000 – Mercury 22 030 112 000 Venus 324 900 616 000 Earth 398 600 925 000Earth’s moon 4903 66 200Mars 42 828 577 000Jupiter 126 686 000 48 200 000Saturn 37 931 000 54 800 000 Uranus 5 794 000 51 800 000Neptune 6 835 100 86 600 000Pluto 830 3 080 000 T able A.3 Some conversion factors 1f t=0.3048 m 1 mile (mi) =1.609 km 1 nautical mile (n mi) =1.151 mi =1.852 km 1 mi/h =0.0004469 km/s 1 lb (mass) =0.4536 kg 1l b( f o r c e ) =4.448 N 1 psi=6895 kPa BAppendix A road map Figure B.1 is a road map through Chapters 1, 2 and 3. Those who from time to time feel they have lost their bearings may find it useful to refer to this flow chart, which shows how the various concepts and results are interrelated. The pivotalinfluence of Sir Isaac Newton is obvious. All of the equations of classical orbitalmechanics (the two-body problem) are derived from those listed here. dA dt=h 2h = r × r υ⊥ =h r υr =µ he sin θ Kepler's second lawConservation of mechanical energy The orbit formula (Kepler's first law)Newton's laws Definition2-body equation of relative motion T =2π µa3 2Kepler's third lawKepler's equations relating true anomaly to timer = − µ r3rυ2 2−µ r= const r =h2 µ1 1 + e cos θF = ma Fg = Gm1m2 r2ˆ ur t =h3dϑ (1 + e cosϑ)2 0θ ∫µ2··· Figure B.1 Logic flow for the major outcomes of Chapters 1, 2 and 3. 585 This page int entionally left blank CAppendix Numerical integration of the n-body equations of motion Appendix outline C.1 Function file accel_3body.m 590 C.2 Script file threebody.m 592 Without loss of generality we shall derive the equations of motion of the three- body system illustrated in Figure C.1. The equations of motion for nbodies can easily be generalized from those of a three-body system. Each mass of a three-body system experiences the force of gravitational attraction from the other members of the system. As shown in Figure C.1, the forces exerted onb o d y1b yb o d i e s2a n d3a r e F 12and F13, respectively. Likewise, body 2 experiences the forces F21and F23whereas the forces F31and F32act on body 3. These gravitational forces can be inferred from Equation 2.6: F12=− F21=Gm 1m2(R2−R1) /bardblR2−R1/bardbl3(C.1a) F13=− F31=Gm 1m3(R3−R1) /bardblR3−R1/bardbl3(C.1b) F23=− F32=Gm 2m3(R3−R2) /bardblR3−R2/bardbl3(C.1c) 587 588 Appendix C Numerical integration of the n-body equations of motion XY ZOR1 R2 R3m1 m2 m3F12 F21F13 F31 F23F32G Inertial frame Figure C.1 Three-body problem. Relative to an inertial frame of reference the accelerations of the bodies are ai=¨Ri i=1, 2, 3 where Riis the absolute position vector of body i. The equation of motion of body 1 is F12+F13=m1a1 Substituting Equations C.1a and C.1b yields a1=Gm 2(R2−R1) /bardblR2−R1/bardbl3+Gm 3(R3−R1) /bardblR3−R1/bardbl3(C.2a) For bodies 2 and 3 we find in a similar fashion that a2=Gm 1(R1−R2) /bardblR1−R2/bardbl3+Gm 3(R3−R2) /bardblR3−R2/bardbl3(C.2b) a3=Gm 1(R1−R3) /bardblR1−R3/bardbl3+Gm 2(R2−R3) /bardblR2−R3/bardbl3(C.2c) The velocities are related to the accelerations by dvi dt=ai i=1, 2, 3 (C.3) and the position vectors are likewise related to the velocities, dRi dt=vi i=1, 2, 3 (C.4) Equations C.2 through C.4 constitute a system of ordinary differential equations (ODEs) in the variable time. Appendix C Numerical integration of the n-body equations of motion 589 Since there are no external forces on the system, the acceleration of the center of mass is zero aG=0 (C.5a) so that dvG dt=0 (C.5b) and dRG dt=vG (C.5c) Given the initial positions Ri0and initial velocities vi0, we must integrate Equation C.3 to find vias a function of time and substitute those results into Equations C.4 to obtain Rias a function of time. The integrations must be done numerically. T o do this using MATLAB, we first resolve all of the vectors into their three components along the XYZ axes of the inertial frame and write them as column vectors, {R1}=  R1X R1Y R1Z  {R2}=  R2X R2Y R2Z  {R3}=  R3X R3Y R3Z  {RG}=  RGX RGY RGZ  (C.6) {v1}=  v1X v1Y v1Z  {v2}=  v2X v2Y v2Z  {v3}=  v3X v3Y v3Z  {vG}=  vGX vGY vGZ  (C.7) According to Equations C.2, {a1}=  a1X a1Y a1Z  =  Gm 2(R2X−R1X) R12+Gm 3(R3X−R1X) R13 Gm 2(R2Y−R1Y) R12+Gm 3(R3Y−R1Y) R13 Gm 2(R2Z−R1Z) R12+Gm 3(R3Z−R1Z) R13  (C.8a) {a 2}=  a2X a2Y a2Z  =  Gm 1(R1X−R2X) R12+Gm 3(R3X−R2X) R13 Gm 1(R1Y−R2Y) R12+Gm 3(R3Y−R2Y) R13 Gm 1(R1Z−R2Z) R12+Gm 3(R3Z−R2Z) R13  (C.8b) {a 3}=  a3X a3Y a3Z  =  Gm 1(R1X−R3X) R12+Gm 2(R2X−R3X) R13 Gm 1(R1Y−R3Y) R12+Gm 2(R2Y−R3Y) R13 Gm 1(R1Z−R3Z) R12+Gm 2(R2Z−R3Z) R13  (C.8c) 590 Appendix C Numerical integration of the n-body equations of motion where R12=/bardbl { R2}−{ R1}/bardbl3R13=/bardbl { R3}−{ R1}/bardbl3R23=/bardbl { R3}−{ R2}/bardbl3(C.9) Next, we form the 24-component column vector {f}=⌊ { R1}{R2}{R3}{RG}{v1}{v2}{v3}{vG}⌋T(C.10) The first derivatives of the components of this vector comprise the column vector /braceleftbiggdf dt/bracerightbigg =⌊ { v1}{v2}{v3}{vG}{a1}{a2}{a3}{0}⌋T(C.11) If the vector { f} is given at time t, then{df/dt}is used to obtain an accurate estimate of { f}a tt i m e t+/Delta1tby means of a procedure such as that due originally to the German mathematicians Carle Runge (1856–1927) and Martin Kutta (1867–1944).Sophisticated Runge–Kutta algorithms ar e implemented in MATLAB in the form of the solvers ode23 andode45 .ode45 is the more accurate of the two and is recommended as a first try for solving most ODEs. For simplicity, we will use MATLAB to solve the three-body problem in the plane. That is, we will restrict ourselves to only the XYcomponents of the vectors R,vand a. The reader can use these scripts as a starting point for investigating more complex n-body problems. The M-function accel_3body.m is used by ode45 to calculate the accelera- tions of each of the masses from Equations C.8. C.1 Function file accel_3body.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ function dfdt = accel_3body(t,f) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ %% This function evaluates the acceleration of each member of a% planar 3-body system at time t from their positions and% velocities at that time.%% G - gravitational constant% (kmˆ3/kg/sˆ2)% m - vector [m1, m2, m3] containing% the masses m1, m2, m3 of the% three bodies (kg)% r1x, r1y; r2x, r2y; r3x, r3y - components of the position% vectors of each mass (km)% v1x, v1y; v2x, v2y; v3x, v3y - components of the velocity% vectors of each mass (km/s)% a1x, a1y; a2x, a2y; a3x, a3y - components of the acceleration % vectors of each mass (km/sˆ2)% rGx, rGy; vGx, vGy; aGx, aGy - components of the position,% velocity and acceleration of% the center of mass C.1 Function file accel_3body.m 591 % t - time (s) % f - column vector containing the % position and velocity % components of the three % masses and the center of % mass at time t % dfdt - column vector containing the % velocity and acceleration % components of the three % masses and the center of % mass at time t %% User M-functions required: none% ------------------------------------------------------------ global G m%...Initialize the 16 by 1 column vector dfdt: dfdt = zeros(16,1); %...For ease of reading the code, assign each component of f %...to a mnemonic variable: r1x = f( 1); r1y = f( 2); r2x = f( 3); r2y = f( 4); r3x = f( 5); r3y = f( 6); rGx = f( 7); rGy = f( 8); v1x = f( 9); v1y = f(10); v2x = f(11); v2y = f(12); v3x = f(13); v3y = f(14); vGx = f(15); vGy = f(16); %...Equations C.9: r12 = norm([r2x - r1x, r2y - r1y])ˆ3;r13 = norm([r3x - r1x, r3y - r1y])ˆ3;r23 = norm([r3x - r2x, r3y - r2y])ˆ3; 592 Appendix C Numerical integration of the n-body equations of motion %...Equations C.8: a1x = G*m(2)*(r2x - r1x)/r12 + G*m(3)*(r3x - r1x)/r13;a1y = G*m(2)*(r2y - r1y)/r12 + G*m(3)*(r3y - r1y)/r13;a2x = G*m(1)*(r1x - r2x)/r12 + G*m(3)*(r3x - r2x)/r23;a2y = G*m(1)*(r1y - r2y)/r12 + G*m(3)*(r3y - r2y)/r23;a3x = G*m(1)*(r1x - r3x)/r13 + G*m(2)*(r2x - r3x)/r23;a3y = G*m(1)*(r1y - r3y)/r13 + G*m(2)*(r2y - r3y)/r23; %...Equation C.5a: a G x=0 ;a G y=0 ; %...Place the evaluated velocity and acceleration components %...into the vector dfdt, to be returned to the calling%...program: dfdt = [v1x; v1y; ... v2x; v2y; ...v3x; v3y; ...vGx; vGy; ...a1x; a1y; ...a2x; a2y; ...a3x; a3y; ...aGx; aGy]; % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ The script threebody.m defines the initial conditions, passes that information to ode45 and finally plots the solutions. The results of this program were used to create Figures 2.5 and 2.6. Similar scripts can obviously be written for the two-body problemand may be used to produce Figures 2.3 and 2.4. C.2 Script file threebody.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ %threebody % ˜˜˜˜˜˜˜˜˜ %% This program presents the graphical solution of the motion of% three bodies in the plane for data provided in the input% definitions below.%% G - gravitational constant (kmˆ3/kg/sˆ2)% t_initial, t_final - initial and final times (s)% m - vector [m1, m2, m3] containing the% masses m1, m2, m3 of the three% bodies (kg) % r0 - 3 by 2 matrix each row of which% contains the initial x and y components% of the position vector of the% respective mass (km) C.2 Script file threebody.m 593 % v0 - 3 by 2 matrix each row of which contains % the initial x and y components of the % velocity of the respective mass (km/s) % rG0 - vector containing the initial x and y % components of the center of mass (km) % vG0 - vector containing the initial x and y % components of the velocity of the % center of mass (km/s) % f0 - column vector of the initial conditions % passed to the Runge-Kutta solver ode45 % t - column vector of times at which the % solution was computed % f - matrix the columns of which contain the % position and velocity components % evaluated at the times t(:): % f(:,1) , f(:,2) = x1(:), y1(:) % f(:,3) , f(:,4) = x2(:), y2(:) % f(:,5) , f(:,6) = x3(:), y3(:) % f(:,7) , f(:,8) = xG(:), yG(:) %% f(:,9) , f(:,10) = v1x(:), v1y(:) % f(:,11), f(:,12) = v2x(:), v2y(:) % f(:,13), f(:,14) = v3x(:), v3y(:) % f(:,15), f(:,16) = vGx(:), vGy(:) %% User M-function required: accel_3body% ------------------------------------------------------------ clear global G mG = 6.67259e-20; %...Input data: t_initial = 0; t_final = 67000;m = [1.e29 1.e29 1.e29];r 0=[ [ 00 ] [300000 0][600000 0]]; v 0=[ [ 0 0 ] [250 250][ 0 0]]; %... %...Initial position and velocity of center of mass: rG0 = m*r0/sum(m);vG0 = m*v0/sum(m); %...Initial conditions must be passed to ode45 in a column %...vector:f0 = [r0(1,:)’; r0(2,:)’; r0(3,:)’; rG0’; ... v0(1,:)’; v0(2,:)’; v0(3,:)’; vG0’] 594 Appendix C Numerical integration of the n-body equations of motion %...Pass the initial conditions and time interval to ode45, %...which calculates the position and velocity at discrete%...times t, returning the solution in the column vector f.%...ode45 uses the m-function ’accel_3body’ to evaluate the%...acceleration at each integration time step.[t,f] = ode45(’accel_3body’, [t_initial t_final], f0); close all %...Plot the motion relative to the inertial frame %...(Figure 2.5):figuretitle(’Figure 2.5: Motion relative to the inertial frame’, ... ’Fontweight’, ’bold’, ’FontSize’, 12) hold on %...x1 vs y1: plot(f(:,1), f(:,2), ’r’, ’LineWidth’, 0.5) %...x2 vs y2: plot(f(:,3), f(:,4), ’g’, ’LineWidth’, 1.0) %...x3 vs y3: plot(f(:,5), f(:,6), ’b’, ’LineWidth’, 1.5) %...xG vs yG: plot(f(:,7), f(:,8), ’--k’, ’LineWidth’, 0.25) xlabel(’X’); ylabel(’Y’) grid onaxis(’equal’) %...Plot the motion relative to the center of mass %...(Figure 2.6):figuretitle(’Figure 2.6: Motion relative to the center of mass’, ... ’Fontweight’, ’bold’, ’FontSize’, 12) hold on %...(x1 - xG) vs (y1 - yG): plot(f(:,1) - f(:,7), f(:,2) - f(:,8), ’r’, ’LineWidth’, 0.5) %...(x2 - xG) vs (y2 - yG): plot(f(:,3) - f(:,7), f(:,4) - f(:,8), ’--g’, ’LineWidth’, 1.0) %...(x3 - xG) vs (y3 - yG): plot(f(:,5) - f(:,7), f(:,6) - f(:,8), ’b’, ’LineWidth’, 1.5) xlabel(’X’); ylabel(’Y’) grid onaxis(’equal’)% ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ DAppendix MATLAB algorithms Appendix outline D.1 Introduction 596 D.2 Algorithm 3.1: solution of Kepler’s equation by Newton’s method 596 D.3 Algorithm 3.2: solution of Kepler’s equation for the hyperbola using Newton’s method 598 D.4 Calculation of the Stumpff functions S(z) and C(z) 600 D.5 Algorithm 3.3: solution of the universal Kepler’s equation using Newton’s method 601 D.6 Calculation of the Lagrange coefficients fand g and their time derivatives 603 D.7 Algorithm 3.4: calculation of the state vector (r, v) given the initial state vector (r 0,v0) and the time lapse /Delta1t 604 D.8 Algorithm 4.1: calculation of the orbital elements from the state vector 606 D.9 Algorithm 4.2: calculation of the state vector from the orbital elements 610 D.10 Algorithm 5.1: Gibbs’ method of preliminary orbit determination 613 D.11 Algorithm 5.2: solution of Lambert’s problem 616D.12 Calculation of Julian day number at 0 hr UT 621D.13 Algorithm 5.3: calculation of local sidereal time 623D.14 Algorithm 5.4: calculation of the state vector from measurements of range, angular position andtheir rates 626 595 596 Appendix D MATLAB algorithms D.15 Algorithms 5.5 and 5.6: Gauss’s method of preliminary orbit determination with iterativeimprovement 631 D.16 Converting the numerical designation of a month or a planet into its name 640 D.17 Algorithm 8.1: calculation of the state vector of a planet at a given epoch 641 D.18 Algorithm 8.2: calculation of the spacecraft trajectory from planet 1 to planet 2 648 D.1 Introduction This appendix lists MATLAB scripts which implement all of the numbered algo- rithms presented throughout the text. The programs use only the most basic features of MATLAB and are liberally commented so as to make reading the codeas easy as possible. T o ‘drive’ the various algorithms, one can use MATLAB to creategraphical user interfaces (GUIs). However, in the interest of simplicity and keep-ing our focus on the algorithms rather than elegant programming techniques, GUIswere not developed. Furthermore, the scripts do not use files to import and exportdata. Data is defined in declaration statements within the scripts. All output is to thescreen, i.e., to the MATLAB command window. It is hoped that interested studentswill embellish these simple scripts or use them as a springboard towards generatingtheir own programs. Each algorithm is illustrated by a MATLAB coding of a related example problem in the text. The actual output of each of these examples is also listed. It would be helpful to have MATLAB documentation at hand. There are a number of practical references on the subject, including Hahn (2002), Kermit and Davis (2002)and Magrab (2000). MATLAB documentation may also be found at The MathWorksweb site (www.mathworks.com). Should it be necessary to do so, it is a fairly simplematter to translate these programs into other software languages. These programs are presented solely as an alternative to carrying out otherwise lengthy hand computations and are intended for academic use only. They are all basedexclusively on the introductory material presented in this text and therefore do notinclude the effects of perturbations of any kind. D.2 Algorithm 3.1: solution of Kepler’s equation by Newton’s method Function file kepler_E.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ functio n E = kepler_E(e, M) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % D.2 Algorithm 3.1: solution of Kepler’s equation by Newton’s method 597 % This function uses Newton’s method to solve Kepler’s % equation E - e*sin(E) = M for the eccentric anomaly,% given the eccentricity and the mean anomaly.%% E - eccentric anomaly (radians)% e - eccentricity, passed from the calling program% M - mean anomaly (radians), passed from the calling program% pi - 3.1415926...%% User M-functions required: none% ------------------------------------------------------------ %...Set an error tolerance: error = 1.e-8; %...Select a starting value for E: i fM<p i E=M+e/2; else E=M-e/2; end %...Iterate on Equation 3.14 until E is determined to within %...the error tolerance:ratio = 1;while abs(ratio) > error ratio = (E - e*sin(E) - M)/(1 - e*cos(E));E=E-ratio; end % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Script file Example_3_02.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % Example_3_02 % ˜˜˜˜˜˜˜˜˜˜˜˜ %% This program uses Algorithm 3.1 and the data of Example 3.2% to solve Kepler’s equation. %% e - eccentricity% M - mean anomaly (rad)% E - eccentric anomaly (rad)%% User M-function required: kepler_E% ------------------------------------------------------------ clear%...Input data for Example 3.2: e = 0.37255;M = 3.6029; %... %...Pass the input data to the function kepler_E, which returns E: E = kepler_E(e, M); 598 Appendix D MATLAB algorithms %...Echo the input data and output to the command window: fprintf('---------------------------------------------------')fprintf('\n Example 3.2\n')fprintf('\n Eccentricity = %g',e)fprintf('\n Mean anomaly (radians) = %g\n',M)fprintf('\n Eccentric anomaly (radians) = %g',E)fprintf('\n-----------------------------------------------\n') % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Output from Example_3_02 ----------------------------------------------------- Example 3.2 Eccentricity = 0.37255 Mean anomaly (radians) = 3.6029 Eccentric anomaly (radians) = 3.47942 ----------------------------------------------------- D.3 Algorithm 3.2: solution of Kepler’s equation for the hyperbola usingNewton’s method Function file kepler_H.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ functio n F = kepler_H(e, M) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ %% This function uses Newton’s method to solve Kepler’s % equation for the hyperbola e*sinh(F )-F=M f o rt h e % hyperbolic eccentric anomaly, given the eccentricity and% the hyperbolic mean anomaly.%% F - hyperbolic eccentric anomaly (radians)% e - eccentricity, passed from the calling program% M - hyperbolic mean anomaly (radians), passed from the% calling program%% User M-functions required: none% ------------------------------------------------------------ %...Set an error tolerance: error = 1.e-8; %...Starting value for F: F=M ; D.3 Algorithm 3.2: solution of Kepler’s equation for the hyperbola 599 %...Iterate on Equation 3.42 until F is determined to within %...the error tolerance:ratio = 1;while abs(ratio) > error ratio = (e*sinh(F)-F-M)/(e*cosh(F) - 1); F=F-ratio; end % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Script file Example_3_05.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % Example_3_05 % ˜˜˜˜˜˜˜˜˜˜˜˜ %% This program uses Algorithm 3.2 and the data of% Example 3.5 to solve Kepler’s equation for the hyperbola.%% e - eccentricity% M - hyperbolic mean anomaly (dimensionless)% F - hyperbolic eccentric anomaly (dimensionless)%% User M-function required: kepler_H% ------------------------------------------------------------ clear%...Input data for Example 3.5: e = 2.7696;M = 40.69;%... %...Pass the input data to the function kepler_H, which returns F: F = kepler_H(e, M); %...Echo the input data and output to the command window: fprintf('---------------------------------------------------') fprintf('\n Example 3.5\n')fprintf('\n Eccentricity = %g',e) fprintf('\n Hyperbolic mean anomaly = %g\n',M)fprintf('\n Hyperbolic eccentric anomaly = %g',F)fprintf('\n-----------------------------------------------\n') % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Output from Example_3_05 ----------------------------------------------------- Example 3.5 Eccentricity = 2.7696 Hyperbolic mean anomaly = 40.69 Hyperbolic eccentric anomaly = 3.46309 ----------------------------------------------------- 600 Appendix D MATLAB algorithms D.4 Calculation of the Stumpff functions S(z)and C(z) The following scripts implement Equations 3.49 and 3.50 for use in other programs. Function file stumpS.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ functio n s = stumpS(z) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % % This function evaluates the Stumpff function S(z) according% to Equation 3.49.%% z - input argument% s - value of S(z)%% User M-functions required: none% ------------------------------------------------------------ i fz>0 s = (sqrt(z) - sin(sqrt(z)))/(sqrt(z))ˆ3; elsei fz<0 s = (sinh(sqrt(-z)) - sqrt(-z))/(sqrt(-z))ˆ3; else s = 1/6; end % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Function file stumpC.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ functio n c = stumpC(z) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ %% This function evaluates the Stumpff function C(z) according % to Equation 3.50.%% z - input argument% c - value of C(z)%% User M-functions required: none% ------------------------------------------------------------ i fz>0 c = (1 - cos(sqrt(z)))/z; elsei fz<0 c = (cosh(sqrt(-z)) - 1)/(-z); else c = 1/2; end % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ D.5 Algorithm 3.3: solution of the universal Kepler’s equation 601 D.5 Algorithm 3.3: solution of the universal Kepler’s equation using Newton’s method Function file kepler_U.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ function x = kepler_U(dt, ro, vro, a) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ %% This function uses Newton’s method to solve the universal% Kepler equation for the universal anomaly.% % mu - gravitational parameter (kmˆ3/sˆ2)% x - the universal anomaly (kmˆ0.5)% dt - time sincex=0( s )% ro - radial position (km) whe nx=0 % vro - radial velocity (km/s) whe nx=0 % a - reciprocal of the semimajor axis (1/km)% z - auxiliary variable (z = a*xˆ2)% C - value of Stumpff function C(z)% S - value of Stumpff function S(z)% n - number of iterations for convergence% nMax - maximum allowable number of iterations%% User M-functions required: stumpC, stumpS% ------------------------------------------------------------global mu %...Set an error tolerance and a limit on the number of % iterations:error = 1.e-8;nMax = 1000; %...Starting value for x: x = sqrt(mu)*abs(a)*dt; %...Iterate on Equation 3.62 until convergence occurs within %...the error tolerance: n= 0 ;ratio = 1;while abs(ratio) > erro r&n< =nMax n =n+1 ;C = stumpC(a*xˆ2);S = stumpS(a*xˆ2);F = ro*vro/sqrt(mu)*xˆ2*C + (1 - a*ro)*xˆ3*S + ro*x-... sqrt(mu)*dt; dFdx = ro*vro/sqrt(mu)*x*(1 - a*xˆ2*S)+... (1 - a*ro)*xˆ2*C+ro; ratio = F/dFdx; x = x - ratio; end %...Deliver a value for x, but report that nMax was reached: 602 Appendix D MATLAB algorithms if n > nMax fprintf('\n **No. iterations of Kepler''s equation')fprintf(' = %g', n)fprintf('\n F/dFdx = %g\n', F/dFdx) end % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Script file Example_3_06.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % Example_3_06 % ˜˜˜˜˜˜˜˜˜˜˜˜ %% This program uses Algorithm 3.3 and the data of Example 3.6% to solve the universal Kepler’s equation.%% mu - gravitational parameter (kmˆ3/sˆ2)% x - the universal anomaly (kmˆ0.5)% dt - time sinc ex=0( s ) % ro - radial position whe nx=0 (km) % vro - radial velocity whe nx=0 (km/s) % a - semimajor axis (km)%% User M-function required: kepler_U% ------------------------------------------------------------ clear global mumu = 398600; %...Input data for Example 3.6: ro = 10000;vro = 3.0752;dt = 3600;a = -19655; %... %...Pass the input data to the function kepler_U, which returns x %...(Universal Kepler’s requires the reciprocal of% semimajor axis):x = kepler_U(dt, ro, vro, 1/a); %...Echo the input data and output the results to the command window: fprintf('---------------------------------------------------')fprintf('\n Example 3.6\n')fprintf('\n Initial radial coordinate (km) = %g',ro)fprintf('\n Initial radial velocity (km/s) = %g',vro)fprintf('\n Elapsed time (seconds) = %g',dt) fprintf('\n Semimajor axis (km) = %g\n',a)fprintf('\n Universal anomaly (kmˆ0.5) = %g',x)fprintf('\n-----------------------------------------------\n') % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ D.6 Calculation of the Lagrange coefficients fandgand their time derivatives 603 Output from Example_3_06 ----------------------------------------------------- Example 3.6 Initial radial coordinate (km) = 10000 Initial radial velocity (km/s) = 3.0752Elapsed time (seconds) = 3600 Semimajor axis (km) = -19655 Universal anomaly (kmˆ0.5) = 128.511 ----------------------------------------------------- D.6 Calculation of the Lagrange coefficients fand gand their time derivatives The following scripts implement Equations 3.66 for use in other programs. Function file f_and_g.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ function [f, g] = f_and_g(x, t, ro, a) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ %% This function calculates the Lagrange f and g coefficients.%% mu - the gravitational parameter (kmˆ3/sˆ2)% a - reciprocal of the semimajor axis (1/km)% ro - the radial position at time t (km)% t - the time elapsed since t (s)% x - the universal anomaly after time t (kmˆ0.5)% f - the Lagrange f coefficient (dimensionless)% g - the Lagrange g coefficient (s)%% User M-functions required: stumpC, stumpS% ------------------------------------------------------------ global muz = a*xˆ2;%...Equation 3.66a: f=1-xˆ2/ro*stumpC(z); %...Equation 3.66b: g=t-1/sqrt(mu)*xˆ3*stumpS(z); % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Function file fDot_and_gDot.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ function [fdot, gdot] = fDot_and_gDot(x, r, ro, a) 604 Appendix D MATLAB algorithms % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ %% This function calculates the time derivatives of the% Lagrange f and g coefficients.%% mu - the gravitational parameter (kmˆ3/sˆ2)% a - reciprocal of the semimajor axis (1/km)% ro - the radial position at time t (km)% t - the time elapsed since initial state vector (s)% r - the radial position after time t (km)% x - the universal anomaly after time t (kmˆ0.5)% fDot - time derivative of the Lagrange f coefficient (1/s)% gDot - time derivative of the Lagrange g coefficient% (dimensionless)% % User M-functions required: stumpC, stumpS% ------------------------------------------------------------ global muz = a*xˆ2;%...Equation 3.66c: fdot = sqrt(mu)/r/ro*(z*stumpS(z) - 1)*x; %...Equation 3.66d: g d o t=1- xˆ2/r*stumpC(z); % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ D.7 Algorithm 3.4: calculation of the state vector ( r,v) given the initial state vector ( r0,v0) and the time lapse /Delta1t Function file rv_from_r0v0.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ function [R,V] = rv_from_r0v0(R0, V0, t) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % This function computes the state vector (R,V) from the% initial state vector (R0,V0) and the elapsed time.%% mu - gravitational parameter (kmˆ3/sˆ2)% R0 - initial position vector (km)% V0 - initial velocity vector (km/s)% t - elapsed time (s)% R - final position vector (km) % V - final velocity vector (km/s)%% User M-functions required: kepler_U, f_and_g, fDot_and_gDot% ------------------------------------------------------------ D.7 Algorithm 3.4: calculation of the state vector 605 global mu %...Magnitudes of R0 and V0: r0 = norm(R0); v0 = norm(V0); %...Initial radial velocity: vr0 = dot(R0, V0)/r0; %...Reciprocal of the semimajor axis (from the energy equation): alpha = 2/r0 - v0ˆ2/mu; %...Compute the universal anomaly: x = kepler_U(t, r0, vr0, alpha); %...Compute the f and g functions: [f, g] = f_and_g(x, t, r0, alpha); %...Compute the final position vector: R = f*R0 + g*V0; %...Compute the magnitude of R: r = norm(R); %...Compute the derivatives of f and g: [fdot, gdot] = fDot_and_gDot(x, r, r0, alpha); %...Compute the final velocity: V = fdot*R0 + gdot*V0; % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Script file Example_3_07.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % Example_3_07 % ˜˜˜˜˜˜˜˜˜˜˜˜ % % This program computes the state vector (R,V) from the% initial state vector (R0,V0) and the elapsed time using the% data in Example 3.7.%% mu - gravitational parameter (kmˆ3/sˆ2)% R0 - the initial position vector (km)% V0 - the initial velocity vector (km/s)% R - the final position vector (km)% V - the final velocity vector (km/s)% t - elapsed time (s)%% User M-functions required: rv_from_r0v0% ------------------------------------------------------------ clear global mumu = 398600; 606 Appendix D MATLAB algorithms %...Input data for Example 3.7: R0 = [ 7000 -12124 0];V0 = [2.6679 4.6210 0];t = 3600;%... %...Algorithm 3.4: [R V] = rv_from_r0v0(R0, V0, t); %...Echo the input data and output the results to the command window: fprintf('---------------------------------------------------')fprintf('\n Example 3.7\n')fprintf('\n Initial position vector (km):')fprintf('\n r0 = (%g, %g, %g)\n', R0(1), R0(2), R0(3))fprintf('\n Initial velocity vector (km/s):') fprintf('\n v0 = (%g, %g, %g)', V0(1), V0(2), V0(3))fprintf('\n\n Elapsed time = %g s\n',t)fprintf('\n Final position vector (km):')fprintf('\n r = (%g, %g, %g)\n', R(1), R(2), R(3))fprintf('\n Final velocity vector (km/s):')fprintf('\n v = (%g, %g, %g)', V(1), V(2), V(3))fprintf('\n-----------------------------------------------\n') % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Output from Example_3_07 ----------------------------------------------------- Example 3.7 Initial position vector (km): r0 = (7000, -12124, 0) Initial velocity vector (km/s): v0 = (2.6679, 4.621, 0) Elapsed time = 3600 sFinal position vector (km): r = (-3297.77, 7413.4, 0) Final velocity vector (km/s): v = (-8.2976, -0.964045, -0) ----------------------------------------------------- D.8 Algorithm 4.1: calculation of the orbital elements from the state vector Function file coe_from_sv.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ function coe = coe_from_sv(R,V) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % D.8 Algorithm 4.1: calculation of the orbital elements from the state vector 607 % This function computes the classical orbital elements (coe) % from the state vector (R,V) using Algorithm 4.1.%% mu - gravitational parameter (kmˆ3/sˆ2)% R - position vector in the geocentric equatorial frame% (km)% V - velocity vector in the geocentric equatorial frame% (km)% r, v - the magnitudes of R and V% vr - radial velocity component (km/s)% H - the angular momentum vector (kmˆ2/s)% h - the magnitude of H (kmˆ2/s)% incl - inclination of the orbit (rad)% N - the node line vector (kmˆ2/s)% n - the magnitude of N % cp - cross product of N and R% RA - right ascension of the ascending node (rad)% E - eccentricity vector% e - eccentricity (magnitude of E)% eps - a small number below which the eccentricity is% considered to be zero% w - argument of perigee (rad)% TA - true anomaly (rad)% a - semimajor axis (km)% pi - 3.1415926...% coe - vector of orbital elements [h e RA incl w TA a]%% User M-functions required: None% ------------------------------------------------------------ global mu; eps = 1.e-10; r = norm(R); v = norm(V); vr = dot(R,V)/r;H = cross(R,V); h = norm(H); %...Equation 4.7: incl = acos(H(3)/h); %...Equation 4.8: N = cross([0 0 1],H);n = norm(N); %...Equation 4.9: if n ∼=0 RA = acos(N(1)/n);if N(2) < 0 RA = 2*pi - RA; end else R A=0 ; end 608 Appendix D MATLAB algorithms %...Equation 4.10: E = 1/mu*((vˆ2 - mu/r)*R - r*vr*V);e = norm(E); %...Equation 4.12 (incorporating the cas ee=0 ) : if n ∼=0 if e > eps w = acos(dot(N,E)/n/e);if E(3) < 0 w=2 * p i-w ; end else w=0 ; end else w=0 ; end %...Equation 4.13a (incorporating the cas ee=0 ) : if e > eps TA = acos(dot(E,R)/e/r);if vr < 0 TA = 2*pi - TA; end else cp = cross(N,R);if cp(3) >= 0 TA = acos(dot(N,R)/n/r); else TA = 2*pi - acos(dot(N,R)/n/r); end end %...Equation 2.61 ( a < 0 for a hyperbola): a = hˆ2/mu/(1 - eˆ2); c o e=[ heR Ai n c lwT Aa ] ; % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Script file Example_4_03.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % Example_4_03 % ˜˜˜˜˜˜˜˜˜˜˜˜ %% This program uses Algorithm 4.1 to obtain the orbital% elements from the state vector provided in Example 4.3.%% pi - 3.1415926...% deg - factor for converting between degrees and radians% mu - gravitational parameter (kmˆ3/sˆ2) % r - position vector (km) in the geocentric equatorial% frame% v - velocity vector (km/s) in the geocentric equatorial% frame D.8 Algorithm 4.1: calculation of the orbital elements from the state vector 609 % coe - orbital elements [h e RA incl w TA a] % where h = angular momentum (kmˆ2/s)% e = eccentricity % RA = right ascension of the ascending node % (rad) % incl = orbit inclination (rad) % w = argument of perigee (rad) % TA = true anomaly (rad) % a = semimajor axis (km) % T - Period of an elliptic orbit (s)%% User M-function required: coe_from_sv% ------------------------------------------------------------ clear global mudeg = pi/180;mu = 398600; %...Input data: r = [ -6045 -3490 2500];v = [-3.457 6.618 2.533];%... %...Algorithm 4.1: coe = coe_from_sv(r,v); %...Echo the input data and output results to the command window: fprintf('---------------------------------------------------')fprintf('\n Example 4.3\n')fprintf('\n Gravitational parameter (kmˆ3/sˆ2) = %g\n', mu)fprintf('\n State vector:\n')fprintf('\n r (km) = [%g %g %g]', ... r(1), r(2), r(3)) fprintf('\n v (km/s) = [%g %g %g]', ... v(1), v(2), v(3)) disp(' ')fprintf('\n Angular momentum (kmˆ2/s) = %g', coe(1))fprintf('\n Eccentricity = %g', coe(2)) fprintf('\n Right ascension (deg) = %g', coe(3)/deg)fprintf('\n Inclination (deg) = %g', coe(4)/deg) fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg)fprintf('\n True anomaly (deg) = %g', coe(6)/deg) fprintf('\n Semimajor axis (km): = %g', coe(7)) %...if the orbit is an ellipse, output its period: if coe(2)<1 T = 2*pi/sqrt(mu)*coe(7)ˆ1.5; % Equation 2.73fprintf('\n Period:')fprintf('\n Seconds = %g', T) fprintf('\n Minutes = %g', T/60) fprintf('\n Hours = %g', T/3600) fprintf('\n Days = %g', T/24/3600) end fprintf('\n-----------------------------------------------\n') % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ 610 Appendix D MATLAB algorithms Output from Example_4_03 ----------------------------------------------------- Example 4.3 Gravitational parameter (kmˆ3/sˆ2) = 398600State vector:r (km) = [-6045 -3490 2500] v (km/s) = [-3.457 6.618 2.533] Angular momentum (kmˆ2/s) = 58311.7 Eccentricity = 0.171212 Right ascension (deg) = 255.279Inclination (deg) = 153.249Argument of perigee (deg) = 20.0683True anomaly (deg) = 28.4456Semimajor axis (km): = 8788.1Period: Seconds = 8198.86Minutes = 136.648Hours = 2.27746Days = 0.0948942 ----------------------------------------------------- D.9 Algorithm 4.2: calculation of the state vector from the orbital elements Function file sv_from_coe.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ function [r, v] = sv_from_coe(coe) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % This function computes the state vector (r,v) from the% classical orbital elements (coe). %% mu - gravitational parameter (kmˆ3; sˆ2)% coe - orbital elements [h e RA incl w TA]% where% h = angular momentum (kmˆ2/s)% e = eccentricity% RA = right ascension of the ascending node (rad)% incl = inclination of the orbit (rad)% w = argument of perigee (rad)% TA = true anomaly (rad)% R3_w - Rotation matrix about the z-axis through the angle w% R1_i - Rotation matrix about the x-axis through the angle i% R3_W - Rotation matrix about the z-axis through the angle RA % Q_pX - Matrix of the transformation from perifocal to% geocentric equatorial frame% rp - position vector in the perifocal frame (km)% vp - velocity vector in the perifocal frame (km/s) D.9 Algorithm 4.2: calculation of the state vector from the orbital elements 611 % r - position vector in the geocentric equatorial frame % (km)% v - velocity vector in the geocentric equatorial frame% (km/s)%% User M-functions required: none% ------------------------------------------------------------ global mu h = coe(1); e = coe(2);RA = coe(3);incl = coe(4); w = coe(5);TA = coe(6); %...Equations 4.37 and 4.38 (rp and vp are column vectors): rp = (hˆ2/mu) * (1/(1 + e*cos(TA))) * (cos(TA)*[1;0;0] ... + sin(TA)*[0;1;0]); vp = (mu/h) * (-sin(TA)*[1;0;0] + (e + cos(TA))*[0;1;0]); %...Equation 4.39: R3_W = [ cos(RA) sin(RA) 0 -sin(RA) cos(RA) 0 0 0 1]; %...Equation 4.40: R1_i = [1 0 0 0 cos(incl) sin(incl)0 -sin(incl) cos(incl)]; %...Equation 4.41: R3_w = [ cos(w) sin(w) 0 -sin(w) cos(w) 0 0 0 1]; %...Equation 4.44: Q_pX = R3_W'*R1_i'*R3_w'; %...Equations 4.46 (r and v are column vectors): r = Q_pX*rp;v = Q_pX*vp; %...Convert r and v into row vectors: r = r';v = v'; % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Script file Example_4_05.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % Example_4_05 612 Appendix D MATLAB algorithms % ˜˜˜˜˜˜˜˜˜˜˜˜ %% This program uses Algorithm 4.2 to obtain the state vector% from the orbital elements provided in Example 4.5.%% pi - 3.1415926...% deg - factor for converting between degrees and radians% mu - gravitational parameter (kmˆ3/sˆ2)% coe - orbital elements [h e RA incl w TA a]% where h = angular momentum (kmˆ2/s)% e = eccentricity% RA = right ascension of the ascending node% (rad)% incl = orbit inclination (rad)% w = argument of perigee (rad) % TA = true anomaly (rad)% a = semimajor axis (km)% r - position vector (km) in geocentric equatorial frame% v - velocity vector (km) in geocentric equatorial frame%% User M-functions required: sv_from_coe% ------------------------------------------------------------ clear global mudeg = pi/180;mu = 398600; %...Input data (angles in degrees): h = 80000;e = 1.4;RA = 40;incl = 30;w = 60;TA = 30;%... coe = [h, e, RA*deg, incl*deg, w*deg, TA*deg];%...Algorithm 4.2 (requires angular elements be in radians): [r, v] = sv_from_coe(coe); %...Echo the input data and output the results to the command window: fprintf('---------------------------------------------------')fprintf('\n Example 4.5\n')fprintf('\n Gravitational parameter (kmˆ3/sˆ2) = %g\n', mu)fprintf('\n Angular momentum (kmˆ2/s) = %g', h)fprintf('\n Eccentricity = %g', e)fprintf('\n Right ascension (deg) = %g', RA)fprintf('\n Argument of perigee (deg) = %g', w)fprintf('\n True anomaly (deg) = %g', TA)fprintf('\n\n State vector:')fprintf('\n r (km) = [%g %g %g]', r(1), r(2), r(3)) fprintf('\n v (km/s) = [%g %g %g]', v(1), v(2), v(3))fprintf('\n-----------------------------------------------\n') % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ D.10 Algorithm 5.1: Gibbs’ method of preliminary orbit determination 613 Output from Example_4_05 ----------------------------------------------------- Example 4.5 Gravitational parameter (kmˆ3/sˆ2) = 398600Angular momentum (kmˆ2/s) = 80000 Eccentricity = 1.4 Right ascension (deg) = 40 Argument of perigee (deg) = 60 True anomaly (deg) = 30 State vector: r (km) = [-4039.9 4814.56 3628.62]v (km/s) = [-10.386 -4.77192 1.74388] ----------------------------------------------------- D.10 Algorithm 5.1: Gibbs’ method of preliminary orbit determination Function file gibbs.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ function [V2, ierr] = gibbs(R1, R2, R3) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ %% This function uses the Gibbs method of orbit determination% to compute the velocity corresponding to the second of% three supplied position vectors.%% mu - gravitational parameter (kmˆ3/sˆ2) % R1, R2, R3 - three coplanar geocentric position vectors% (km) % r1, r2, r3 - the magnitudes of R1, R2 and R3 (km)% c12, c23, c31 - three independent cross products among % R1, R2 and R3 % N, D, S - vectors formed from R1, R2 and R3 during% the Gibbs’ procedure % tol - tolerance for determining if R1, R2 and R3 % are coplanar % ierr -=0i fR 1 ,R 2 ,R 3a r efound to be coplanar % = 1 otherwise % V2 - the velocity corresponding to R2 (km/s) %% User M-functions required: none% ----------------------------------------------------------- global mu tol = 1e-4;ierr = 0; 614 Appendix D MATLAB algorithms %...Magnitudes of R1, R2 and R3: r1 = norm(R1);r2 = norm(R2);r3 = norm(R3); %...Cross products among R1, R2 and R3: c12 = cross(R1,R2);c23 = cross(R2,R3);c31 = cross(R3,R1); %...Check that R1, R2 and R3 are coplanar; if not set error flag: if abs(dot(R1,c23)/r1/norm(c23)) > tol ierr = 1; end %...Equation 5.13: N = r1*c23 + r2*c31 + r3*c12; %...Equation 5.14: D = c12 + c23 + c31; %...Equation 5.21: S = R1*(r2 - r3) + R2*(r3 - r1) + R3*(r1 - r2); %...Equation 5.22: V2 = sqrt(mu/norm(N)/norm(D))*(cross(D,R2)/r2 + S); % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Script file Example_5_01.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % Example_5_01 % ˜˜˜˜˜˜˜˜˜˜˜˜ %% This program uses Algorithm 5.1 (Gibbs’ method) and% Algorithm 4.1 to obtain the orbital elements from the data % provided in Example 5.1.%% deg - factor for converting between degrees and% radians% pi - 3.1415926...% mu - gravitational parameter (kmˆ3/sˆ2)% r1, r2, r3 - three coplanar geocentric position vectors (km)% ierr - 0 if r1, r2, r3 are found to be coplanar% 1 otherwise% v2 - the velocity corresponding to r2 (km/s)% coe - orbital elements [h e RA incl w TA a]% where h = angular momentum (kmˆ2/s)% e = eccentricity % RA = right ascension of the ascending% node (rad)% incl = orbit inclination (rad)% w = argument of perigee (rad) D.10 Algorithm 5.1: Gibbs’ method of preliminary orbit determination 615 % TA = true anomaly (rad) % a = semimajor axis (km) % T - period of elliptic orbit (s) %% User M-functions required: gibbs, coe_from_sv% ------------------------------------------------------------ clear deg = pi/180;global mu %...Input data for Example 5.1: mu = 398600;r1 = [-294.32 4265.1 5986.7];r2 = [-1365.4 3637.6 6346.8]; r3 = [-2940.3 2473.7 6555.8];%... %...Echo the input data to the command window: fprintf('---------------------------------------------------')fprintf('\n Example 5.1: Gibbs Method\n')fprintf('\n\n Input data:\n')fprintf('\n Gravitational parameter (kmˆ3/sˆ2) = %g\n', mu)fprintf('\n r1 (km) = [%g %g %g]', r1(1), r1(2), r1(3))fprintf('\n r2 (km) = [%g %g %g]', r2(1), r2(2), r2(3))fprintf('\n r3 (km) = [%g %g %g]', r3(1), r3(2), r3(3))fprintf('\n\n');%...Algorithm 5.1:[v2, ierr] = gibbs(r1, r2, r3); %...If the vectors r1, r2, r3, are not coplanar, abort: if ierr == 1 fprintf('\n These vectors are not coplanar.\n\n')return end %...Algorithm 4.1 coe = coe_from_sv(r2,v2); h = coe(1); e = coe(2); RA = coe(3);incl = coe(4);w = coe(5);TA = coe(6);a = coe(7); %...Output the results to the command window: fprintf(' Solution:')fprintf('\n');fprintf('\n v2 (km/s) = [%g %g %g]', v2(1), v2(2), v2(3))fprintf('\n\n Orbital elements:');fprintf('\n Angular momentum (kmˆ2/s) = %g', h)fprintf('\n Eccentricity = %g', e) fprintf('\n Inclination (deg) = %g', incl/deg) fprintf('\n RA of ascending node (deg) = %g', RA/deg)fprintf('\n Argument of perigee (deg) = %g', w/deg)fprintf('\n True anomaly (deg) = %g', TA/deg) 616 Appendix D MATLAB algorithms fprintf('\n Semimajor axis (km) = %g', a) %...If the orbit is an ellipse, output the period:i fe<1 T = 2*pi/sqrt(mu)*coe(7)ˆ1.5;fprintf('\n Period (s) = %g', T) endfprintf('\n-----------------------------------------------\n') % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Output from Example_5_01 ----------------------------------------------------- Example 5.1: Gibbs Method Input data: Gravitational parameter (kmˆ3/sˆ2) = 398600 r1 (km) = [-294.32 4265.1 5986.7] r2 (km) = [-1365.4 3637.6 6346.8]r3 (km) = [-2940.3 2473.7 6555.8] Solution: v2 (km/s) = [-6.2176 -4.01237 1.59915]Orbital elements: Angular momentum (kmˆ2/s) = 56193Eccentricity = 0.100159Inclination (deg) = 60.001RA of ascending node (deg) = 40.0023Argument of perigee (deg) = 30.1093True anomaly (deg) = 49.8894Semimajor axis (km) = 8002.14Period (s) = 7123.94 ----------------------------------------------------- D.11 Algorithm 5.2: solution of Lambert’s problem Function file lambert.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ function [V1, V2] = lambert(R1, R2, t, string) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ %% This function solves Lambert’s problem.%% mu - gravitational parameter (kmˆ3/sˆ2) % R1, R2 - initial and final position vectors (km)% r1, r2 - magnitudes of R1 and R2% t - the time of flight from R1 to R2% (a constant) (s) D.11 Algorithm 5.2: solution of Lambert’s problem 617 % V1, V2 - initial and final velocity vectors (km/s) % c12 - cross product of R1 into R2% theta - angle between R1 and R2% string - 'pro' if the orbit is prograde% 'retro' if the orbit is retrograde % A - a constant given by Equation 5.35 % z - alpha*xˆ2, where alpha is the reciprocal of the % semimajor axis and x is the universal anomaly % y(z) - a function of z given by Equation 5.38% F(z,t) - a function of the variable z and constant t,% given by Equation 5.40 % dFdz(z) - the derivative of F(z,t), given by% Equation 5.43 % ratio - F/dFdz% tol - tolerance on precision of convergence % nmax - maximum number of iterations of Newton’s% procedure % f, g - Lagrange coefficients% gdot - time derivative of g% C(z), S(z) - Stumpff functions% dum - a dummy variable%% User M-functions required: stumpC and stumpS% ----------------------------------------------------------- global mu global r1 r2 A %...Magnitudes of R1 and R2: r1 = norm(R1);r2 = norm(R2); c12 = cross(R1, R2); theta = acos(dot(R1,R2)/r1/r2); %...Determine whether the orbit is prograde or retrograde: if strcmp(string, 'pro') if c12(3) <= 0 theta = 2*pi - theta; end elseif strcmp(string,'retro') if c12(3) >= 0 theta = 2*pi - theta; end else string = 'pro'fprintf('\n ** Prograde trajectory assumed.\n') end %...Equation 5.35: A = sin(theta)*sqrt(r1*r2/(1 - cos(theta))); %...Determine approximately where F(z,t) changes sign, and %...use that value of z as the starting value for Equation 5.45: z = -100;while F(z,t) < 0 z=z+0.1; end 618 Appendix D MATLAB algorithms %...Set an error tolerance and a limit on the number of iterations: tol = 1.e-8;nmax = 5000; %...Iterate on Equation 5.45 until z is determined to within %...the error tolerance:ratio = 1;n= 0 ;while (abs(ratio) > tol) & (n <= nmax) n =n+1 ;ratio = F(z,t)/dFdz(z);z = z - ratio; end %...Report if the maximum number of iterations is exceeded: if n >= nmax fprintf('\n\n **Number of iterations exceeds')fprintf(' %g \n\n ', nmax) end %...Equation 5.46a: f = 1 - y(z)/r1; %...Equation 5.46b: g = A*sqrt(y(z)/mu); %...Equation 5.46d: g d o t=1- y(z)/r2; %...Equation 5.28: V1 = 1/g*(R2 - f*R1); %...Equation 5.29: V2 = 1/g*(gdot*R2 - R1); return % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜% Subfunctions used in the main body: %...Equation 5.38: function dum = y(z) global r1 r2 Adum = r1 + r2 + A*(z*S(z) - 1)/sqrt(C(z)); return %...Equation 5.40: function dum = F(z,t) global mu Adum = (y(z)/C(z))ˆ1.5*S(z) + A*sqrt(y(z)) - sqrt(mu)*t; return %...Equation 5.43: function dum = dFdz(z) global A i fz= =0 dum = sqrt(2)/40*y(0)ˆ1.5 + A/8*(sqrt(y(0)) ... + A*sqrt(1/2/y(0))); D.11 Algorithm 5.2: solution of Lambert’s problem 619 else dum = (y(z)/C(z))ˆ1.5*(1/2/z*(C(z) - 3*S(z)/2/C(z)) ... + 3*S(z)ˆ2/4/C(z)) ...+ A/8*(3*S(z)/C(z)*sqrt(y(z)) ...+ A*sqrt(C(z)/y(z))); end return %...Stumpff functions: function dum = C(z) dum = stumpC(z); returnfunction dum = S(z) dum = stumpS(z); return % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Script file Example_5_02.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % Example_5_02 % ˜˜˜˜˜˜˜˜˜˜˜˜ %% This program uses Algorithm 5.2 to solve Lambert’s problem% for the data provided in Example 5.2.%% deg - factor for converting between degrees and radians% pi - 3.1415926...% mu - gravitational parameter (kmˆ3/sˆ2)% r1, r2 - initial and final position vectors (km)% dt - time between r1 and r2 (s)% string - = 'pro' if the orbit is prograde% = 'retro' if the orbit is retrograde % v1, v2 - initial and final velocity vectors (km/s)% coe - orbital elements [h e RA incl w TA a]% where h = angular momentum (kmˆ2/s) % e = eccentricity % RA = right ascension of the ascending node % (rad) % incl = orbit inclination (rad) % w = argument of perigee (rad) % TA = true anomaly (rad) % a = semimajor axis (km) % TA1 - Initial true anomaly% TA2 - Final true anomaly% T - period of an elliptic orbit%% User M-functions required: lambert, coe_from_sv% ----------------------------------------------------------- clear global mudeg = pi/180;mu = 398600; 620 Appendix D MATLAB algorithms %...Input data from Example 5.2: r1 = [ 5000 10000 2100];r2 = [-14600 2500 7000];dt = 3600;string = 'pro';%... %...Algorithm 5.2: [v1, v2] = lambert(r1, r2, dt, string); %...Algorithm 4.1 (using r1 and v1): coe = coe_from_sv(r1, v1);%...Save the initial true anomaly:TA1 = coe(6); %...Algorithm 4.1 (using r2 and v2): coe = coe_from_sv(r2, v2);%...Save the final true anomaly:TA2 = coe(6); %...Echo the input data and output the results to the command window: fprintf('---------------------------------------------------')fprintf('\n Example 5.2: Lambert''s Problem\n')fprintf('\n\n Input data:\n');fprintf('\n Gravitational parameter (kmˆ3/sˆ2) = %g\n', mu)fprintf('\n r1 (km) = [%g %g %g]', ... r1(1), r1(2), r1(3)) fprintf('\n r2 (km) = [%g %g %g]', ... r2(1), r2(2), r2(3)) fprintf('\n Elapsed time (s) = %g', dt);fprintf('\n\n Solution:\n') fprintf('\n v1 (km/s) = [%g %g %g]', ... v1(1), v1(2), v1(3)) fprintf('\n v2 (km/s) = [%g %g %g]', ... v2(1), v2(2), v2(3)) fprintf('\n\n Orbital elements:') fprintf('\n Angular momentum (kmˆ2/s) = %g', coe(1))fprintf('\n Eccentricity = %g', coe(2)) fprintf('\n Inclination (deg) = %g', coe(4)/deg)fprintf('\n RA of ascending node (deg) = %g', coe(3)/deg)fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg)fprintf('\n True anomaly initial (deg) = %g', TA1/deg)fprintf('\n True anomaly final (deg) = %g', TA2/deg)fprintf('\n Semimajor axis (km) = %g', coe(7))fprintf('\n Periapse radius (km) = %g', ... coe(1)ˆ2/mu/(1 + coe(2))) if coe(2)<1 T = 2*pi/sqrt(mu)*coe(7)ˆ1.5;fprintf('\n Period:')fprintf('\n Seconds = %g', T)fprintf('\n Minutes = %g', T/60) fprintf('\n Hours = %g', T/3600)fprintf('\n Days = %g', T/24/3600) endfprintf('\n-----------------------------------------------\n') % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ D.12 Calculation of Julian day number at 0 hr UT 621 Output from Example_5_02 ----------------------------------------------------- Example 5.2: Lambert’s Problem Input data: Gravitational parameter (kmˆ3/sˆ2) = 398600 r1 (km) = [5000 10000 2100] r2 (km) = [-14600 2500 7000] Elapsed time (s) = 3600 Solution: v1 (km/s) = [-5.99249 1.92536 3.24564] v2 (km/s) = [-3.31246 -4.19662 -0.385288] Orbital elements: Angular momentum (kmˆ2/s) = 80466.8Eccentricity = 0.433488 Inclination (deg) = 30.191 RA of ascending node (deg) = 44.6002Argument of perigee (deg) = 30.7062True anomaly initial (deg) = 350.83True anomaly final (deg) = 91.1223Semimajor axis (km) = 20002.9Periapse radius (km) = 11331.9Period: Seconds = 28154.7 Minutes = 469.245 Hours = 7.82075 Days = 0.325865 ----------------------------------------------------- D.12 Calculation of Julian day number at 0 hr UT The following script implements Equation 5.48 for use in other programs. Function file J0.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ function j0 = J0(year, month, day) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ %% This function computes the Julian day number at 0 UT for any % year between 1900 and 2100 using Equation 5.48.%% j0 - Julian day at 0 hr UT (Universal Time)% year - range: 1901 - 2099% month - range:1-1 2% day - range:1-3 1 622 Appendix D MATLAB algorithms % % User M-functions required: none% ------------------------------------------------------------ j0 = 367*year - fix(7*(year + fix((month + 9)/12))/4) ... + fix(275*month/9) + day + 1721013.5; % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Script file Example_5_04.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % Example_5_04 % ˜˜˜˜˜˜˜˜˜˜˜˜ %% This program computes J0 and the Julian day number using the% data in Example 5.4.%% year - range: 1901 - 2099% month - range :1-1 2 % day - range :1-3 1 % hour - range :0-2 3 (Universal Time) % minute - range :0-6 0 % second - range :0-6 0 % ut - universal time (hr)% j0 - Julian day number at 0 hr UT% jd - Julian day number at specified UT%% User M-function required: J0% ------------------------------------------------------------ clear%...Input data from Example 5.4: year = 2004;month = 5;day = 12; hour = 14; minute = 45;second = 30;%... ut = hour + minute/60 + second/3600; %...Equation 5.48: j0 = J0(year, month, day); %...Equation 5.47: jd = j0 + ut/24; %...Echo the input data and output the results to the command window: fprintf('---------------------------------------------------')fprintf('\n Example 5.4: Julian day calculation\n')fprintf('\n Input data:\n');fprintf('\n Year = %g', year) D.13 Algorithm 5.3: calculation of local sidereal time 623 fprintf('\n Month = %g', month) fprintf('\n Day = %g', day) fprintf('\n Hour = %g', hour) fprintf('\n Minute = %g', minute) fprintf('\n Second = %g\n', second) fprintf('\n Julian day number = %11.3f', jd); fprintf('\n-----------------------------------------------\n') % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Output from Example_5_04 ----------------------------------------------------- Example 5.4: Julian day calculation Input data: Year = 2004 Month = 5 Day = 12 Hour = 14 Minute = 45 Second = 30 Julian day number = 2453138.115 ----------------------------------------------------- D.13 Algorithm 5.3: calculation of local sidereal time Function file LST.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ function lst = LST(y, m, d, ut, EL) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜% % This function calculates the local sidereal time.%% lst - local sidereal time (degrees)% y - year% m - month%d -d a y% ut - Universal Time (hours)% EL - east longitude (degrees)% j0 - Julian day number at 0 hr UT% j - number of centuries since J2000% g0 - Greenwich sidereal time (degrees) at 0 hr UT% gst - Greenwich sidereal time (degrees) at the specified UT %% User M-function required: J0% ------------------------------------------------------------ %...Equation 5.48; 624 Appendix D MATLAB algorithms j0 = J0(y, m, d); %...Equation 5.49: j = (j0 - 2451545)/36525; %...Equation 5.50: g0 = 100.4606184 + 36000.77004*j + 0.000387933*jˆ2 ... - 2.583e-8*jˆ3; %...Reduce g0 so it lies in the rang e 0 - 360 degrees g0 = zeroTo360(g0); %...Equation 5.51: gst = g0 + 360.98564724*ut/24; %...Equation 5.52: lst = gst + EL; %...Reduce lst to the rang e 0 - 360 degrees: lst = lst - 360*fix(lst/360); return % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % Subfunction used in the main body:% ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ functio n y = zeroTo360(x) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ %% This subfunction reduces an angle to the range% 0 - 360 degrees.%% x - The angle (degrees) to be reduced% y - The reduced value%% ------------------------------------------------------------if (x >= 360) x=x- fix(x/360)*360; elseif (x < 0) x=x- (fix(x/360) - 1)*360; endy=x ;return % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Script file Example_5_06.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % Example_5_06 % ˜˜˜˜˜˜˜˜˜˜˜˜ %% This program uses Algorithm 5.3 to obtain the local sidereal D.13 Algorithm 5.3: calculation of local sidereal time 625 % time from the data provided in Example 5.6. %% lst - local sidereal time (degrees)% EL - east longitude of the site (west longitude is% negative): % degrees (0 - 360) % minutes (0 - 60) % seconds (0 - 60) % WL - west longitude% year - range: 1901 - 2099% month - range:1-1 2% day - range:1-3 1% ut - universal time% hour (0 - 23) % minute (0 - 60) % second (0 - 60) %% User M-function required: LST% ------------------------------------------------------------ clear%...Input data for Example 5.6:% East longitude: degrees = 139; minutes = 47;seconds = 0; % Date: year = 2004;month = 3;day = 3; % Universal time: hour = 4;minute = 30;second = 0; %... %...Convert negative (west) longitude to east longitude: if degrees < 0 degrees = degrees + 360; end %...Express the longitudes as decimal numbers: EL = degrees + minutes/60 + seconds/3600;WL = 360 - EL; %...Express universal time as a decimal number: ut = hour + minute/60 + second/3600; %...Algorithm 5.3: lst = LST(year, month, day, ut, EL); %...Echo the input data and output the results to the command window: fprintf('---------------------------------------------------')fprintf('\n Example 5.6: Local sidereal time calculation\n') 626 Appendix D MATLAB algorithms fprintf('\n Input data:\n'); fprintf('\n Year = %g', year)fprintf('\n Month = %g', month)fprintf('\n Day = %g', day)fprintf('\n UT (hr) = %g', ut)fprintf('\n West Longitude (deg) = %g', WL)fprintf('\n East Longitude (deg) = %g', EL)fprintf('\n\n'); fprintf(' Solution:') fprintf('\n'); fprintf('\n Local Sidereal Time (deg) = %g', lst)fprintf('\n Local Sidereal Time (hr) = %g', lst/15) fprintf('\n-----------------------------------------------\n') % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Output from Example_5_06 ----------------------------------------------------- Example 5.6: Local sidereal time calculation Input data: Year = 2004 Month = 3Day = 3UT (hr) = 4.5West Longitude (deg) = 220.217East Longitude (deg) = 139.783 Solution: Local Sidereal Time (deg) = 8.57688 Local Sidereal Time (hr) = 0.571792 ----------------------------------------------------- D.14 Algorithm 5.4: calculation of the state vector from measurements of range,angular position and their rates Function file rv_from_observe.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ function [r,v] = rv_from_observe(rho, rhodot, A, Adot, a,... adot, theta, phi, H) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ %% This function calculates the geocentric equatorial position% and velocity vectors of an object from radar observations of% range, azimuth, elevation angle and their rates. %% deg - conversion factor between degrees and radians% pi - 3.1415926...% D.14 Algorithm 5.4: calculation of the state vector 627 % Re - equatorial radius of the earth (km) % f - earth’s flattening factor% wE - angular velocity of the earth (rad/s)% omega - earth’s angular velocity vector (rad/s) in the% geocentric equatorial frame %% theta - local sidereal time (degrees) of tracking site% phi - geodetic latitude (degrees) of site% H - elevation of site (km)% R - geocentric equatorial position vector (km) of% tracking site % Rdot - inertial velocity (km/s) of site% rho - slant range of object (km)% rhodot - range rate (km/s)% A - azimuth (degrees) of object relative to observation % site % Adot - time rate of change of azimuth (degrees/s)% a - elevation angle (degrees) of object relative to% observation site % adot - time rate of change of elevation angle (degrees/s)% dec - topocentric equatorial declination of object (rad)% decdot - declination rate (rad/s)% h - hour angle of object (rad)% RA - topocentric equatorial right ascension of object% (rad) % RAdot - right ascension rate (rad/s)%% Rho - unit vector from site to object% Rhodot - time rate of change of Rho (1/s)% r - geocentric equatorial position vector of object (km)% v - geocentric equatorial velocity vector of object (km)%% User M-functions required: none% ------------------------------------------------------------ global f Re wE deg = pi/180;omega = [0 0 wE]; %...Convert angular quantities from degrees to radians: A = A *deg;Adot = Adot *deg;a = a *deg;adot = adot *deg;theta = theta*deg;phi = phi *deg; %...Equation 5.56: R = [(Re/sqrt(1-(2*f - f*f)*sin(phi)ˆ2) + H) ... *cos(phi)*cos(theta), ... (Re/sqrt(1-(2*f - f*f)*sin(phi)ˆ2) + H) ...*cos(phi)*sin(theta), ...(Re*(1 - f)ˆ2/sqrt(1-(2*f - f*f) ... *sin(phi)ˆ2) + H)*sin(phi)]; %...Equation 5.66: Rdot = cross(omega, R); 628 Appendix D MATLAB algorithms %...Equation 5.83a: dec = asin(cos(phi)*cos(A)*cos(a) + sin(phi)*sin(a)); %...Equation 5.83b: h = acos((cos(phi)*sin(a) - sin(phi)*cos(A)*cos(a))/cos(dec));i f( A>0 )&( A<p i ) h=2 * p i-h ; end %...Equation 5.83c: RA = theta - h; %...Equations 5.57: Rho = [cos(RA)*cos(dec) sin(RA)*cos(dec) sin(dec)]; %...Equation 5.63: r = R + rho*Rho; %...Equation 5.84: decdot = (-Adot*cos(phi)*sin(A)*cos(a) ... + adot*(sin(phi)*cos(a) ...- cos(phi)*cos(A)*sin(a)))/cos(dec); %...Equation 5.85: RAdot = wE ... + (Adot*cos(A)*cos(a) - adot*sin(A)*sin(a) ...+ decdot*sin(A)*cos(a)*tan(dec)) ... /(cos(phi)*sin(a) - sin(phi)*cos(A)*cos(a)); %...Equations 5.69 and 5.72: Rhodot = [-RAdot*sin(RA)*cos(dec) - decdot*cos(RA)*sin(dec),... RAdot*cos(RA)*cos(dec) - decdot*sin(RA)*sin(dec),...decdot*cos(dec)]; %...Equation 5.64: v = Rdot + rhodot*Rho + rho*Rhodot; % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Script file Example_5_10.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % Example_5_10 % ˜˜˜˜˜˜˜˜˜˜˜˜ %% This program uses Algorithms 5.4 and 4.1 to obtain the% orbital elements from the observational data provided in% Example 5.10.%% deg - conversion factor between degrees and radians% pi - 3.1415926... % mu - gravitational parameter (kmˆ3/sˆ2)%% Re - equatorial radius of the earth (km)% f - earth’s flattening factor% wE - angular velocity of the earth (rad/s) D.14 Algorithm 5.4: calculation of the state vector 629 % omega - earth’s angular velocity vector (rad/s) in the % geocentric equatorial frame %% rho - slant range of object (km)% rhodot - range rate (km/s)% A - azimuth (deg) of object relative to observation% site % Adot - time rate of change of azimuth (deg/s)% a - elevation angle (deg) of object relative to% observation site % adot - time rate of change of elevation angle% (degrees/s) %% theta - local sidereal time (deg) of tracking site% phi - geodetic latitude (deg) of site % H - elevation of site (km)%% r - geocentric equatorial position vector of object (km)% v - geocentric equatorial velocity vector of object (km)%% coe - orbital elements [h e RA incl w TA a]% where h = angular momentum (kmˆ2/s) % e = eccentricity % RA = right ascension of the ascending node % (rad) % incl = inclination of the orbit (rad) % w = argument of perigee (rad) % TA = true anomaly (rad) % a = semimajor axis (km) % rp - perigee radius (km)% T - period of elliptical orbit (s)%% User M-functions required: rv_from_observe, coe_from_sv% ------------------------------------------------------------ clear global f Re wE mu deg = pi/180; f = 1/298.256421867; Re = 6378.13655;wE = 7.292115e-5;mu = 398600.4418; %...Input data for Example 5.10: rho = 2551;rhodot = 0;A = 90;Adot = 0.1130;a = 30;adot = 0.05651;theta = 300;phi = 60; H= 0 ;%...%...Algorithm 5.4:[r,v] = rv_from_observe(rho, rhodot, A, Adot, a, adot, theta, ... phi, H); 630 Appendix D MATLAB algorithms %...Algorithm 4.1: coe = coe_from_sv(r,v); h = coe(1); e = coe(2);RA = coe(3);incl = coe(4);w = coe(5);TA = coe(6);a = coe(7); %...Equation 2.40 rp = hˆ2/mu/(1 + e); %...Echo the input data and output the solution to % the command window:fprintf('---------------------------------------------------')fprintf('\n Example 5.10')fprintf('\n\n Input data:\n')fprintf('\n Slant range (km) = %g', rho)fprintf('\n Slant range rate (km/s) = %g', rhodot)fprintf('\n Azimuth (deg) = %g', A)fprintf('\n Azimuth rate (deg/s) = %g', Adot)fprintf('\n Elevation (deg) = %g', a)fprintf('\n Elevation rate (deg/s) = %g', adot)fprintf('\n Local sidereal time (deg) = %g', theta)fprintf('\n Latitude (deg) = %g', phi)fprintf('\n Altitude above sea level (km) = %g', H)fprintf('\n\n') fprintf(' Solution:') fprintf('\n\n State vector:\n') fprintf('\n r (km) = [%g, %g, %g]', ... r(1), r(2), r(3)) fprintf('\n v (km/s) = [%g, %g, %g]', ... v(1), v(2), v(3)) fprintf('\n\n Orbital elements:\n') fprintf('\n Angular momentum (kmˆ2/s) = %g', h)fprintf('\n Eccentricity = %g', e) fprintf('\n Inclination (deg) = %g', incl/deg)fprintf('\n RA of ascending node (deg) = %g', RA/deg)fprintf('\n Argument of perigee (deg) = %g', w/deg)fprintf('\n True anomaly (deg) = %g\n', TA/deg)fprintf('\n Semimajor axis (km) = %g', a)fprintf('\n Perigee radius (km) = %g', rp)%...If the orbit is an ellipse, output its period:i fe<1 T = 2*pi/sqrt(mu)*aˆ1.5;fprintf('\n Period:')fprintf('\n Seconds = %g', T)fprintf('\n Minutes = %g', T/60)fprintf('\n Hours = %g', T/3600) fprintf('\n Days = %g', T/24/3600) endfprintf('\n-----------------------------------------------\n') % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ D.15 Algorithms 5.5 and 5.6: Gauss method with iterative improvement 631 Output from Example_5_10 ----------------------------------------------------- Example 5.10 Input data:Slant range (km) = 2551 Slant range rate (km/s) = 0Azimuth (deg) = 90 Azimuth rate (deg/s) = 0.113 Elevation (deg) = 5168.62 Elevation rate (deg/s) = 0.05651Local sidereal time (deg) = 300Latitude (deg) = 60 Altitude above sea level (km) = 0 Solution: State vector: r (km) = [3830.68, -2216.47, 6605.09] v (km/s) = [1.50357, -4.56099, -0.291536] Orbital elements: Angular momentum (kmˆ2/s) = 35621.4 Eccentricity = 0.619758 Inclination (deg) = 113.386 RA of ascending node (deg) = 109.75Argument of perigee (deg) = 309.81True anomaly (deg) = 165.352 Semimajor axis (km) = 5168.62 Perigee radius (km) = 1965.32 Period: Seconds = 3698.05 Minutes = 61.6342 Hours = 1.02724 Days = 0.0428015 ----------------------------------------------------- D.15 Algorithms 5.5 and 5.6: Gauss’s method of preliminary orbit determination withiterative improvement Function file gauss.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ function [r, v, r_old, v_old] = ... gauss(Rho1, Rho2, Rho3, R1, R2, R3, t1, t2, t3) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % This function uses the Gauss method with iterative% improvement (Algorithms 5.5 and 5.6) to calculate the state 632 Appendix D MATLAB algorithms % vector of an orbiting body from angles-only observations at % three closely-spaced times.%% mu - the gravitational parameter (kmˆ3/sˆ2)% t1, t2, t3 - the times of the observations (s)% tau, tau1, tau3 - time intervals between observations (s)% R1, R2, R3 - the observation site position vectors% at t1, t2, t3 (km)% Rho1, Rho2, Rho3 - the direction cosine vectors of the% satellite at t1, t2, t3% p1, p2, p3 - cross products among the three direction% cosine vectors% Do - scalar triple product of Rho1, Rho2 and% Rho3% D - Matrix of the nine scalar triple products % of R1, R2 and R3 with p1, p2 and p3% E - dot product of R2 and Rho2% A, B - constants in the expression relating% slant range to geocentric radius% a,b,c - coefficients of the 8th order polynomial% in the estimated geocentric radius x% x - positive root of the 8th order polynomial% rho1, rho2, rho3 - the slant ranges at t1, t2, t3% r1, r2, r3 - the position vectors at t1, t2, t3 (km)% r_old, v_old - the estimated state vector at the end of% Algorithm 5.5 (km, km/s)% rho1_old,% rho2_old, and% rho3_old - the values of the slant ranges at t1, t2,% t3 at the beginning of iterative% improvement (Algorithm 5.6) (km)% diff1, diff2,% and diff3 - the magnitudes of the differences between% the old and new slant ranges at the end% of each iteration% tol - the error tolerance determining% convergence% n - number of passes through the% iterative improvement loop% nmax - limit on the number of iterations % ro, vo - magnitude of the position and% velocity vectors (km, km/s)% vro - radial velocity component (km)% a - reciprocal of the semimajor axis (1/km)% v2 - computed velocity at time t2 (km/s)% r, v - the state vector at the end of% Algorithm 5.6 (km, km/s)%% User M-functions required: kepler_U, f_and_g% User subfunctions required: posroot% ------------------------------------------------------------ global mu%...Equations 5.98: tau1 = t1 - t2;tau3 = t3 - t2; D.15 Algorithms 5.5 and 5.6: Gauss method with iterative improvement 633 %...Equation 5.101: tau = tau3 - tau1; %...Independent cross products among the direction cosine vectors: p1 = cross(Rho2,Rho3);p2 = cross(Rho1,Rho3);p3 = cross(Rho1,Rho2); %...Equation 5.108: Do = dot(Rho1,p1); %...Equations 5.109b, 5.110b and 5.111b: D = [[dot(R1,p1) dot(R1,p2) dot(R1,p3)] [dot(R2,p1) dot(R2,p2) dot(R2,p3)] [dot(R3,p1) dot(R3,p2) dot(R3,p3)]]; %...Equation 5.115b: E = dot(R2,Rho2); %...Equations 5.112b and 5.112c: A = 1/Do*(-D(1,2)*tau3/tau + D(2,2) + D(3,2)*tau1/tau);B = 1/6/Do*(D(1,2)*(tau3ˆ2 - tauˆ2)*tau3/tau ... + D(3,2)*(tauˆ2 - tau1ˆ2)*tau1/tau); %...Equations 5.117: a = -(Aˆ2 + 2*A*E + norm(R2)ˆ2);b = -2*mu*B*(A + E);c = -(mu*B)ˆ2; %...Calculate the roots of Equation 5.116 using MATLAB’s % polynomial ‘roots’ solver:Roots = roots([10a00b00c]); %...Find the positive real root: x = posroot(Roots); %...Equations 5.99a and 5.99b: f1 = 1 - 1/2*mu*tau1ˆ2/xˆ3;f3 = 1 - 1/2*mu*tau3ˆ2/xˆ3; %...Equations 5.100a and 5.100b: g1 = tau1 - 1/6*mu*(tau1/x)ˆ3;g3 = tau3 - 1/6*mu*(tau3/x)ˆ3; %...Equation 5.112a: r h o 2=A+mu*B/xˆ3; %...Equation 5.113: rho1 = 1/Do*((6*(D(3,1)*tau1/tau3 + D(2,1)*tau/tau3)*xˆ3 ... + mu*D(3,1)*(tauˆ2 - tau1ˆ2)*tau1/tau3) .../(6*xˆ3 + mu*(tauˆ2 - tau3ˆ2)) - D(1,1)); %...Equation 5.114: rho3 = 1/Do*((6*(D(1,3)*tau3/tau1 - D(2,3)*tau/tau1)*xˆ3 ... + mu*D(1,3)*(tauˆ2 - tau3ˆ2)*tau3/tau1) .../(6*xˆ3 + mu*(tauˆ2 - tau3ˆ2)) - D(3,3)); 634 Appendix D MATLAB algorithms %...Equations 5.86: r1 = R1 + rho1*Rho1;r2 = R2 + rho2*Rho2;r3 = R3 + rho3*Rho3; %...Equation 5.118: v2 = (-f3*r1 + f1*r3)/(f1*g3 - f3*g1); %...Save the initial estimates of r2 and v2: r_old = r2;v_old = v2; %...End of Algorithm 5.5 %...Use Algorithm 5.6 to improve the accuracy of the initial estimates. %...Initialize the iterative improvement loop and set error tolerance: rho1_old = rho1; rho2_old = rho2; rho3_old = rho3;diff1 = 1; diff2 = 1; diff3 = 1;n= 0 ;nmax = 1000;tol = 1.e-8; %...Iterative improvement loop: while ((diff1 > tol) & (diff2 > tol) & (diff3 > tol)) ... & (n < nmax) n = n+1; %...Compute quantities required by universal kepler’s equation: ro = norm(r2);vo = norm(v2);vro = dot(v2,r2)/ro;a = 2/ro - voˆ2/mu; %...Solve universal Kepler’s equation at times tau1 and tau3 % for universal anomalies x1 and x3: x1 = kepler_U(tau1, ro, vro, a);x3 = kepler_U(tau3, ro, vro, a); %...Calculate the Lagrange f and g coefficients at times tau1 and tau3: [ff1, gg1] = f_and_g(x1, tau1, ro, a); [ff3, gg3] = f_and_g(x3, tau3, ro, a); %...Update the f and g functions at times tau1 and tau3 by % averaging old and new: f1 = (f1 + ff1)/2;f3 = (f3 + ff3)/2;g1 = (g1 + gg1)/2;g3 = (g3 + gg3)/2; %...Equations 5.96 and 5.97: c1 = g3/(f1*g3 - f3*g1);c3 = -g1/(f1*g3 - f3*g1); %...Equations 5.109a, 5.110a and 5.111a: rho1 = 1/Do*( -D(1,1) + 1/c1*D(2,1) - c3/c1*D(3,1));rho2 = 1/Do*( -c1*D(1,2) + D(2,2) - c3*D(3,2));rho3 = 1/Do*(-c1/c3*D(1,3) + 1/c3*D(2,3) - D(3,3)); D.15 Algorithms 5.5 and 5.6: Gauss method with iterative improvement 635 %...Equations 5.86: r1 = R1 + rho1*Rho1;r2 = R2 + rho2*Rho2;r3 = R3 + rho3*Rho3; %...Equation 5.118: v2 = (-f3*r1 + f1*r3)/(f1*g3 - f3*g1); %...Calculate differences upon which to base convergence: diff1 = abs(rho1 - rho1_old);diff2 = abs(rho2 - rho2_old);diff3 = abs(rho3 - rho3_old); %...Update the slant ranges: rho1_old = rho1; rho2_old = rho2; rho3_old = rho3; end%...End iterative improvement loop fprintf('\n( **Number of Gauss improvement iterations') fprintf(' = %g)\n\n', n) if n >= nmax fprintf('\n\n **Number of iterations exceeds %g \n\n ', nmax); end %...Return the state vector for the central observation: r = r2;v = v2; return % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % Subfunction used in the main body: % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ function x = posroot(Roots) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ %% This subfunction extracts the positive real roots from% those obtained in the call to MATLAB's 'roots' function. % If there is more than one positive root, the user is% prompted to select the one to use.%% x - the determined or selected positive root % Roots - the vector of roots of a polynomial% posroots - vector of positive roots%% User M-functions required: none% ------------------------------------------------------------ %...Construct the vector of positive real roots: posroots = Roots(find(Roots>0 & ˜imag(Roots))); npositive = length(posroots); %...Exit if no positive roots exist: if npositive == 0 636 Appendix D MATLAB algorithms fprintf('\n\n ** There are no positive roots. \n\n') return end %...If there is more than one positive root, output the %...roots to the command window and prompt the user to%...select which one to use:if npositive == 1 x = posroots; else fprintf('\n\n ** There are two or more positive roots.\n')for i = 1:npositive fprintf('\n root #%g = %g', i, posroots(i)) end fprintf('\n\n Make a choice:\n')nchoice = 0;while nchoic e<1| nchoice > npositive nchoice = input(' Use root #? '); endx = posroots(nchoice);fprintf('\n We will use %g .\n', x) end return % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Script file Example_5_11.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % Example_5_11 % ˜˜˜˜˜˜˜˜˜˜˜˜ %% This program uses Algorithms 5.5 and 5.6 (Gauss’s method) to% compute the state vector from the data provided in% Example 5.11.%% deg - factor for converting between degrees and % radians% pi - 3.1415926...% mu - gravitational parameter (kmˆ3/sˆ2)% Re - earth’s radius (km)% f - earth’s flattening factor% H - elevation of observation site (km)% phi - latitude of site (deg)% t - vector of observation times t1, t2, t3 (s)% ra - vector of topocentric equatorial right% ascensions at t1, t2, t3 (deg)% dec - vector of topocentric equatorial right% declinations at t1, t2, t3 (deg)% theta - vector of local sidereal times for t1, t2, t3 % (deg)% R - matrix of site position vectors at t1, t2, t3% (km)% rho - matrix of direction cosine vectors at t1, D.15 Algorithms 5.5 and 5.6: Gauss method with iterative improvement 637 % t2, t3 % fac1, fac2 - common factors% r_old, v_old - the state vector without iterative improvement% (km, km/s) % r, v - the state vector with iterative improvement % (km, km/s) % coe - vector of orbital elements for r, v: % [h, e, RA, incl, w, TA, a] % where h = angular momentum (kmˆ2/s) % e = eccentricity % incl = inclination (rad) % w = argument of perigee (rad) % TA = true anomaly (rad) % a = semimajor axis (km) % coe_old - vector of orbital elements for r_old, v_old %% User M-functions required: gauss, coe_from_sv% ------------------------------------------------------------ clear global mu deg = pi/180; mu = 398600;Re = 6378;f = 1/298.26; %...Input data: H= 1 ;phi = 40*deg;t = [ 0 118.104 237.577];ra = [ 43.5365 54.4196 64.3178]*deg;dec = [-8.78334 -12.0739 -15.1054]*deg;theta = [ 44.5065 45.000 45.4992]*deg;%... %...Equations 5.56 and 5.57: fac1 = Re/sqrt(1-(2*f - f*f)*sin(phi)ˆ2);fac2 = (Re*(1-f)ˆ2/sqrt(1-(2*f - f*f)*sin(phi)ˆ2) + H) ... *sin(phi); for i = 1:3 R(i,1) = (fac1 + H)*cos(phi)*cos(theta(i));R(i,2) = (fac1 + H)*cos(phi)*sin(theta(i));R(i,3) = fac2;rho(i,1) = cos(dec(i))*cos(ra(i));rho(i,2) = cos(dec(i))*sin(ra(i));rho(i,3) = sin(dec(i)); end %...Algorithms 5.5 and 5.6: [r, v, r_old, v_old] = gauss(rho(1,:), rho(2,:), rho(3,:), ... R(1,:), R(2,:), R(3,:), ... t(1), t(2), t(3)); %...Algorithm 4.1 for the initial estimate of the state vector % and for the iteratively improved one: 638 Appendix D MATLAB algorithms coe_old = coe_from_sv(r_old,v_old); coe = coe_from_sv(r,v); %...Echo the input data and output the solution to % the command window:fprintf('---------------------------------------------------')fprintf('\n Example 5.11: Orbit determination by the Gauss method\n') fprintf('\n Radius of earth (km) = %g', Re)fprintf('\n Flattening factor = %g', f)fprintf('\n Gravitational parameter (kmˆ3/sˆ2) = %g', mu)fprintf('\n\n Input data:\n');fprintf('\n Latitude (deg) = %g', phi/deg);fprintf('\n Altitude above sea level (km) = %g', H);fprintf('\n\n Observations:') fprintf('\n Time (s) Right ascension (deg) Declination (deg)') fprintf(' Local sidereal time (deg)')for i = 1:3 fprintf('\n %9.4g %17.4f %19.4f %23.4f', ... t(i), ra(i)/deg, dec(i)/deg, theta(i)/deg) end fprintf('\n\n Solution:\n')fprintf('\n Without iterative improvement...\n') fprintf('\n');fprintf('\n r (km) = [%g, %g, %g]', ... r_old(1), r_old(2), r_old(3)) fprintf('\n v (km/s) = [%g, %g, %g]', ... v_old(1), v_old(2), v_old(3)) fprintf('\n'); fprintf('\n Angular momentum (kmˆ2/s) = %g', coe_old(1)) fprintf('\n Eccentricity = %g', coe_old(2))fprintf('\n RA of ascending node (deg) = %g', coe_old(3)/deg)fprintf('\n Inclination (deg) = %g', coe_old(4)/deg)fprintf('\n Argument of perigee (deg) = %g', coe_old(5)/deg)fprintf('\n True anomaly (deg) = %g', coe_old(6)/deg)fprintf('\n Semimajor axis (km) = %g', coe_old(7))fprintf('\n Periapse radius (km) = %g', coe_old(1)ˆ2 ... /mu/(1 + coe_old(2))) %...If the orbit is an ellipse, output the period:if coe_old(2)<1 T = 2*pi/sqrt(mu)*coe_old(7)ˆ1.5;fprintf('\n Period:')fprintf('\n Seconds = %g', T)fprintf('\n Minutes = %g', T/60)fprintf('\n Hours = %g', T/3600)fprintf('\n Days = %g', T/24/3600) end fprintf('\n\n With iterative improvement...\n') fprintf('\n');fprintf('\n r (km) = [%g, %g, %g]', ... r(1), r(2), r(3)) fprintf('\n v (km/s) = [%g, %g, %g]', ... v(1), v(2), v(3)) fprintf('\n'); D.15 Algorithms 5.5 and 5.6: Gauss method with iterative improvement 639 fprintf('\n Angular momentum (kmˆ2/s) = %g', coe(1)) fprintf('\n Eccentricity = %g', coe(2)) fprintf('\n RA of ascending node (deg) = %g', coe(3)/deg)fprintf('\n Inclination (deg) = %g', coe(4)/deg) fprintf('\n Argument of perigee (deg) = %g', coe(5)/deg)fprintf('\n True anomaly (deg) = %g', coe(6)/deg) fprintf('\n Semimajor axis (km) = %g', coe(7)) fprintf('\n Periapse radius (km) = %g', coe(1)ˆ2 ... /mu/(1 + coe(2))) %...If the orbit is an ellipse, output the period:if coe(2)<1 T = 2*pi/sqrt(mu)*coe(7)ˆ1.5;fprintf('\n Period:')fprintf('\n Seconds = %g', T) fprintf('\n Minutes = %g', T/60) fprintf('\n Hours = %g', T/3600) fprintf('\n Days = %g', T/24/3600) endfprintf('\n-----------------------------------------------\n') % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Output from Example_5_11 ( **Number of Gauss improvement iterations = 14) ----------------------------------------------------- Example 5.11: Orbit determination by the Gauss method Radius of earth (km) = 6378 Flattening factor = 0.00335278 Gravitational parameter (kmˆ3/sˆ2) = 398600 Input data:Latitude (deg) = 40 Altitude above sea level (km) = 1 Observations: Right Local Time (s) Ascension (deg) Declination (deg) Sidereal time (deg) 0 43.5365 -8.7833 44.5065 118.1 54.4196 -12.0739 45.0000 237.6 64.3178 -15.1054 45.4992 Solution:Without iterative improvement... r (km) = [5659.03, 6533.74, 3270.15] v (km/s) = [-3.90774, 5.05735, -2.22224] Angular momentum (kmˆ2/s) = 62426.4 Eccentricity = 0.084887 RA of ascending node (deg) = 270.375Inclination (deg) = 29.8362 Argument of perigee (deg) = 87.6835True anomaly (deg) = 46.9821 Semimajor axis (km) = 9847.83 Periapse radius (km) = 9011.88 640 Appendix D MATLAB algorithms Period: Seconds = 9725.73Minutes = 162.095Hours = 2.70159Days = 0.112566 With iterative improvement... r (km) = [5662.04, 6537.95, 3269.05]v (km/s) = [-3.88542, 5.12141, -2.2434] Angular momentum (kmˆ2/s) = 62816.7 Eccentricity = 0.0999909RA of ascending node (deg) = 269.999Inclination (deg) = 30.001 Argument of perigee (deg) = 89.9723True anomaly (deg) = 45.0284Semimajor axis (km) = 9999.48Periapse radius (km) = 8999.62Period: Seconds = 9951.24Minutes = 165.854Hours = 2.76423Days = 0.115176 ----------------------------------------------------- D.16 Converting the numerical designation of a month or a planet into its name The following simple script can be used in programs that input the numerical values for a month and/or a planet. Function file month_planet_names.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ function [month, planet] = month_planet_names(month_id, planet_id) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ %% This function returns the name of the month and the planet% corresponding, respectively, to the numbers ‘‘month_id’’ and% ‘‘planet_id’’.%% month - name of the month% planet - name of the planet% months - a vector containing the names of the 12 months% planets - a vector containing the names of the 9 planets% month_id - the month number (1 - 12) % planet_id - the planet number (1 - 9)%% User M-functions required: none% ------------------------------------------------------------ D.17 Algorithm 8.1: calculation of the state vector of a planet at a given epoch 641 months = ['January ' 'February ''March ''April ''May ''June ''July ''August ''September''October ''November ''December ']; planets = ['Mercury' 'Venus ' 'Earth ''Mars ''Jupiter''Saturn ''Uranus ''Neptune''Pluto ']; month = months (month_id, 1:9); planet = planets(planet_id, 1:7); % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ D.17 Algorithm 8.1: calculation of the state vector of a planet at a given epoch Function file planet_elements_and_sv.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ function [coe, r, v, jd] = planet_elements_and_sv ... (planet_id, year, month, day, hour, minute, second) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % % This function calculates the orbital elements and the state% vector of a planet from the date (year, month, day)% and universal time (hour, minute, second).%% mu - gravitational parameter of the sun (kmˆ3/sˆ2)% deg - conversion factor between degrees and radians% pi - 3.1415926...%% coe - vector of heliocentric orbital elements% [h e RA incl w TA a w_ha tLME ] , % where % h = angular momentum (kmˆ2/s) % e = eccentricity % RA = right ascension (deg) % incl = inclination (deg) % w = argument of perihelion (deg) % TA = true anomaly (deg) 642 Appendix D MATLAB algorithms % a = semimajor axis (km) % w_hat = longitude of perihelion% ( = RA + w) (deg)% L = mean longitud e ( = w_hat + M) (deg) % M = mean anomaly (deg)% E = eccentric anomaly (deg)%% planet_id - planet identifier:% 1 = Mercury% 2 = Venus% 3 = Earth% 4 = Mars% 5 = Jupiter% 6 = Saturn% 7 = Uranus % 8 = Neptune% 9 = Pluto%% year - range: 1901 - 2099% month - range :1-1 2 % day - range :1-3 1 % hour - range :0-2 3 % minute - range :0-6 0 % second - range :0-6 0 %% j0 - Julian day number of the date at 0 hr UT% ut - universal time in fractions of a day% jd - julian day number of the date and time%% J2000_coe - row vector of J2000 orbital elements from% Table 8.1% rates - row vector of Julian centennial rates from% Table 8.1% t0 - Julian centuries between J2000 and jd% elements - orbital elements at jd%% r - heliocentric position vector% v - heliocentric velocity vector%% User M-functions required: J0, kepler_E, sv_from_coe % User subfunctions required: planetary_elements, zero_to_360% ------------------------------------------------------------ global mu deg = pi/180; %...Equation 5.48: j0 = J0(year, month, day); ut = (hour + minute/60 + second/3600)/24; %...Equation 5.47 jd = j0 + ut;%...Obtain the data for the selected planet from Table 8.1: [J2000_coe, rates] = planetary_elements(planet_id); %...Equation 8.104a: t0 = (jd - 2451545)/36525; D.17 Algorithm 8.1: calculation of the state vector of a planet at a given epoch 643 %...Equation 8.104b: elements = J2000_coe + rates*t0; a = elements(1); e = elements(2); %...Equation 2.61: h = sqrt(mu*a*(1 - eˆ2)); %...Reduce the angular elements to within the range 0 - 360 degrees: incl = elements(3);RA = zero_to_360(elements(4));w_hat = zero_to_360(elements(5));L = zero_to_360(elements(6)); w = zero_to_360(w_hat - RA);M = zero_to_360((L - w_hat)); %...Algorithm 3.1 (for which M must be in radians) E = kepler_E(e, M*deg); %...Equation 3.10 (converting the result to degrees): TA = zero_to_360... (2*atan(sqrt((1 + e)/(1 - e))*tan(E/2))/deg); coe = [h e RA incl w TA a w_hat L M E/deg];%...Algorithm 4.2 (for which all angles must be in radians): [r, v] = sv_from_coe([h e RA*deg incl*deg w*deg TA*deg]); return% ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜% Subfunctions used in the main body:% ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ function [J2000_coe, rates] = planetary_elements(planet_id) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ %% This function extracts a planet’s J2000 orbital elements and% centennial rates from Table 8.1.%% planet_id - 1 through 9, for Mercury through Pluto%% J2000_elements-9b y6matrix of J2000 orbital elements for% the nine planets Mercury through Pluto. The % columns of each row are: % a = semimajor axis (AU) % e = eccentricity % i = inclination (degrees) % RA = right ascension of the ascending % node (degrees) % w_hat = longitude of perihelion (degrees) % L = mean longitude (degrees) % 644 Appendix D MATLAB algorithms % cent_rates - 9 by 6 matrix of the rates of change of the % J2000_elements per Julian century (Cy).% Using ''dot'' for time derivative, the% columns of each row are:% a_dot (AU/Cy)% e_dot (1/Cy)% i_dot (arcseconds/Cy)% RA_dot (arcseconds/Cy)% w_hat_dot (arcseconds/Cy)% Ldot (arcseconds/Cy)%% J2000_coe - row vector of J2000_elements corresponding% to ''planet_id'', with au converted to km% rates - row vector of cent_rates corresponding% to ''planet_id'', with au converted to km % and arcseconds converted to degrees%% au - astronomical unit (km)%% User M-functions required: none% ------------------------------------------------------------ J2000_elements = ... [ 0.38709893 0.20563069 7.00487 48.33167 77.45645 252.25084 0.72333199 0.00677323 3.39471 76.68069 131.53298 181.979731.00000011 0.01671022 0.00005 -11.26064 102.94719 100.464351.52366231 0.09341233 1.85061 49.57854 336.04084 355.453325.20336301 0.04839266 1.30530 100.55615 14.75385 34.404389.53707032 0.05415060 2.48446 113.71504 92.43194 49.94432 19.19126393 0.04716771 0.76986 74.22988 170.96424 313.2321830.06896348 0.00858587 1.76917 131.72169 44.97135 304.8800339.48168677 0.24880766 17.14175 110.30347 224.06676 238.92881]; cent_rates = ... [ 0.00000066 0.00002527 -23.51 -446.30 573.57 538101628.29 0.00000092 -0.00004938 -2.86 -996.89 -108.80 210664136.06 -0.00000005 -0.00003804 -46.94 -18228.25 1198.28 129597740.63-0.00007221 0.00011902 -25.47 -1020.19 1560.78 68905103.78 0.00060737 -0.00012880 -4.15 1217.17 839.93 10925078.35 -0.00301530 -0.00036762 6.11 -1591.05 -1948.89 4401052.95 0.00152025 -0.00019150 -2.09 -1681.4 1312.56 1542547.79 -0.00125196 0.00002514 -3.64 -151.25 -844.43 786449.21-0.00076912 0.00006465 11.07 -37.33 -132.25 522747.90]; J2000_coe = J2000_elements(planet_id,:); rates = cent_rates(planet_id,:); %...Convert from AU to km: au = 149597871;J2000_coe(1) = J2000_coe(1)*au;rates(1) = rates(1)*au; %...Convert from arcseconds to fractions of a degree: rates(3:6) = rates(3:6)/3600; D.17 Algorithm 8.1: calculation of the state vector of a planet at a given epoch 645 return % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ function y = zero_to_360(x) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ %% This function reduces an angle to lie in the range% 0 - 360 degrees.%% x - the original angle in degrees% y - the angle reduced to the range 0 - 360 degrees% % User M-functions required: none% ------------------------------------------------------------ i fx> =3 6 0 x=x-fix(x/360)*360; elseifx<0 x=x-(fix(x/360) - 1)*360; endy=x ;return % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Script file Example_8_07.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % Example_8_07 % ˜˜˜˜˜˜˜˜˜˜˜˜ %% This program uses Algorithm 8.1 to compute the orbital% elements and state vector of the earth at the date and time% specified in Example 8.7. To obtain the same results for % Mars, set planet_id = 4.%% mu - gravitational parameter of the sun (kmˆ3/sˆ2)% deg - conversion factor between degrees and radians% pi - 3.1415926...%% coe - vector of heliocentric orbital elements% [h e RA incl w TA a w_ha tLME ] , % where % h = angular momentum (kmˆ2/s) % e = eccentricity % RA = right ascension (deg) % incl = inclination (deg) % w = argument of perihelion (deg) % TA = true anomaly (deg) % a = semimajor axis (km) 646 Appendix D MATLAB algorithms % w_hat = longitude of perihelion % ( = RA + w) (deg)% L = mean longitud e ( = w_hat + M) (deg) % M = mean anomaly (deg)% E = eccentric anomaly (deg)%% r - heliocentric position vector (km)% v - heliocentric velocity vector (km/s)%% planet_id - planet identifier:% 1 = Mercury% 2 = Venus% 3 = Earth% 4 = Mars% 5 = Jupiter % 6 = Saturn% 7 = Uranus% 8 = Neptune% 9 = Pluto%% year - range: 1901 - 2099% month - range :1-1 2 % day - range :1-3 1 % hour - range :0-2 3 % minute - range :0-6 0 % second - range :0-6 0 %% User M-functions required: planet_elements_and_sv,% month_planet_names% ------------------------------------------------------------ global mu mu = 1.327124e11;deg = pi/180; %...Input data planet_id = 3;year = 2003;month = 8;day = 27; hour = 12;minute = 0;second = 0;%... %...Algorithm 8.1: [coe, r, v, jd] = planet_elements_and_sv ... (planet_id, year, month, day, hour, minute, second); %...Convert the planet_id and month numbers into names for output: [month_name, planet_name] = month_planet_names(month, ... planet_id); %...Echo the input data and output the solution to % the command window:fprintf('---------------------------------------------------')fprintf('\n Example 8.7') D.17 Algorithm 8.1: calculation of the state vector of a planet at a given epoch 647 fprintf('\n\n Input data:\n'); fprintf('\n Planet: %s', planet_name)fprintf('\n Year : %g', year)fprintf('\n Month : %s', month_name)fprintf('\n Day : %g', day)fprintf('\n Hour : %g', hour)fprintf('\n Minute: %g', minute)fprintf('\n Second: %g', second)fprintf('\n\n Julian day: %11.3f', jd) fprintf('\n\n'); fprintf(' Orbital elements:')fprintf('\n'); fprintf('\n Angular momentum (kmˆ2/s) = %g', coe(1)); fprintf('\n Eccentricity = %g', coe(2)); fprintf('\n Right ascension of the ascending node')fprintf(' (deg) = %g', coe(3));fprintf('\n Inclination to the ecliptic (deg) = %g', coe(4));fprintf('\n Argument of perihelion (deg) = %g', coe(5));fprintf('\n True anomaly (deg) = %g', coe(6)); fprintf('\n Semimajor axis (km) = %g', coe(7)); fprintf('\n');fprintf('\n Longitude of perihelion (deg) = %g', coe(8)); fprintf('\n Mean longitude (deg) = %g', coe(9)); fprintf('\n Mean anomaly (deg) = %g', coe(10)); fprintf('\n Eccentric anomaly (deg) = %g', coe(11)); fprintf('\n\n'); fprintf(' State vector:')fprintf('\n'); fprintf('\n Position vector (km) = [%g %g %g]', ... r(1), r(2), r(3)) fprintf('\n Magnitude = %g\n', norm(r)) fprintf('\n Velocity (km/s) = [%g %g %g]', ... v(1), v(2), v(3)) fprintf('\n Magnitude = %g', norm(v)) fprintf('\n-----------------------------------------------\n') % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Output from Example_8_07 ----------------------------------------------------- Example 8.7 Input data: Planet: Earth Year : 2003Month : August 648 Appendix D MATLAB algorithms Day : 27 Hour : 12Minute: 0Second: 0 Julian day: 2452879.000 Orbital elements: Angular momentum (kmˆ2/s) = 4.4551e+09 Eccentricity = 0.0167088Right ascension of the ascending node (deg) = 348.554Inclination to the ecliptic (deg) = -0.000426218Argument of perihelion (deg) = 114.405True anomaly (deg) = 230.812 Semimajor axis (km) = 1.49598e+08 Longitude of perihelion (deg) = 102.959 Mean longitude (deg) = 335.267Mean anomaly (deg) = 232.308Eccentric anomaly (deg) = 231.558 State vector: Position vector (km) = [1.35589e+08 -6.68029e+07 286.909] Magnitude = 1.51152e+08Velocity (km/s) = [12.6804 26.61 -0.000212731]Magnitude = 29.4769 ----------------------------------------------------- D.18 Algorithm 8.2: calculation of the spacecraft trajectory from planet 1to planet 2 Function file interplanetary.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ function [planet1, planet2, trajectory] = interplanetary ... (depart, arrive) % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ %% This function determines the spacecraft trajectory from the% sphere of influence of planet 1 to that of planet 2 using% Algorithm 8.2.%% mu - gravitational parameter of the sun (kmˆ3/sˆ2)% dum - a dummy vector not required in this procedure% % planet_id - planet identifier:% 1 = Mercury% 2 = Venus% 3 = Earth% 4 = Mars D.18 Algorithm 8.2: calculation of the spacecraft trajectory 649 % 5 = Jupiter % 6 = Saturn % 7 = Uranus % 8 = Neptune % 9 = Pluto %% year - range: 1901 - 2099% month - range:1-1 2% day - range:1-3 1% hour - range:0-2 3% minute - range:0-6 0% second - range:0-6 0%% jd1, jd2 - Julian day numbers at departure and arrival% tof - time of flight from planet 1 to planet 2 (s) %% Rp1, Vp1 - state vector of planet 1 at departure (km, km/s)% Rp2, Vp2 - state vector of planet 2 at arrival (km, km/s)% R1, V1 - heliocentric state vector of spacecraft at% departure (km, km/s) % R2, V2 - heliocentric state vector of spacecraft at% arrival (km, km/s) %% depart - [planet_id, year, month, day, hour, minute,% second] at departure % arrive - [planet_id, year, month, day, hour, minute,% second] at arrival % planet1 - [Rp1, Vp1, jd1] % planet2 - [Rp2, Vp2, jd2]% trajectory - [V1, V2]%% User M-functions required: planet_elements_and_sv, lambert% ------------------------------------------------------------ global muplanet_id = depart(1); year = depart(2);month = depart(3);day = depart(4); hour = depart(5);minute = depart(6);second = depart(7); %...Use Algorithm 8.1 to obtain planet 1's state vector (don't %...need its orbital elements [''dum'']):[dum, Rp1, Vp1, jd1] = planet_elements_and_sv ... (planet_id, year, month, day, hour, minute, second); planet_id = arrive(1); year = arrive(2);month = arrive(3);day = arrive(4); hour = arrive(5);minute = arrive(6);second = arrive(7); %...Likewise use Algorithm 8.1 to obtain planet 2’s state vector: 650 Appendix D MATLAB algorithms [dum, Rp2, Vp2, jd2] = planet_elements_and_sv ... (planet_id, year, month, day, hour, minute, second); tof = (jd2 - jd1)*24*3600;%...Patched conic assumption: R1 = Rp1;R2 = Rp2; %...Use Algorithm 5.2 to find the spacecraft’s velocity at % departure and arrival, assuming a prograde trajectory:[V1, V2] = lambert(R1, R2, tof, 'pro'); planet1 = [Rp1, Vp1, jd1]; planet2 = [Rp2, Vp2, jd2]; trajectory = [V1, V2]; % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Script file Example_8_08.m % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ % Example_8_08 % ˜˜˜˜˜˜˜˜˜˜˜˜ %% This program uses Algorithm 8.2 to solve Example 8.8.%% mu - gravitational parameter of the sun (kmˆ3/sˆ2)% deg - conversion factor between degrees and radians% pi - 3.1415926...%% planet_id - planet identifier:% 1 = Mercury% 2 = Venus% 3 = Earth% 4 = Mars% 5 = Jupiter% 6 = Saturn% 7 = Uranus% 8 = Neptune % 9 = Pluto% planet_name - name of the planet%% year - range: 1901 - 2099% month - range :1-1 2 % month_name - name of the month% day - range :1-3 1 % hour - range :0-2 3 % minute - range :0-6 0 % second - range :0-6 0 %% depart - [planet_id, year, month, day, hour, minute,% second] at departure % arrive - [planet_id, year, month, day, hour, minute,% second] at arrival%% planet1 - [Rp1, Vp1, jd1] D.18 Algorithm 8.2: calculation of the spacecraft trajectory 651 % planet2 - [Rp2, Vp2, jd2] % trajectory - [V1, V2]%% coe - orbital elements [h e RA incl w TA] % where % h = angular momentum (kmˆ2/s) % e = eccentricity % RA = right ascension of the ascending % node (rad) % incl = inclination of the orbit (rad) % w = argument of perigee (rad) % TA = true anomaly (rad) % a = semimajor axis (km) %% jd1, jd2 - Julian day numbers at departure and arrival % tof - time of flight from planet 1 to planet 2 % (days) %% Rp1, Vp1 - state vector of planet 1 at departure% (km, km/s) % Rp2, Vp2 - state vector of planet 2 at arrival% (km, km/s) % R1, V1 - heliocentric state vector of spacecraft at% departure (km, km/s) % R2, V2 - heliocentric state vector of spacecraft at% arrival (km, km/s) %% vinf1, vinf2 - hyperbolic excess velocities at departure% and arrival (km/s) %% User M-functions required: interplanetary, coe_from_sv,% month_planet_names % ------------------------------------------------------------ clear global mumu = 1.327124e11;deg = pi/180; %...Data for planet 1: planet_id = 3; % (earth)year = 1996;month = 11;day = 7;hour = 0;minute = 0;second = 0;%... depart = [planet_id year month day hour minute second];%...Data for planet 2: planet_id = 4; % (Mars)year = 1997;month = 9;day = 12; 652 Appendix D MATLAB algorithms hour = 0; minute = 0;second = 0;%... arrive = [planet_id year month day hour minute second];[planet1, planet2, trajectory] = interplanetary ... (depart, arrive); R1 = planet1(1,1:3);Vp1 = planet1(1,4:6);jd1 = planet1(1,7); R2 = planet2(1,1:3); Vp2 = planet2(1,4:6); jd2 = planet2(1,7); V1 = trajectory(1,1:3); V2 = trajectory(1,4:6); tof = jd2 - jd1;%...Use Algorithm 4.1 to find the orbital elements of the % spacecraft trajectory based on [Rp1, V1]...coe = coe_from_sv(R1, V1);% ... and [R2, V2]coe2 = coe_from_sv(R2, V2); %...Equations 8.102 and 8.103: vinf1 = V1 - Vp1;vinf2 = V2 - Vp2; %...Echo the input data and output the solution to % the command window:fprintf('---------------------------------------------------')fprintf('\n Example 8.8')fprintf('\n\n Departure:\n');[month_name, planet_name] = month_planet_names(depart(3), ... depart(1)); fprintf('\n Planet: %s', planet_name)fprintf('\n Year : %g', depart(2))fprintf('\n Month : %s', month_name)fprintf('\n Day : %g', depart(4))fprintf('\n Hour : %g', depart(5))fprintf('\n Minute: %g', depart(6))fprintf('\n Second: %g', depart(7))fprintf('\n\n Julian day: %11.3f\n', jd1)fprintf('\n Planet position vector (km) = [%g %g %g]', ... R1(1), R1(2), R1(3)) fprintf('\n Magnitude = %g\n', norm(R1))fprintf('\n Planet velocity (km/s) = [%g %g %g]', ... Vp1(1), Vp1(2), Vp1(3)) fprintf('\n Magnitude = %g\n', norm(Vp1)) D.18 Algorithm 8.2: calculation of the spacecraft trajectory 653 fprintf('\n Spacecraft velocity (km/s) = [%g %g %g]', ... V1(1), V1(2), V1(3)) fprintf('\n Magnitude = %g\n', norm(V1)) fprintf('\n v-infinity at departure (km/s) = [%g %g %g]', ... vinf1(1), vinf1(2), vinf1(3)) fprintf('\n Magnitude = %g\n', norm(vinf1)) fprintf('\n\n Time of flight = %g days\n', tof)fprintf('\n\n Arrival:\n'); [month_name, planet_name] = month_planet_names(arrive(3), ... arrive(1)); fprintf('\n Planet: %s', planet_name)fprintf('\n Year : %g', arrive(2))fprintf('\n Month : %s', month_name)fprintf('\n Day : %g', arrive(4))fprintf('\n Hour : %g', arrive(5))fprintf('\n Minute: %g', arrive(6))fprintf('\n Second: %g', arrive(7))fprintf('\n\n Julian day: %11.3f\n', jd2)fprintf('\n Planet position vector (km) = [%g %g %g]', ... R2(1), R2(2), R2(3)) fprintf('\n Magnitude = %g\n', norm(R1)) fprintf('\n Planet velocity (km/s) = [%g %g %g]', ... Vp2(1), Vp2(2), Vp2(3)) fprintf('\n Magnitude = %g\n', norm(Vp2)) fprintf('\n Spacecraft Velocity (km/s) = [%g %g %g]', ... V2(1), V2(2), V2(3)) fprintf('\n Magnitude = %g\n', norm(V2)) fprintf('\n v-infinity at arrival (km/s) = [%g %g %g]', ... vinf2(1), vinf2(2), vinf2(3)) fprintf('\n Magnitude = %g', norm(vinf2)) fprintf('\n\n\n Orbital elements of flight trajectory:\n')fprintf('\n Angular momentum (kmˆ2/s) = %g', coe(1)) fprintf('\n Eccentricity = %g', coe(2)) fprintf('\n Right ascension of the ascending node')fprintf(' (deg) = %g', coe(3)/deg)fprintf('\n Inclination to the ecliptic (deg) = %g', ... coe(4)/deg) fprintf('\n Argument of perihelion (deg) = %g', ... coe(5)/deg) fprintf('\n True anomaly at departure (deg) = %g', ... coe(6)/deg) fprintf('\n True anomaly at arrival (deg) = %g\n', ... coe2(6)/deg) fprintf('\n Semimajor axis (km) = %g', coe(7)) 654 Appendix D MATLAB algorithms if coe(2) < 1 fprintf('\n Period (days) = %g', ... 2*pi/sqrt(mu)*coe(7)ˆ1.5/24/3600) endfprintf('\n-----------------------------------------------\n') % ˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜˜ Output from Example_8_08 ----------------------------------------------------- Example 8.8 Departure: Planet: Earth Year : 1996Month : NovemberDay : 7Hour : 0Minute: 0Second: 0 Julian day: 2450394.500Planet position vector (km) = [1.04994e+08 1.04655e+08 988.331] Magnitude = 1.48244e+08 Planet velocity (km/s) = [-21.515 20.9865 0.000132284] Magnitude = 30.0554 Spacecraft velocity (km/s) = [-24.4282 21.7819 0.948049] Magnitude = 32.7427 v-infinity at departure (km/s) = [-2.91321 0.79542 0.947917] Magnitude = 3.16513 Time of flight = 309 daysArrival: Planet: Mars Year : 1997Month : SeptemberDay : 12Hour : 0Minute: 0Second: 0 Julian day: 2450703.500 Planet position vector (km) = [-2.08329e+07 -2.18404e+08 -4.06287e+06] Magnitude = 1.48244e+08 Planet velocity (km/s) = [25.0386 -0.220288 -0.620623] Magnitude = 25.0472 D.18 Algorithm 8.2: calculation of the spacecraft trajectory 655 Spacecraft Velocity (km/s) = [22.1581 -0.19666 -0.457847] Magnitude = 22.1637 v-infinity at arrival (km/s) = [-2.88049 0.023628 0.162776] Magnitude = 2.88518 Orbital elements of flight trajectory: Angular momentum (kmˆ2/s) = 4.84554e+09 Eccentricity = 0.205785 Right ascension of the ascending node (deg) = 44.8942Inclination to the ecliptic (deg) = 1.6621 Argument of perihelion (deg) = 19.9738 True anomaly at departure (deg) = 340.039 True anomaly at arrival (deg) = 199.695 Semimajor axis (km) = 1.84742e+08 Period (days) = 501.254 ------------------------------------------------------------- This page int entionally left blank EAppendix Gravitational potential energy of a sphere Figure E.1 shows a point mass mwith cartesian coordinates ( x,y,z)a sw e l la sa system of Npoint masses m1,m2,m3,..., mN.T h e ith one of these particles has mass miand coordinates ( xi,yi,zi). The total mass of the Nparticles is M, M=N/summationdisplay i=1mi (E.1) mi (xi, yi, zi) m (x, y, z)ri zx y Figure E.1 A system of point masses and a neighboring test mass m. 657 658 Appendix E Gravitational potential energy of a sphere θ yx zφr′ RdM mr R0 C Figure E.2 Sphere with a spherically symmetric mass distribution. The position vector drawn from mitomisriand the unit vector in the direction ofriis ˆui=ri ri The gravitational force exerted on mbymiis opposite in direction to ri, and is given by Fi=−Gmm i r2 iˆui=−Gmm i r3 iri The potential energy of this force is Vi=− Gmm i ri(E.2) The total gravitational potential energy of the system due to the gravitational attraction of all of the Nparticles is V=N/summationdisplay i=1Vi (E.3) Therefore, the total force of gravity Fon the mass mis F=− ∇ V=−/parenleftbigg∂V ∂xˆi+∂V ∂yˆj+∂V ∂zˆk/parenrightbigg (E.4) Consider the solid sphere of mass Mand radius R0illustrated in Figure E.2. Instead of a discrete system as above, we have a continuum with mass density /rho1. Each ‘particle’ Appendix E Gravitational potential energy of a sphere 659 is a differential element dM=/rho1dvof the total mass M. Equation E.1 becomes M=/integraldisplay/integraldisplay/integraldisplay v/rho1dv (E.5) where dvis the volume element and vis the total volume of the sphere. In this case, Equation E.2 becomes dV=− GmdM r=− Gm/rho1dv r where ris the distance from the differential mass dM to the finite point mass m. Equation E.3 is replaced by V=− Gm/integraldisplay/integraldisplay/integraldisplay v/rho1dv r(E.6) Let the mass of the sphere have a spherically symmetric distribution, which means that the mass density /rho1depends only on r/prime, the distance from the center Cof the sphere. An element of mass dMhas spherical coordinates ( r/prime,θ,φ), where the angle θis measured in the xyplane of a cartesian coordinate system with origin at C,a s shown in Figure E.2. In spherical coordinates the volume element is dv=r/prime2sinφdφdr/primedθ (E.7) Therefore Equation E.5 becomes M=/integraldisplay2π θ=0/integraldisplayR0 r/prime=0/integraldisplayπ φ=0/rho1r/prime2sinφdφdr/primedθ=/parenleftbigg/integraldisplay2π 0dθ/parenrightbigg/parenleftbigg/integraldisplayπ 0sinφdφ/parenrightbigg/parenleftbigg/integraldisplayR0 0/rho1r/prime2dr/prime/parenrightbigg =(2π)(2)/parenleftbigg/integraldisplayR0 0/rho1r/prime2dr/prime/parenrightbigg so that the mass of the sphere is given by M=4π/integraldisplayR0 r/prime=0/rho1r/prime2dr/prime(E.8) Substituting Equation E.7 into Equation E.6 yields V=− Gm/integraldisplay2π θ=0/integraldisplayR0 r/prime=0/integraldisplayπ φ=0/rho1r/prime2sinφdφdr/primedθ r =− 2πGm/bracketleftbigg/integraldisplayR0 0/parenleftbigg/integraldisplayπ 0sinφdφ r/parenrightbigg /rho1r/prime2dr/prime/bracketrightbigg (E.9) The distance ris found by using the law of cosines, r=(R2+r/prime2−2r/primeRcosφ)1 2 where Ris the distance from the center of the sphere to the mass m. Differentiating this equation with respect to φ, holding r/primeconstant, yields dr dφ=1 2(R2+r/prime2−2r/primeRcosφ)−1 2(2r/primeRsinφdφ)=r/primeRsinφ r 660 Appendix E Gravitational potential energy of a sphere so that sinφdφ=rdr r/primeR It follows that /integraldisplayπ φ=0sinφdφ r=1 r/primeR/integraldisplayR+r/prime R−r/primedr=2 R Substituting this result along with Eq uation E.8 into Equation E.9 yields V=−GMm R We conclude that the gravitational potential energy, and hence (from Equation E.4) the gravitational force, of a sphere with a spherically symmetric mass distribution M is the same as that of a point mass Mlocated at the center of the sphere. Index Absolute acceleration angular 402–8, 436–40, 484–6 nutation dampers 496point masses 20–9rigid-body kinematics 401–8, 410, 411 two-body motion 35 Absolute angular momentum 411–14 Absolute angular velocity gyroscopic attitude control 521 nutation dampers 496rigid-body dynamics 402–7, 451 torque-free motion 479–80 Absolute position vectors 20–9Absolute velocity close-proximity circular orbits 340 rigid-body kinematics 401–8, 451 two-body motion 35two-impulse maneuvers 331vectors 20–9 Acceleration see also absolute...; angular...; relative Coriolis 21five-term 21, 23gravitational 7–10, 177gyroscopic attitude control 521–5 oblateness 177–8point masses 2–7, 16–18, 20–9 preliminary orbit determination 228relative motion and rendezvous 317–20 restricted three-body motion 91 rocket vehicle dynamics 553three-body systems 590–4 Advance of perigee 178–80, 184Aiming radius hyperbolic trajectories 71–2, 75, 382 planetary rendezvous 370–1, 373 Altitude equation 554gravity-gradient stabilization 534 perigee 64–5, 208–10, 211–12preliminary orbit determination 208–10,211–12, 231–5 rocket performance 554, 558–9 Sun-synchronous three dimensional orbits 181 two-body motion 53–4 Amplitude 478Angles see also flight path... auxiliary 111–15azimuth 227–8, 231–6, 626–31 dihedral 290, 293elevation 227–8, 232, 235–6, 626–31 Euler’s 158–9, 448–59, 480to periapse 373, 374–5phase 350–3preliminary orbit determination 228–50 of rotation 282–5spin 508 tilt 446turn 70, 75, 369–70, 378, 382 wobble 481–2 Angular acceleration absolute 402–8, 436–40, 484–6 gyroscopic attitude control 529–30 point masses 16–18, 20–9relative 317, 319, 437rendezvous 317, 319rigid-body kinematics 401–8satellite attitude dynamics 513, 529–30, 484–6 torque-free motion 484–6yo-yo despin 513 Angular momentum chase maneuvers 286–9conservation of 79–80double-gimbaled control moment gyros 528–9 Hohmann transfers 261hyperbolic trajectories 74–5, 130 Lagrange coefficients 78–89moments of inertia 414–15, 420–1, 423–5 orbit formulas 42–50plane change maneuvers 302–3 planetary departure 361–2planetary flyby 379point masses 13–15preliminary orbit determination 195–201,204–5, 236–8 rigid-body dynamics 414–15, 420–1, 423–5, 435–40 661 662 Index Angular momentum (continued) satellite attitude dynamics coning maneuvers 503–5dual-spin spacecraft 491–2gyroscopic control 517, 518–22, 524–5 nutation dampers 498–9thrusters 506–9torque-free motion 476–7, 481–2, 484–6, 488 yo-yo despin 509–10, 511 spinning tops 447 three dimensional orbits 158–60, 161–2 torque-free motion 476–7, 481–2, 484–6, 488 two-body motion 42–50, 74–5, 78–89 Angular position 232, 629–34Angular velocity close-proximity circular orbits 338 Euler angles 453–6Euler’s equations 436–40moments of inertia 420–1pitch 462–3point masses 16–18, 21, 25–7relative motion and rendezvous 316–17, 319 rigid-body dynamics 420–1, 436–40, 453–6 roll 462–3satellite attitude dynamics dual-spin spacecraft 493–5 gravity-gradient stabilization 535–6 gyroscopic control 516–17, 521, 523–5 thrusters 506torque-free motion 477–80, 484, 486–9 yo-yo despin 511–13 spinning tops 444two-body motion 47–8yaw 462–3 Angular-impulse 13–15, 413–14 Apoapse 56, 290–303, 373Apogee kick 258–60radius 62, 70–1, 183–7towards the sun 117–19velocity 62 Applied torque 413–14Approach trajectories 368–75, 379–86, 397 see also two-body motion Apse lines 260, 273–85Arcseconds 387, 388Areal velocity 44Argument of perigee oblateness 178–81, 183–7orbital elements 159, 161, 163 Arrival phase 391, 393–4Astronomical units 387, 388Attitude dynamics seesatellite... Auxiliary angles 111–15Axial bearing loads 459Axial torques 523–5Axis of rotation 150Axisymmetric dual-spin 491–5, 518–21, 529–30 Axisymmetric tops 443–8Azimuth angles 227–8, 231–6, 626–31averaged radius 62plane change maneuvers 294–5 Bearing forces 456–9 Bent rods 431–5Bessel functions 121–2Bi-elliptic Hohmann transfers 264–8 Bias values 526Bivariate functions 570Body cones 482Body frames 18Burnout Jacobi constant 100–1rocket vehicle dynamics 558–9, 561, 564–78 sensitivity analysis 366–8Capture orbits 372–3, 375, 396–7 Capture radius 372Cartesian coordinates elliptical orbits 58equation of a parabola 68hyperbolic trajectories 72–3rotation 169–72three dimensional orbits 164–5 Cassini gravity assist maneuvers 386 Celestial bodies 149–54Center of mass inertial frames 34–7moving reference frames 38–42 rigid-body dynamics Euler angles 454–6Euler’s equations 435–40moments of inertia 417–18, 420–1, 423–5, 430–5 parallel axis theorem 430–5rotational motion 412–14translational motion 408–10 two-body motion 34–7, 38–42 Chase maneuvers 285–9, 322–40 Chasles’ theorem 399–400Circular orbits close-proximity relative motion 338–40 Hohmann transfers 257, 264parking 362, 394–6position as a function of time 108–9 rigid-body kinetic energy 442–3 two-body motion 51–5 Classical orbital elements 159–61, 175, 199–201,607–14 Clohessy–Wiltshire (CW) frames equations 324–30, 336–7, 338, 340 Index 663 gravity-gradient stabilization 533–4 matrices 329, 333, 336 Close-proximity circular orbits 338–40 Co-moving reference frames 316–22, 324–30 Coaxial elliptic orbits 260, 273–4 Common apse lines 273–9Common focus 290Conics 359–60, 391–7Coning maneuvers 503–5, 507Conservation of... angular momentum 79–80energy 65, 509–10momentum 79–80, 509–10 Constant amplitude 478Continuous three dimensional bodies 408–10 Control moment gyros 526–30Coordinate systems 218–28 see also Cartesian...; topocentric... polar 48 Coordinate transformations geocentric equatorial 172–6, 186–7, 224–8 perifocal frames 172–6, 186–7rotation 169–72three dimensional orbits 164–76, 186–7 topocentric 224–8 Coplanar orbits 257, 297–8, 340 Cord lengths 509–16Cord unwind rates 512–13Coriolis acceleration 21Cosine vectors 242–3, 244Cruise phase 391–2, 393–6Curvature of the earth 553–4Curvilinear motion 1–7CW seeClohessy–Wiltshire frames Dark side approaches 379–84, 385Declination preliminary orbit determination 222–3,225–6, 230–2 state vectors 155–8three dimensional orbits 149–54 Delta-H requirements 504–5, 507–9 Delta-v requirements bi-elliptic Hohmann transfers 264–8 chase maneuvers 285–9Hohmann transfers 257–73, 348–50 impulsive orbital maneuvers 256–7, 330–7 interplanetary trajectories 348–50, 362, 364–6 non-Hohmann transfers 273–82, 394–7 phasing maneuvers 268–73plane change maneuvers 290–303 planetary rendezvous 362, 364–6, 371–5 rocket vehicle dynamics 551–2 two-impulse maneuvers 330–7 Departure trajectories 360–6, 391, 393–6 Despin mechanisms 509–16Diagonal moment of inertia matrices 425–8 Dihedral angles 290, 293Direct ascent trajectories 363Direction cosine vectors 242–3, 244 Distances between planets 389–91 Double-gimbaled control moment gyros 526–30 Downrange equations 554Drag force 553, 558Dual-spin satellites 518–21, 529–30 Dual-spin spacecraft 491–5Earth centered inertial frames 23–9earth orbits 52–3, 149, 258–60 earth satellites 149, 183–7earth-moon systems 98–101earth’s curvature 553–4earth’s gravitational parameter 52 earth’s oblateness 177–87earth’s shadow 117–19earth’s sphere of influence 358–9 low earth orbits 52–3, 297–8, 300 East longitude 218–21, 222–3East-North-Zenith (ENZ) frame 223 Easterly launches 294–5Eccentric anomaly hyperbolic trajectories 126–33 Kepler’s equation 113–17, 130, 596–600 MATLAB algorithms 115, 130, 596–600 oblateness 184orbit equation 135position as a function of time 111–15, 126–33, 135 Eccentricity chase maneuvers 286–9elliptical orbits 55–65 hyperbolic trajectories 125–6interplanetary trajectories 361–2, 387, 388 limiting values 120–1non-Hohmann transfers 282–5 orbit formulas 46–7orbital elements 158–9, 160–1, 163 plane change maneuvers 302–3 planetary departure 361–2planetary ephemeris 387, 388planetary flyby 379 664 Index Eccentricity (continued) planetary rendezvous 369position as a function of time 113, 120–1, 125–6 preliminary orbit determination 219 Ecliptic plane 150Effective exhaust velocity 556–7Eigenvalues 426–8Eigenvectors 426–8Elevation angles 227–8, 232, 235–6, 626–31 Elliptical orbits Hohmann transfers 257–68non-Hohmann trajectories 396–7 position as a function of time 109–23, 134–5 two-body motion 55–65 Empty masses 560, 561, 566–9Energy circular orbits 52conservation of 65, 509–10dissipation 495–503elliptical orbits 59Hohmann transfers 257hyperbolic trajectories 73kinetic 441–3, 488–91, 493–4, 509–12 law 50–1, 52, 59, 73non-Hohmann transfers 275–6 orbital elements 158–9plane change maneuvers 293 position as a function of time 135 potential 36–7, 658–61sinks 492–5three dimensional orbits 158–9 ENZ seeEast-North-Zenith Ephemeris 152–3, 387–91Epochs 388, 641–8Equations of motion double-gimbaled control moment gyros 528–9 dual-spin spacecraft 492inertial frames 34–7integration 587–94 interplanetary trajectories 356–7 linearization of relative motion 322–4 numerical integration 587–94relative 37–42, 322–4rocket vehicle dynamics 552–5 rotational 410–14, 517–21satellite attitude dynamics 496–503 translational 408–10 Equations of parabolas 68Equatorial frames see also geocentric... plane change maneuvers 293–4, 301–2 state vectors 154–8three dimensional orbits 150topocentric coordinates 221–3, 225–7 Equilibrium points 92–6Escape velocity 66, 73Euler, Leonhard 16Euler rotations 448–59Euler’s angles 158–9, 448–59, 480 Euler’s equation rigid-body dynamics 435–40satellite attitude dynamics 478, 485–7, 525, 533 Excess speed 73–4 Excess velocity 360–6, 368–75, 392–4 Exhaust 555–7Extremum 571–2 Field-free space restricted staging 560–70 Five-term acceleration 21, 23Flattening seeoblateness Flight path angles elliptical orbits 63–4hyperbolic flyby 381Newton’s law of gravitation 9–10non-Hohmann transfers 274–5 parabolic trajectories 66rocket vehicle dynamics 552–5 Flight time 265–8Floor 120Flow rates 558–9Fluids 410Flyby 375–86Flywheels 516–30Forces see also gravitational... bearing 456–9drag 553, 558gyroscopic 456–9lifting 554net 27–9nutation dampers 497point masses 7–15sphere of influence 355–9units of 10–15 Free-fall 9–10 Gases 410 Gauss’s method of preliminary orbit determination235–50, 631–41 GEO seegeostationary equatorial orbits Geocentric... latitude 219–21orbits 115–17, 130–3, 442–3position vectors 194–201right ascension-declination 149–54, 155–8 satellites 276–9 Geocentric equatorial frames coordinate transformations 172–6, 186–7, 224–8 MATLAB algorithms 232, 626–31 orbital elements 158–64, 175–6 perifocal frame 172–6, 186–7state vectors 154–8, 175–6 Index 665 topocentric transformations 224–8 transformations 172–6, 186–7, 224–8 Geodetic latitude 220–1Geostationary equatorial orbits (GEO) 53–6 phasing maneuvers 271–3plane change maneuvers 293–4, 297–8, 300 Geosynchronous dual-spin communication satellites 494 Gibb’s method 194–201, 614–18Gimbals 406–8, 526–30Gradient operator 36–7Gravitation acceleration 7–10, 177attraction 33–105geocentric right ascension-declination151–2 point masses 7–10potential energy 658–61restricted three-body motion 91–2 satellite attitude dynamics 530–1 sphere of influence 355–9 Gravity assist maneuvers 385–6Gravity gradient stabilization 530–42 Gravity turn trajectories 552–5 Greenwich sidereal time 214, 216, 218 Ground track 296–7Guided missiles 554Gyros gyroscope equation 447gyroscopic attitude control 516–30 gyroscopic forces 456–9gyroscopic moment 447motors 439–40rotors 406–8, 420–1satellite attitude dynamics 491–5Heliocentric trajectories 359 approach velocity 368post-flyby 375–86speed 360, 363velocity 368, 375–86 High-energy precession rates 446–7 Hohmann transfers bi-elliptic transfers 264–8common apse line 274interplanetary trajectories 348–50, 391–7 non-Hohmann trajectories 391–7 orbital maneuvers 257–73, 274 phasing maneuvers 268–73plane change maneuvers 297–9, 300–1 planetary rendezvous 368–9, 373 Horizon coordinate system 223–8 Hyperbolas 130, 598–600Hyperbolic trajectories approach 368–75, 397departure 360–6excess velocity 360–6, 368–86, 392–4 flyby 375–86position as a function of time 125–35 rotations 370–1two-body motion 69–76 Identity matrices 167 Impulse angular 13–15, 413–14coning maneuvers 503–5rendezvous maneuvers 257–73, 330–7 rocket vehicle dynamics 552, 557–9, 562–4, 570–8 Impulsive orbital maneuvers 255–73 Inclinationdouble-gimbaled control moment gyros 528 plane change maneuvers 294–301 planetary ephemeris 387–8Sun-synchronous orbits 181three dimensional orbits 159, 160, 162 Inertia see also moments of inertia angular velocity 89, 402–3, 535–6 equations of two-body motion 34–7 gravity-gradient stabilization 531–2, 535–8 matrices 416, 421–8, 519–25rigid-body dynamics 414–35, 457 tensors 421–8, 434torque-free motion stability 491 velocity 89, 91, 402–3, 535–6 Insertion points 293–4Integration, equations of motion 587–94 Intercept trajectories 285–9Intermediate-axis spinners 502–3 Interplanetary dual-spin spacecraft 494 Interplanetary trajectories 347–98 ephemeris 387–91 flyby 375–86Hohmann transfers 348–50method of patched conics 359–60 non-Hohmann 391–7patched conics 359–60planetary departure 360–6planetary ephemeris 387–91planetary flyby 375–86planetary rendezvous 368–75rendezvous 349–54, 368–75sensitivity analysis 366–8sphere of influence 354–9three dimensional orbits 149 666 Index Iterations 242–3, 245–50, 631–40 Jacobi constant 96–101 Julian centuries 388–91Julian days (JD) 214–18 numbers 214–18, 388–91, 621–3, 641–8 Jupiter’s right ascension 225–6 Kepler, Johannes 44 Kepler’s equation Bessel functions 121–2eccentric anomaly 113–17, 130, 596–600 hyperbola eccentric anomaly 115, 130, 598–600 hyperbolic trajectories 128–30 MATLAB algorithms 115, 130, 596–600, 601–3 Newton’s method 596–600, 601–3 position as a function of time 1, 34–5, 113–17, 121–2,128–30, 134–44 universal variables 134–5, 136–44 Kepler’s second law 44Kilograms 10–15Kinematics 2–7, 400–8Kinetic energy 441–3, 488–91, 493–4, 509–12 Lagrange coefficients MATLAB algorithms 603–5position as a function of time 141–4 preliminary orbit determination 204,207–10, 237–9, 249 two-body motion 78–89 Lagrange multiplier method 570–8 Lagrange points 92–6Lambert’s problemchase maneuvers 285, 288–9 MATLAB algorithms 208, 616–22 patched conics 391–7preliminary orbit determination 202–13,616–22 Laplace limit 120–1Latitude 54–5, 218–23, 231, 294–7 Latus rectum 49, 302–3Launch azimuth 294–7Launch vehicle boost trajectories 552–5 Leading-side flyby 375–6, 378–9LEO seelow-earth orbits Libration points 92–6Lifting forces 554Limiting values 120–1Linear momentum 412Linearized equations of relative motion 322–4 Local horizon 49Local sidereal time 214, 216–18, 623–6 Longitude of perihelion 388Low earth orbits (LEO) 52–3, 297–8, 300 Low-energy precession rates 446–7 Lunar trajectories 359 Major-axis spinners 502–3, 541 Mars missions 354–5Mass gravitational potential energy 657–60 moments of inertia 422nutation dampers 496–503point masses 7–15ratios 557–9, 560–1, 564rocket vehicle dynamics 573–8 MATLAB algorithms 595–656 acceleration 590–4angular position 232, 626–31chase maneuvers 287–8classical orbital elements 159–61, 175, 606–13 eccentric anomaly 115, 130, 138–9, 596–600 epochs 388, 641–8Gauss’s method of preliminary orbitdetermination 242–3,245–50, 631–41 geocentric equatorial position 232, 626–31 Gibbs method of preliminary orbit determination613–16 hyperbola eccentric anomaly 130, 598–600 Julian day number 388, 621–3, 641–8 Kepler’s equation 115, 130, 138–9, 596–603 Lagrange coefficients 603–5Lambert’s problem 208, 616–22 local sidereal time 217, 623–7month identity conversions 640–1 Newton’s method 115, 130, 138–9, 596–603 non-Hohmann trajectories 393 numerical designation conversions 640–1 orbital elements from the state vector 159–61, 175,606–13 planet identity conversions 640–1 planet state vector calculation 388, 641–8 planetary ephemeris 388–9position as a function of time 142, 604–6 preliminary orbit determination 198, 208,217, 232, 242–50, 613–40 range 232, 626–31sidereal time 217, 623–6 Index 667 spacecraft trajectories 393, 648–55 sphere of influence 393, 648–55 state vectors 159–61, 175, 232, 604–13, 626–31 Stumpff functions 600–1three-body systems 589–94time lapse 604–6transformation matrices 175universal anomaly 138–9, 601–3 universal Kepler’s equation 138–9, 601–3 Universal Time 388, 641–8 Matrices see also transformation... Clohessy–Wiltshire frames 329, 333, 336 diagonal 425–8direction cosines 166–72, 174–6, 186–7 identity matrices 167inertia 416, 421–8, 519–25moments of inertia 421–8, 519–25 orthogonal 320, 421–8, 449–50 rotation 460–3unit 167 Mean... anomaly 110–15, 124–6, 134–5, 159 distance 61 longitude 388motion 110, 184, 326, 338 Mercator projections 296–7Method of patched conics 359–60, 391–7 Minor-axis spinners 502–3, 541Missiles 554Molniya orbit 182–3Moments 410–14, 435–40, 454–6 Moments of inertia gravity-gradient stabilization 531–2 matrices 421–8, 519–25parallel axis theorem 428–35 principal 419, 426–8, 431–6, 457 rigid-body dynamics 414–40, 457 torque-free motion stability 491 Momentum see also angular... absolute angular 411–14conservation of 509–10exchange systems 406–8, 420–1, 439–40, 491–5 linear 412rigid-body rotational motion 412 rocket vehicle dynamics 555–7 yo-yo despin 509–10 Month identity conversions 640–1 Moon ephemeris 152–3Moving reference frames 37–42, 316–22, 324–30 Moving vectors 15–20Multi-stage vehicles 552, 562, 563–78 Mutual gravitational attraction 33–105 see also two-body motion nbody equations of motion 587–94 Net forces 27–9Net moments 437–40, 454–6Newton’s law of gravitation 7–10, 355–9 Newton’s laws of motion 10–15, 409 Newton’s method Kepler’s equation 596–600, 601–3 MATLAB algorithms 138–9, 596–600, 601–3 preliminary orbit determination 206, 207, 209roots 114–15 universal Kepler’s equation 138–9, 601–3 Newton’s second law of motion 10–15, 409 Node regression 178–80Non-coplanar orbits 290–303Non-Hohmann transfers 273–85, 391–7 Non-rotating inertial frames 23–9 Numerical designation conversions 640–1 Numerical integration, equations of motion587–94 Nutation dampers 495–503, 509double-gimbaled control moment gyros 527 rigid-body dynamics 451–4spinning tops 445torque-free motion 476–7 Oblateness preliminary orbit determination 219 satellite attitude dynamics 481–2, 494, 495–6 spinner stability 475, 491three dimensional orbits 177–87 One-dimensional momentum analysis 555–7 Optimal staging 570–8Orbit formulas 42–50, 135Orbit rotation 302–3Orbital elements geocentric equatorial frame 158–64 interplanetary trajectories 387, 388, 392 non-Hohmann trajectories 392 oblateness 184–7planet state vectors 388, 641–8 668 Index Orbital elements (continued) planetary flyby 379preliminary orbit determination 199–201,208–11, 232–5, 250 state vectors 158–64, 175, 607–14 three dimensional orbits 158–64 Orbital maneuvers 255–314 apse line rotation 279–85bi-elliptic Hohmann transfers 264–8 chase maneuvers 285–9common apse line 273–9Hohmann transfers 257–73, 274 impulsive 255–314non-Hohmann transfers 273–85 phasing maneuvers 268–73plane change 290–303two-impulse rendezvous 330–7 Orbital parameters 286–9Orbiting Solar Observatory (OSO-1) 491–2 Orientation delta-v maneuver 276–9, 280–2 gravity-gradient stabilization 540–2 rigid-body dynamics 448 Orthogonal transformation matrices 320, 421–8,449–50 Orthogonal unit vectors 5–7Orthonormal basis vectors 165Overall payload fractions 564 Parabolic trajectories 65–9, 124–5 Parallel axis theorem 428–35 Parallel staging 563Parallelepipeds 456–9, 540–2Parameter of the orbit 49Parking orbits 360–6, 394–6Particles 1–7 Passive altitude stabilization 534Passive energy dissipation 495–503 Patched conics 359–60, 391–7Payloads masses 560, 564–70ratios 560–1, 564, 567–8velocity 566–7 Periapse angle to 373, 374–5orbit formulas 49plane change maneuvers 290–303 radius 360–2, 369, 370, 372–3speed 362time since 108–9two-body motion 49, 55–6 Perifocal frame 76–8, 172–6, 186–7 Perigee advance 178–80, 184altitude 64–5, 208–10, 211–12argument of 159, 161, 163, 178–81, 183–7 location 364–6orbit equation 68–9passage 115–17radius 71, 75, 183–7time since 131, 158–9, 184–5, 208–11, 287–8 time to 211, 212–13towards the sun 117–19velocity 61–2 Perihelion radius 384, 385Period of orbit circular orbits 51, 53elliptical orbits 59, 65orbital elements 158–9rendezvous opportunities 351–2, 354 restricted three-body motion 89 Perturbations gravitation 151–2oblateness 177–8sphere of influence 357–8torque-free motion stability 488 Phase angles 350–3Phasing maneuvers 268–73, 350Physical data 583–4Pitch 459–63, 533, 534–42Pitchover 554Pivots 514, 526–30Plane change maneuvers 290–303 Planetary... see also interplanetary trajectories departure 360–6ephemeris 387–91flyby 375–86rendezvous 368–75 Planets geocentric right ascension-declination152–4 identity conversions 644–5state vectors 645–53 Planning Hohmann transfers 262–4 Point masses 1–32 absolute vectors 20–9force 7–15gravitational potential energy 661–4 kinematics 2–7mass 7–15moments of inertia 417–18moving vector time derivatives 15–20 Newton’s law of gravitation 7–10 Newton’s law of motion 10–15 relative motion 20–9 relative vectors 20–9 Polar coordinates 48Position errors 366–8 Position as a function of time 107–47 circular orbits 108–9elliptical orbits 109–23, 134–5 Index 669 hyperbolic trajectories 125–35 MATLAB algorithms 142, 601–6 parabolic trajectories 124–5universal variables 134–44 Position vectors absolute 20–9equatorial frames 175geocentric 175, 194–201Gibb’s method 194–201gravitational potential energy 657 gravity-gradient stabilization 530–1, 536 inertial frames 34–7Lagrange coefficients 78–89, 141–4 MATLAB algorithms 159–61, 175, 232, 604–13,626–31, 641–8 nutation dampers 496orbit formulas 47–9perifocal frame 76–7point masses 2–7, 20–9preliminary orbit determination 218–19,223–4, 228, 236–8,242–3, 247–9 restricted three-body motion 90–1 rigid-body dynamics 400–8, 410–14 satellite attitude dynamics 496, 510, 530–1, 536 three dimensional geocentric orbits 156–8 two-body motion 34–7, 47–9, 78–89 two-impulse maneuvers 330, 336 yo-yo despin 510 Post-flyby orbits 379–86Potential energy 36–7, 657–60Pound 10Powered ascent phase 293Pre-flyby ellipse 380–1Precessiondouble-gimbaled control moment gyros 527 nutation dampers 497–8rigid-body dynamics 451–4satellite attitude dynamics 480–4, 497–8, 508 spinning tops 444–8thrusters 508torque-free motion 480–4 Preliminary orbit determination 193–254 angle measurements 228–50 Gauss’s method 235–50, 631–41 Gibbs method 194–201, 613–16 Lagrange coefficients 204, 207–10, 237–9, 249 Lambert’s problem 202–13, 616–21 MATLAB algorithms 198, 208, 217, 232, 242–50,613–41 range measurements 228–35sidereal time 213–18topocentric coordinate systems 218–28 Primed systems 165, 168, 424–5Principal directions 425–8, 431–5 Principal moments of inertia 419, 426–8, 431–6, 457 Prograde... coasting flights 393–4 precession 481–4trajectories 203 Prolate bodies 481–2, 494Propellant field-free space restricted staging 560–70 Lagrange multiplier method 573–8 mass 256–7, 364–6rocket vehicle dynamics 555–9, 560–70, 573–8 thrust equation 555–7 Propellers 403–4Propulsion 551–79r-bars 316 Radar observations 232, 626–31 Radial distances 85–8Radial release 514, 515–16Radius aiming 71–2, 75, 370–1, 373, 382 apoapse 373azimuth 62capture 372earth’s sphere of influence 358–9 gravitational potential energy 658–60 periapse 360–2, 369, 370, 372–3 perigee 71, 75, 183–7perihelion 384, 385true-anomaly-averaged 61, 62–3 Range measurements 228–35, 626–31 Rates of precession 444–8, 451–4 Rates of spin 444–8, 451–4Regulus 153–4Relative acceleration angular 437point masses 23, 25–6relative motion and rendezvous 317–20 rigid-body kinematics 401–8two-body motion 38 Relative angular... acceleration 437momentum 42–4velocity 350–1 Relative linear momentum 412Relative motion 315–40 Clohessy–Wiltshire equations 324–30, 336–7 close-proximity circular orbits 338–40 co-moving reference frames 316–22, 324–30 670 Index Relative motion (continued) linearization of equations of relative motion 322–4 point masses 20–9restricted three-body motion 37, 38, 91 two-impulse maneuvers 330–7 Relative position point masses 22, 24–5preliminary orbit determination 230–1 sphere of influence 356two-body motion 37 Relative vectors 20–9, 37, 230–1, 356 Relative velocity Clohessy–Wiltshire equations 328–9 close-proximity circular orbits 338–40 point masses 22, 25relative motion and rendezvous 317–20 rigid-body kinematics 401–8two-body motion 38two-impulse maneuvers 330, 332–3 Rendezvous 315–40 Clohessy–Wiltshire equations 324–30, 336–7 close-proximity circular orbits 338–40 co-moving reference frames 316–22, 324–30 equations of relative motion 322–4 Hohmann transfers 262–4interplanetary trajectories 349–54, 368–75 relative motion equations 322–4 two-impulse maneuvers 330–7 Restricted staging 560–70Restricted three-body motion 89–101 Retrofire 262–4Retrograde orbits 203, 295–6, 481–4 Right ascension oblateness 178–81, 185–6planetary ephemeris 388preliminary orbit determination 222–3,225–6, 230–2 state vectors 155–8three dimensional orbits 149–54, 159, 160, 162 Rigid-body dynamics 399–463 Chasles’ theorem 399–400equations of rotational motion 410–14 equations of translational motion 408–10 Euler angles 448–59Euler’s equations 435–40inertia 414–35kinematics 400–8kinetic energy 441–3moments of inertia 414–35moving vector time derivatives 15–16 parallel axis theorem 428–35pitch 459–63plane change maneuvers 302–3 roll 459–63rotation of the ellipse 302–3rotational motion 410–14satellite attitude dynamics 498–503 spinning tops 443–8translational motion 408–10yaw 459–63 Rocket equation 552Rocket vehicle dynamics 551–79 equations of motion 552–5field-free space restricted staging 560–70 impulsive orbital maneuvers 256–7 Lagrange multiplier method 570–8 motors 256–7optimal staging 570–8 restricted staging 560–70rocket performance 555–60staging 560–78thrust equation 555–7 Rods 418–19, 430–5Roll 459–63, 533, 534–42Roots 426, 427, 434Rotating platforms 447–8, 456–9 Rotation axis of 150 Cartesian coordinate systems 169–72 coordinate transformations 169–72 geocentric equatorial frames 173–6 matrices 460–3perifocal frames 173–6three dimensional orbits 150, 169–72 true anomaly 284–5 Rotational... equations of motion 410–14, 517–21 kinetic energy 488–91, 493–4, 509–12 motion equations 410–14, 517–21 Rotationally symmetric satellites 477 Round-trip missions 353–5 Routh–Hurwitz stability criteria 501–3 Satellite attitude dynamics 475–550 axisymmetric dual-spin satellites 518–21, 529–30 coning maneuvers 503–5, 507control thrusters 504–9 despin mechanisms 509–16dual-spin spacecraft 491–5gravity-gradient stabilization 530–42 Index 671 gyroscopic attitude control 516–30 gyrostats 491–5nutation dampers 495–503, 509 passive energy dissipaters 495–503 rigid-body dynamics 399thrusters 504–9torque-free motion 476–86, 487–91, 518–21, 529–30 yo-yo despin 509–16 Satellites dual-spin 518–21, 529–30earth 149, 183–7geocentric 276–9orientation 540–2 Saturation 526Second order differential equations 326–8 Second zonal harmonics 177Semi-latus rectum 49Semimajor axis elliptical orbits 62, 65equation 134–5hyperbolic trajectories 75phasing maneuvers 269planetary ephemeris 387, 388three dimensional orbits 158–9, 184 Semiminor axis equation 135Sensitivity analysis 366–8Series two-stage rockets 562, 563–70 SEZ seeSouth-East-Zenith Shafts on rotating platforms 456–9 Sidereal time 213–18, 231, 623–6 Single stage rockets 566–7Single-spin stabilized spacecraft 492–5 Slant ranges 239–41, 246, 249Slug 11–12Sounding rockets 552, 554–5, 558–9 South-East-Zenith (SEZ) frame 223Space cones 482 Spacecraft trajectories 393, 648–55 Specific energy circular orbits 52elliptical orbits 59Hohmann transfers 257hyperbolic trajectories 73non-Hohmann transfers 275–6 three dimensional orbits 158–9 Specific impulse impulsive orbital maneuvers 256–7 rocket vehicle dynamics 552, 557, 559, 562, 564,570–8 Speed circular orbits 53–4elliptical orbits 63excess 73–4hyperbolic trajectories 133parabolic trajectories 65planetary departure 362yo-yo despin 511, 515–16 Sphere of influence 354–9, 366–75, 392–4, 648–55 Spheres 657–60Spherically symmetric distribution 657–60 Spin accelerations 521–5angles 508 rates 451–4, 480, 527stabilized spacecraft 475 Spinning rotors 447–8Spinning tops 443–8Stability dual-spin spacecraft 492–5gravity-gradient stabilization 533–4 nutation dampers 500–3spinning satellites 475torque-free motion 487–91 Stable pitch oscillation frequency 537 Staging 552, 560–78Stars 152–3 State vectors geocentric equatorial frame 154–8, 175–6 MATLAB algorithms 159–61, 175, 232, 604–13,626–31, 641–8 non-Hohmann trajectories 393–4 orbital elements 159–61, 175, 606–13 planetary ephemeris 387–9 preliminary orbit determination 228–9,232, 237–9, 244–50 three dimensional orbits 154–64, 175–6, 184 two-impulse maneuvers 330–2 Step mass 573–5, 576–8Strap-on boosters 563Structural ratios 560–1, 564, 568, 570–8 Stumpff functions 135–6, 142, 204–7, 600–1 Sun-synchronous orbits 180–7Sunlit side approaches 379, 382, 384–6 Synodic period 351–2, 354 Tandem two-stage rockets 562, 563–70 Tangential release 514–16Target vehicles 322–40T ension 514, 515–16Three dimensional curvilinear motion 1–7 Three dimensional orbits 149–91 celestial sphere 149–54coordinate transformations 164–76 declination 149–54earth’s oblateness 177–87geocentric equatorial frame 154–8, 172–6 672 Index Three dimensional orbits (continued) geocentric right ascension-declination149–54 oblateness 177–87orbital elements 158–64patched conics 391–7perifocal frame transformations 172–6 right ascension 149–54state vectors 154–64 Three-body systems 41–2, 355–9, 587–94 Three-stage launch vehicles 577–8 Thrust equation 555–7Thrust-to-weight ratio 558Thrusters 504–9Tilt angles 446Time see also position as a function of time dependent vectors 18–20derivatives Lagrange coefficients 80–3, 85–7, 603–5 moving vectors 15–20relative motion 25–9 Hohmann transfers 265–8lapse 601–6manned Mars missions 354–5to perigee 211, 212–13satellite attitude dynamics 505, 515–16 since periapse 108–9since perigee 131, 158–9, 184–5, 208–11, 287–8 Titan II 562T opocentric coordinates 218–28, 230–5 To rqu e axial 523–5free motion 476–86, 487–91, 518–21, 529–30 rigid-body dynamics 413–14satellite attitude dynamics 513–14, 521–5, 533Trailing-side flyby 375, 376 Transfer ellipses 297–8, 348–50Transfer times 203, 204, 352, 354 Transformation matrices MATLAB algorithms 175moments of inertia 421–8orthogonal 320, 421–8, 449–50 pitch 460–3relative motion and rendezvous 320 rigid-body dynamics 421–8, 449–50, 452, 456 roll 460–3satellite attitude dynamics 536 three dimensional orbits 166–72, 174–6, 186–7 topocentric horizon system 225–8 torque-free motion 484–6two-impulse maneuvers 330–2 yaw 460–3 Translational motion equations 408–10 Transverse bearing loads 459True anomalies averaged orbital radius 61, 62–3 elliptical orbits 110, 135hyperbolic flyby 382hyperbolic trajectories 69, 75, 125–6, 132–3 Lagrange coefficients 80–2, 83–5 non-Hohmann transfers 279–85 parabolic trajectories 68–9, 124–5 plane change maneuvers 301position as a function of time 108–9, 110, 124–6,132–3, 135, 139–42 preliminary orbit determination 202–13 rendezvous opportunities 350three dimensional orbits 158–9, 161, 163–4, 184 time since periapse 108–9universal variables 139–42 Turn angles 70, 75, 369–70, 378, 382 Two-body motion angular momentum 42–50energy law 50–1equations of motion 34–42equations of relative motion 37–42 hyperbolic trajectories 69–76inertial frame equations of motion 34–7 Lagrange coefficients 78–89mutual gravitational attraction 33–105 orbit formulas 42–50parabolic trajectories 65–9perifocal frame 76–8restricted three-body motion 89–101 three dimensional orbits 149 Two-impulse maneuvers 257–73, 330–7 Two-stage rockets 562, 563–70 Unit matrices 167 Unit triads 168–9Unit vectors gravitational potential energy 657 Lagrange coefficients 78–9moments of inertia 422point masses 5–7, 21three dimensional orbits 164–72 Units of force 10–15Universal anomaly 134–44, 601–3 Universal Kepler’s equation 138–9, 601–3 Universal Time (UT) 213–18, 388, 621–8, 641–8 Universal variables 134–44Unprimed systems 168, 424–5 Index 673 UTseeUniversal time Vectors 33–105 see also position...; state...; two-body motion;unit...; velocity... direction cosine 242–3, 244eigenvectors 426–8moving 15–20orthogonal unit 5–7orthonormal basis 165preliminary orbit determination 194–201 relative 20–9, 37, 230–1, 356time dependent 18–20time derivatives 15–20weight 444–5 Velocity see also delta-v... errors 366–8escape 66, 73excess 360–6, 368–75, 392–4geocentric orbits 156–8Hohmann transfers 261 non-Hohmann transfers 274–5 plane change maneuvers 290–1, 301–2 relative motion and rendezvous 316–20 rocket vehicle dynamics 560–70 Vec tors absolute 20–9geocentric equatorial frame 175 Lagrange coefficients 78–89, 141–4 MATLAB algorithms 159–61, 175, 604–13, 626–31,641–8 perifocal frame 76–7point masses 2–7, 16–18, 20–9 preliminary orbit determination 194–201,203–4, 228, 241–2, 250restricted three-body motion 91 rotations 292–3, 299satellite attitude dynamics 496, 510–11 two-impulse maneuvers 330 Venus ephemeris 152–3Venus flyby 379–80Vernal equinox 150–4Visible surface areas 54–5 Wait time 353–4 Weight 7–10, 36Weight vectors 444–5Wobble angles 481–2 Y aw 403, 459–63, 533–42 Y o-yo despin 509–16 Zonal variation 177–87 This page int entionally left blank A road map dA dt=h 2h = r × r υ⊥ =h r υr =µ he sin θ Kepler's second lawConservation of mechanical energy The orbit formula (Kepler's first law)Newton's laws Definition2-body equation of relative motion T =2π µa3 2Kepler's third lawKepler's equations relating true anomaly to timer = − µ r3rυ2 2−µ r= const r =h2 µ1 1 + e cos θF = ma Fg = Gm1m2 r2ˆ ur t =h3dϑ (1 + e cosϑ)2 0θ ∫µ2··· See Appendix B (p. 585) for more information.