Appendix F v2

Appendix F v2 PhL 12.15.16 D.6 Equation of Motion for the Dumbbell Satellite in Frame S' We now place the dumbbell satellite in a more general orientation: Description of the Figure Half the battle is having a clear picture of what is going on and we shall expend many words to describe the above picture. It shows the dumbbell satellite in orbit with a completely arbitrary orientation. The two masses m1 and m2 are connected by a massless stick of length r'1+r'2 = s which we have not drawn. If we find that this stick is always in tension for some situation, we can replace it in that situation with a non-stretching massless tether. If the stick were to go into compression, the replacement tether would lose its linear shape and we don't want to deal with that problem. As in our earlier pictures, Frame S is an inertial frame whose center is the center of the Earth and which is fixed with respect to the stars. Frame S' rotates as the satellite center of mass (Frame S' origin) moves around its orbit. At time t = 0, the axes of both Frames align with each other. At any time, axes y,z and y',z' are in the plane of paper. The axis z always points to the center of mass point which is the origin of Frame S'. Thus b = b = b' at any time. The axes x,x' always point direction out of the plane of paper (normal to paper). The orbital rotation vector ω0 also points out of the plane of paper and recall that ω0 = 2π/T0 where T0 is about 88 minutes for a low-Earth orbit, although we have in mind an orbit at any altitude. Frame S' has spherical coordinates r',θ',φ' define in the usual "physics" convention as shown. Angle θ' is the polar angle down from the "vertical" z' axis, φ' is the azimuthal angle measured from the x' axis toward the t' axis. Mass m1 has Frame S' position r'1 and Frame S position r1.These two vectors as drawn lie on the viewer's side of the plane of paper and we show them in red. The corresponding vectors r'2 and r2 lie to the back of the plane of paper and are shown in blue. So red is near, blue is far. There are two large non-right triangles in the picture and neither lies in the plane of paper. The triangle on the right side has internal angles α1 and β1 as shown, while that on the left side has angles α2 and β2. The particle of mass m1 has r'1 = (r'1,θ',φ') as its spherical coordinates. Because the Frame S' origin is the center of mass, the particle of mass m2 has r'1 = (r'2,π-θ',φ'+π). The vector r'2 is collinear with r'1 but points in the opposite direction from r'1. The relation m1r'1 = m2r'2 (see below) shows how the masses and the vector lengths are related. The particle of mass m1 is constrained to lie on a sphere of radius r'1, while the particle of mass m2 is constrained to lie on a sphere of radius r'2. The picture assumes m1 < m2 so r'1 > r'2. If m1 lies at a point on its sphere, m2 lies on the inverse point but on its sphere, as the spherical coordinates above show. If the masses are the same, the two spheres coincide. Due to these constraints on the vectors r'1 and r'2, the corresponding velocity vectors v'1 and v'2 of the two masses must be tangential to their respective spheres. Thus for example v'1 = v'1θ ' + v'1φ ' , whereas r'1 = r'1 ' . This is an example of a problem where it would probably be better to use the "swap notation" to get rid of the sea of primes we shall encounter, but we shall plod on with all the primes. The circled dot at the bottom is the center of the Earth (mass ME) and we have drawn the Earth surface in green. The blue circle is the orbit of the center of mass of the satellite (Frame S' origin) and it lies in the plane of paper. Note that ω0 is for the orbit and has nothing whatsoever to do with the rotation of the Earth. The above picture is the same whether or not the Earth rotates at its 24 hour ω1 rate about some obscure 1 axis (not shown!). Triangle Trig The internal angles of the red triangle (not in the plane of paper) are β1, α1 and π-θ' at the origin. Thus we know from one of the three laws of cosines that r12 = b2 + r'12 - 2br'1cos(π-θ') = b2 + r'12 + 2br'1cosθ' The internal angles of the blue triangle (not in the plane of paper) are β2, α2 and θ' at the origin. Thus we know that r22 = b2 + r'22 - 2br'2cos θ' The law of sines for the two triangles tells us that = = = - = = Summary: r12 = b2 + r'12 + 2br'1cosθ' r22 = b2 + r'22 - 2br'2cos θ' = = - = = + Basic Dumbbell Facts and Angular Momentum Recalling (D.2.8), m1r'1 = - m2r'2 m1r'1 = m2r'2 , r'2/r'1 = m1/m2 , '2 = – '1 (D.2.8) we apply time derivative ∂S' once and then again to obtain two new sets of facts, m1v'1 = - m2v'2 m1v'1 = m2v'2 , v'2/v'1 = m1/m2 , '2 = – '1 (D.6.2) m1a'1 = - m2a'2 m1a'1 = m2a'2 , a'2/a'1 = m1/m2 , '2 = – '1 (D.6.3) For angular momentum we take our reference point to be c' = 0 in Frame S' and thus c = b in Frame S. Then m1v'1 = - m2v'2 p'1 = -p'2 L'(0) = m1r'1x v'1 + m2r'2x v'2 (D.6.4) = m1r'1x (v'1 – v'2) = m1r'1x (v'1 + (m1/m2)v'1) = m1( 1+ (m1/m2)) r'1x v'1 M L L/T = ML2/T = (m1/m2) M r'1x v'1 . Now we know that v = vθ + vφ so v'1 = v'θ1' + v'φ1 ' Then r'1x v'1 = r'1'1x[v'θ1' + v'φ1 '] = r'1(v'θ1' – v'φ1') Then since vθ = r vφ = r sinθ we have v'θ1 = r'1 ' v'φ1 = r'1' sinθ' Thus L'(0) = (m1/m2) M r'1(v'θ1' – v'φ1') = (m1/m2) M r'1( r'1 '' – r'1' sinθ'') = (m1/m2) M r'12( '' – ' sinθ'') // dim = ML2/T The dumbbell has no angular momentum around the 1' axis which seems very reasonable since it consists of two point masse aligned with 1' I guess I will want to do everything in sphericals and then look at the torque equation? Sound very complicated. But I think these are the natural coordinates to use!!!! Taking a ∂S' time derivative gives this rate of change of angular momentum in Frame S' '(0) = m1 [ r'1x a'1 + v'1x v'1 ] + m2 [ r'2x a'2 + v'2x v'2 ] = m1 r'1x a'1+ m2 r'2x a'2 . = m1r'1x (a'1- a'2) = m1( 1+ (m1/m2)) r'1xa'1 =(m1/m2) M r'1xa'1 a'1 = a'r1 '1 + a'θ1'1 + a'φ1 '1 = a'r1 '1 + a'θ1' + a'φ1 ' r'1xa'1 = r'1 '1x [ a'r1 '1 + a'θ1' + a'φ1 ' ] = r'1(a'θ1 ' – a'φ1) ' So we find that '(0) = (m1/m2) M r'1 [ a'θ1 ' – a'φ1 ' ] We know that aθ = r - r 2 sinθ cosθ = r( - 2 sinθ cosθ) aφ = 2 r cosθ + rsinθ = r(2cosθ + sinθ ) Therefore a'θ1 = r'1(' - '2 sinθ' cosθ') a'φ1 = r'1(2''cosθ' + 'sinθ' ) And then '(0) = (m1/m2) M r'12 [ (' - '2 sinθ' cosθ') ' – (2''cosθ' + 'sinθ' ) ' ] // ML2/T2 Here are the conclusions so far: L'(0) = (m1/m2) M r'12( '' – ' sinθ'') '(0) = (m1/m2) M r'12 [ (' - '2 sinθ' cosθ') ' – (2''cosθ' + 'sinθ' ) ' ] Obviously if L' has no r' component, so L'r ≡ 0, then one must also have 'r ≡ 0. So far so good. The equation of motion for the satellite within Frame S' is given by (11.3.4), N'(0)eff = '(0) Our next task then is to compute the total effective torque on the satellite in Frame S' which from (11.3.5) is N'(0)eff = N(b) + N'(0)fict (11.3.5) (D.6.9) Here N(b) is the Frame S torque on the satellite relative to point b in Frame S, and N'(0)fict is the fictitious torque that arises because Frame S' is a rotating frame of reference. ok to here D.7 The true torque on the Dumbbell Satellite in Frame S The total torque in Frame S on the satellite is given by N(b) = r'1 x F1 + r'2 x F2 L M L/T2 = ML2/T2 (D.7.1) where all four of these vectors are shown in the figure. Recall from (D.2.13) that F1 = - (GMEm1/r12) 1 = - (GMEm1/r13) r1 F2 = - (GMEm2/r22) 2 = - (GMEm2/r23) r2 . (D.2.13) We really should include the stick tension/compression T since these are supposed to be total forces, but since r'1 x T = r'2 x T = 0 so we can ignore T. Therefore r'1 x F1 = r'1 x [ - (GMEm1/r13) r1 ] = - (GMEm1/r13) r'1 x r1 r'2 x F2 = r'2 x [ - (GMEm2/r23) r2 ] = - (GMEm2/r23) r'2 x r2 (D.7.2) Now we know that r1 = b + r'1 so that r'1 x r1 = r'1 x (b + r'1) = r'1 x b = r'1b( '1 x ') Now from App E we know thast = cosθ - sinθ so x = x ( cosθ - sinθ ) = - sinθ so therefore '1x ' = - sinθ' ' and we then find that r'1 x r1 = - r'1bsinθ' ' Next, r2 = b + r'2 r'2 x r2 = r'2 x (b + r'2) = r'2 x b = r'2b( '2 x ') = - r'2b( '1 x ') = (- r'2b) ( - sinθ' ') = + r'2bsinθ' ' Therefore: r'1 x F1 = - (GMEm1/r13) r'1 x r1 = - (GMEm1/r13)(- r'1bsinθ' ') = GMEm1b (r'1/r13) sinθ' ' r'2 x F2 = - (GMEm2/r23) ( r'2bsinθ' ') = - GMEm2b (r'2/r23) sinθ' ' = - GMEm1b (r'1/r23) sinθ' ' Then N(b) = r'1 x F1 + r'2 x F2 LML/T2 = ML2/T2 = GMEm1b (r'1/r13) sinθ' ' - GMEm1b (r'1/r23) sinθ' ' = GMEm1b r'1sinθ' [ (1/r13) - (1/r23)] ' force x L where r12 = b2 + r'12 + 2br'1cosθ' r22 = b2 + r'22 - 2br'2cos θ' How do I interpret this result that the torque only has one component and it is about the ' axis ?? If you were look at φ' = π/2, then we have a planar picture and the torque clearly is about ' and has no components in the other two directions since everything is in a plane. In that case ' = - ' so it makes sense. Quick checks: If the satellite is vertically aligned, θ' = 0,π, then sinθ'=0 and N(b) = 0 as expected (no moment arms). If the satellite is horizontally aligned and m1= m2, then r1= r2 so N(b) = 0 (balanced moment arms). If m2> m1, then r'2< r'1 so for horizontal alignment one has r2<r1 so [ (1/r13) - (1/r23) ] < 0. Then if m1 is on the right, we have θ' = π/2 and sinθ' = 1 and then (D.7.7) has N(b) = -(positive)'. But for this orientation ' = - ' so we get N(b) = (positive)' (D.7.8) In this case the lever arms balance in the sense that m2r'2 = m1r'1, but m2 is closer to Earth center so it feels the stronger force and so we expect N(b) = (positive)' Following needs repair, will have sinθ ↔ cosθ, but the result is invariant under this change! If we now assume as before that r1,r2,b >> r'1, r'2 we can approximate [ (1/r23) - (1/r13) ] by adding on to the Maple code shown in (D.3.5) to get where recall e1 = ε1 ≡ (r'1/b). This time there is a leading linear term in ε1 so then (1/r23) - (1/r13) ≈ -3 (r'1/b) = 3r'1sinθ'/ (μ2b4) (D.7.9) Then I think the result will be (1/r23) - (1/r13) ≈ 3r'1cosθ'/ (μ2b4) so then N(b) = GMEm1b r'1sinθ' [ (1/r13) - (1/r23)] ' ≈ - GMEm1b r'1sinθ' ( 3r'1cosθ'/ (μ2b4) ) ' = - 3GME(m1/μ2) b-3r'12sinθ'cosθ' ' To this order of approximation, the torque N(b) vanishes when the satellite is horizontally aligned as well as when it is vertically aligned, and this is due to r1 ≈ r2 in the horizontal case. D.8 The fictitious torque on the Dumbbell Satellite in Frame S' Recall the general fictitious torque expression given in (11.3.10), acting on a single particle of mass m, N'(c')fict = - (r'-c') x [ mS +mω x (ω x r') + 2m ω x v' + m x r'] + m(' + ω x c' + S) x ( v' + ω x r' + S) – m' x v' . (11.3.10) (D.8.1) Here ω is the angular rotation rate of the satellite about the Earth. Our application has the torque center at c' = 0 (and c = b) so this simplifies somewhat to N'(0)fict = - mr' x [ S +ω x (ω x r') + 2ω x v' + x r'] + S x ( mv' + ω x [mr']) . (D.8.2) frame centrifugal Coriolis Euler Recall from (8.1.8) that the square bracket in the above is - F'fict/m so we can trace the origin of the terms. Now write this separately for each of the two masses of our satellite, N'(0)f,1 = - m1r'1 x [ S +ω x (ω x r'1) + 2ω x v1' + x r'1] + S x ( m1v'1 + ω x[ m1r'1]) N'(0)f,2 = - m2r'2 x [ S +ω x (ω x r'2) + 2ω x v'2 + x r'2] + S x ( m2v'2 + ω x [m2r'2]) . (D.8.3) In the second line, use (D.2.8) to replace m2r'2 = - m1r'1 and (D.6.2) to replace m2v'2 = - m1v'1 : N'(0)f,2 = + m1r'1 x [ S +ω x (ω x r'2) + 2ω x v'2 + x r'2] + S x ( -m1v'1 + ω x [-m1r'1]) . (D.8.4) Next, add the two torques to get the total fictitious torque on the satellite, N'(0)fict = N'(0)f,1 + N'(0)f,2 = - m1r'1 x [ S +ω x (ω x r'1) + 2ω x v1' + x r'1] + S x ( m1v'1 + ω x[ m1r'1]) + m1r'1 x [ S +ω x (ω x r'2) + 2ω x v'2 + x r'2] + S x ( -m1v'1 + ω x [-m1r'1]) = - m1r'1 x { ω x (ω x [r'1- r'2]) + 2ω x [v1'-v'2] + x [r'1- r'2] } . (D.8.5) centrifugal Coriolis Euler = - (m1/m2)M r'1 x [ ω x (ω x r'1) + 2ω x v1' + x r'1 ] All terms involving S and S have cancelled out. For the Earth orbit we have = 0 and ω = ω'. Now consider this vector identity" C x [A x (A x C)] = C x [ (AC)A - A2C ] = (AC) C x A = – (AC) A x C Then r'1 x [ ω x (ω x r'1) ] = – (ωr'1) (ω x r'1) = - ω2r'12 ( ' '1)( ' x '1) = - ω2r'12 sinθ'cosφ'( ' x '1) Now ' x '1 = [cosφ'sinθ' '1 + cosφ'cosθ' ' - sinφ' '] x '1 = [ cosφ'cosθ' ' - sinφ' '] x '1 = [ - cosφ'cosθ' ' - sinφ' '] = - cosφ'cosθ' ' - sinφ' ' Therefore r'1 x [ ω x (ω x r'1) ] = ω2r'12 sinθ'cosφ' [ cosφ'cosθ' ' + sinφ' ' ] Next, the Coriolis term involves, r'1 x (ω x v1') = (r'1v1')ω - (r'1ω)v1' // A x (B x C) = (AC)B - (AB)C But v1' is tangent to the r'1 sphere so (r'1v1') = 0. Then r'1 x (ω x v1') = - (r'1ω)v1' = - r'1ω( '1')v1' = - r'1ωsinθ'cosφ'v1' We now have a complicated expression for N'(0)fict : N'(0)fict = - (m1/m2)M r'1 x [ ω x (ω x r'1) + 2ω x v1' ] = - (m1/m2)M { r'1 x [ω x (ω x r'1)] + 2 r'1 x (ω x v1') } = - (m1/m2)M { ω2r'12 sinθ'cosφ' [ cosφ'cosθ' ' + sinφ' ' ] - 2r'1ωsinθ'cosφ'v1' } Now use v1' = v'θ1' + v'φ1 ' = r'1 '' + r'1 ' sinθ' ' and we then have = - (m1/m2)M { ω2r'12 sinθ'cosφ' [ cosφ'cosθ' ' + sinφ' ' ] - 2r'12ωsinθ'cosφ'[ '' + ' sinθ' '] } = - (m1/m2)M * {A ' + B '} where A = ω2r'12 sinθ'cosφ'sinφ' - 2r'12ωsinθ'cosφ'' = ωr'12 sinθ' cosφ' ( ωsinφ' -2' ) B = ω2r'12 sinθ'cosφ' cosφ'cosθ' - 2r'12ωsinθ'cosφ' ' sinθ' = ωr'12sinθ'cosφ' ( ω cosφ'cosθ' - 2' sinθ' ) Restate the result: N'(0)fict = - (m1/m2)M {A ' + B '} where A = ωr'12 sinθ' cosφ' ( ωsinφ' -2' ) B = ωr'12sinθ'cosφ' ( ω cosφ'cosθ' - 2' sinθ' ) Do it again Sam N'(0)fict = - (m1/m2)Mωr'12sinθ'cosφ' { ( ωsinφ' -2' ) ' + ( ω cosφ'cosθ' - 2' sinθ' ) ' } and again N'(0)fict = - (m1/m2)Mωr'12sinθ'cosφ' [ ( ωsinφ' -2' ) ' + ( ω cosφ'cosθ' - 2' sinθ' ) '] Not too bad! Last time I did this I got the complete result being 0, so what happened? Well set φ' = π/2 to get the planar situation I treated last time. Then sinφ' = 1 and cosφ' = 0 and you see N'(0)fict = 0 !! So this agrees with what I did previously for my special orientation of the satellite (which therefore missed most of the action!) How might one interpret this simple result? We examine the pieces of N'(0)fict= 0: (a) the "frame effects" due to the motion of b (S and S) are equal and opposite for the two masses because the origin of Frame S' is at the center of mass causing m2r'2 = - m1r'1, as shown in (D.8.5). (b) within Frame S', the centrifugal acceleration term ω x (ω x r'1) = -ω2r'1 tries to push m1 to a larger radius. But r'1 is constrained to lie on a sphere of radius r'1 so m1 cannot go to a larger radius. This centrifugal acceleration is neutralized by part of the tension in the stick which we avoided talking about. This centrifugal term, by the way, involves "the short vector" r'1 and not the long vector r1. We discussed this situation in Section 8.2 for an Earth-based Frame S'. In our current context, the picture that corresponds to Fig (8..2.8) is the following (D.8.11) The arrow in each location during the orbit represents the vector r'1 where we assume that other effects are turned off so r'1 stays fixed in Frame S'. When these arrows are transferred to the picture on the right with common tails, we see that r'1 does in fact go around in a circle of radius r'1 and that is why the corresponding centrifugal force acting on m1 is -ω2r'1 . (c) Within Frame S', the Coriolis force - 2m1 ω x v1' tries to deflect mass m1 "to the right" in Fig (D.6.1). But again, r'1 is constrained to lie on a sphere of radius r'1 so m1 cannot deflect to a different radius. This Coriolis force is neutralized by the rest of the tension in the stick. D.9 The total torque on the Dumbbell Satellite in Frame S' After much effort, we have arrived at this set of results for the satellite : L'(0) = (m1/m2) M r'12( '' – ' sinθ'') '(0) = (m1/m2) M r'12 [ (' - '2 sinθ' cosθ') ' – (2''cosθ' + 'sinθ' ) ' ] N(b) = GMEm1b r'1sinθ' [ (1/r13) - (1/r23)] ' r12 = b2 + r'12 + 2br'1cosθ' r22 = b2 + r'22 - 2br'2cos θ' N'(0)fict = - (m1/m2)Mωr'12sinθ'cosφ' [ ( ωsinφ' -2' ) ' + ( ω cosφ'cosθ' - 2' sinθ' ) '] It has taken a LONG TIME to get this point on 12.16.16, Now use the orbit equation (8.6.7) applied to the current situation, GME = ω2b3 large b means small ω Then we have N(b) = ω2b4m1r'1sinθ' [ (1/r13) - (1/r23)] ' T-2L5M/L3 = ML2/T2 ok N'(0)fict = - (m1/m2)Mωr'12sinθ'cosφ' [ ( ωsinφ' -2' ) ' + ( ω cosφ'cosθ' - 2' sinθ' ) '] = MT-1L2 * T-1 What is the relative size of these terms? Maybe write this way: N(b) = ω2b4m1r'1sinθ' [ (1/r13) - (1/r23)] ' T-2L5M/L3 = ML2/T2 ok N'(0)fict = - (m1/μ2)ω2r'12 sinθ'cosφ' [ ( sinφ' - 2('/ω) ) ' + ( cosφ'cosθ' - 2('/ω) sinθ' ) '] Magnitudes? N(b)2 = (ω2b4m1r'1sinθ')2 [ (1/r13) - (1/r23)]2 N'(0)fict2 = [(m1/μ2)ω2r'12 sinθ'cosφ']2 { ( sinφ' - 2('/ω) )2+ ( cosφ'cosθ' - 2('/ω) sinθ' )2 } The ratio is then N'(0)fict/ N(b) = / [ (1/r13) - (1/r23)] = / [ (1/r13) - (1/r23)] So two glaring issues: dealing with the difference in the denom, the with ω ratios in the num. So I don't really have much to say yet about the relative size of these terms. The Equations of Motion Start with '(0) = (m1/m2) M r'12 [ (' - '2 sinθ' cosθ') ' – (2''cosθ' + 'sinθ' ) ' ] N(b) = ω2b4m1r'1sinθ' [ (1/r13) - (1/r23)] ' ≈ ω2b4m1r'1sinθ' [ (1/r13) - (1/r23)] ' ≈ ω2b4m1r'1sinθ' [ -3r'1cosθ'/ (μ2b4)] ' ≈ -3ω2(m1/μ2)r'12sinθ' cosθ' ' N'(0)fict = - (m1/m2)Mωr'12sinθ'cosφ' [ ( ωsinφ' -2' ) ' + ( ω cosφ'cosθ' - 2' sinθ' ) '] Is there some limit where one or the other torque dominates? Write again as N(b) ≈ -3ω2(m1/μ2)r'12sinθ' cosθ' ' N'(0)fict = - (m1/μ2)ωr'12sinθ'cosφ' [ ( ωsinφ' -2' ) ' + ( ω cosφ'cosθ' - 2' sinθ' ) '] The magnitude ratio factor is [ -3ω2(m1/μ2)r'12sinθ' cosθ' / - (m1/μ2)ωr'12sinθ'cosφ' ] = [ 3ωcosθ' / cosφ' ] and then Nfict/Ntrue = (cosφ'/3ωcosθ') = (cosφ'/3cosθ') I want an argument that makes this small. Only if φ' = π/2 and it is planar is this small! Case 1: vertically aligned. Then cosθ' = 1 and sinθ' = 0 and the above gives Nfict/Ntrue = (cosφ'/3ω) Not sure what that means. Limit #1: Let b → ∞ so that ω → 0. Then both torques vanish and the result is no torque at all. But in the limit not quite there we might say N(b) ≈ -3ω2(m1/μ2)r'12sinθ' cosθ' ' N'(0)fict = - (m1/μ2)ωr'12sinθ'cosφ' [ ( -2' ) ' + ( - 2' sinθ' ) '] and then the fictional force dominates! We then have in the ω→0 limit, '(0) = (m1/μ2) r'12 [ (' - '2 sinθ' cosθ') ' – (2''cosθ' + 'sinθ' ) ' ] N'tot ≈ - (m1/μ2)ωr'12sinθ'cosφ' [ ( -2' ) ' + ( - 2' sinθ' ) '] This gives in this limit two equations of motion (m1/μ2) r'12 [ (' - '2 sinθ' cosθ') ] = - (m1/μ2)ωr'12sinθ'cosφ' ( - 2' sinθ' ) ' - (m1/μ2) r'12 (2''cosθ' + 'sinθ' ) = - (m1/μ2)ωr'12sinθ'cosφ' ( -2' ) ' or [ (' - '2 sinθ' cosθ') ] = -ωsinθ'cosφ' ( - 2' sinθ' ) ' - (2''cosθ' + 'sinθ' ) = - ωsinθ'cosφ' ( -2' ) ' or ' - '2 sinθ' cosθ' = +2ωsin2θ'cosφ' ' ' 2''cosθ' + 'sinθ' = - ωsinθ'cosφ' ( 2' ) ' or ' -sinθ' cosθ' '2 - 2ωsin2θ'cosφ' ' = 0 sinθ' ' + 2cosθ''' + 2 ωsinθ'cosφ'' = 0 They are very ugly, but for the very first time ever I have something I can write down!! The equations are horribly coupled. Maybe in my limit I can ignore the last terms as ω→0, huge high orbit. Then ' -sinθ' cosθ' '2 = 0 ' + 2cotθ''' = 0 but these equations are inscrutable. Suppose instead I assume close to vertical. Then ' -sinθ' cosθ' '2 - 2ωsin2θ'cosφ' ' = 0 sinθ' ' + 2cosθ''' + 2 ωsinθ'cosφ'' = 0 ' - θ' '2 - 2ωθ'2cosφ' ' = 0 θ' ' + 2 '' + 2 ωθ'cosφ'' = 0 Suppose we try ansatz that φ = constant. Then we get ' = 0 2 ωθ'cosφ'' = 0 It then just sits there with ' = ' = 0, not very exciting. Let's instead try the large-ω limit and see what happens there. Start over with N(b) ≈ -3ω2(m1/μ2)r'12sinθ' cosθ' ' N'(0)fict = - (m1/μ2)ωr'12sinθ'cosφ' [ ( ωsinφ' -2' ) ' + ( ω cosφ'cosθ' - 2' sinθ' ) '] Now just assume N(b) dominates because it has ω2 sitting there. Then we get '(0) = (m1/m2) M r'12 [ (' - '2 sinθ' cosθ') ' – (2''cosθ' + 'sinθ' ) ' ] N ≈ -3ω2(m1/μ2)r'12sinθ' cosθ' ' This says that we don't change the ang mom about the ' which is the "dipping" axis. The two equations of motion now are (m1/m2) M r'12 [ (' - '2 sinθ' cosθ') = -3ω2(m1/μ2)r'12sinθ' cosθ' (2''cosθ' + 'sinθ' ) = 0 or (' - '2 sinθ' cosθ') = -3ω2sinθ' cosθ' 2''cosθ' + 'sinθ' = 0 Now do ansatz where φ = constant. Then 2nd equation is happy and first says ' = -3ω2sinθ' cosθ' NOW also assume small angle to get ' = -3ω2θ' or ' +3ω2θ' = 0 Finally we get some oscillation which involves ω and involves a factor of 3. What did my PDF say about this? Here from p 124, So they are saying ωlibration = ω and that agrees with my equation, hurray!!! So you have to ignore the fictitious force to get this result! What is the condition for doing that? I show above that φ' = π/2 would do it, the planar case only! What is the Ref 2 above? D.10 Force analysis of the Dumbbell Satellite in Frame S' We use the same picture and picture description of Section D.6 and trig relations stated there. The only forces on a dumbbell mass are gravity and stick tension T. From ** we then write F1 = - (GMEm1/r13) r1 - T '1 = - (GMEm1/r13)( b + r'1) - T '1 F2 = - (GMEm2/r23) r2 - T '2 = - (GMEm2/r23) ( b + r'1) + T '1 These are forces in Frame S. Without going to Frame S' we could write Fi = miai to get - (GMEm1/r13)( b + r'1) - T '1 = m1a1 - (GMEm2/r23)( b + r'1) + T '1 = m2a2 or [- (GMEm1/r13) - T] '1 + [ - (GMEm1/r13)] b = m1a1 [- (GMEm2/r23) + T] '1 + [ - (GMEm2/r23)] b = m2a2 = - m1a1 But we know from Appendix E that b = b = b = b' = bcosθ' '1 - bsinθ' ' so the above equations are then [- (GMEm1/r13) - T] '1 + [ - (GMEm1/r13)] ( bcosθ' '1 - bsinθ' ') = m1a1 [- (GMEm2/r23) + T] '1 + [ - (GMEm2/r23)] ( bcosθ' '1 - bsinθ' ') = m2a2 or [- (GMEm1/r13) - T - (GMEm1/r13)] bcosθ'] '1 + [ (GMEm1/r13) bsinθ'] ' = m1a1 [- (GMEm2/r23) + T - (GMEm2/r23) bcosθ'] '1 + [ (GMEm2/r23) bsinθ'] ' = m2a2 or [- (GMEm1/r13){1+bcosθ'} - T ] '1 + [ (GMEm1/r13) bsinθ'] ' = m1a1 [- (GMEm2/r23){1+bcosθ'} + T ] '1 + [ (GMEm2/r23) bsinθ'] ' = m2a2 These equations are fine, but I don't know really what to do next. They are a correct statement of Newton's Law in Frame S, so no fictitious forces are required. Well we could say b + r'1 = r1 Now take time derivatives in Frame S S + ∂Sr'1 = 1 = v1 // naturals on the right S + ∂2Sr'1 = 1 = 1 = a1 // naturals on the right We have here some "cross derivatives" to worry about. But let's use as is to get [- (GMEm1/r13){1+bcosθ'} - T ] '1 + [ (GMEm1/r13) bsinθ'] ' = m1S + m1(∂2Sr'1) [- (GMEm2/r23){1+bcosθ'} + T ] '1 + [ (GMEm2/r23) bsinθ'] ' = m2S + m2(∂2Sr'2) Basically I am just bumbling along the Frame S' pathway ignoring what I already know. So let's now do things in Frame S' and use are earlier results to get a solution! We have done much of the work already. We know that F1 = [- (GMEm1/r13){1+bcosθ'} - T ] '1 + [ (GMEm1/r13) bsinθ'] ' F2 = [- (GMEm2/r23){1+bcosθ'} + T ] '1 + [ (GMEm2/r23) bsinθ'] ' We must now compute the fictitious force on each mass. F'fict,1 = – m1S – m1ω x (ω x r'1) – 2m1 ω x v'1 – m1 x r'1 F'fict,2 = – m2S – m2ω x (ω x r'2) – 2m2 ω x v'2 – m2 x r'2 Use the simple CMS relations to rewrite the second line m2v'2 = - m1v'1 F'fict,1 = – m1S – m1ω x (ω x r'1) – 2m1 ω x v'1 – m1 x r'1 F'fict,2 = – m2S + m1ω x (ω x r'1) + 2m1 ω x v'1 + m1 x r'1 Now compute A x (A x C) = (AC)A - A2C ω x (ω x r'1) = (ωr'1)ω - ω2 r'1 = (ωr'1)ω ' - ω2r'1 '1 We want to vectors expressed in Frame S' unit vectors and the above is a start. Now since our setup is a Special Case #1 problem, we can write S = x b + ω x (ω x b) = x b + (ωb)ω ' - ω2b But (ωb) = 0 from Fig 1, and = = ' so S = x b - ω2b ' At this point let's set = 0 because the satellite orbit is assumed unperturbed. Then F'fict,1 = + m1ω2b ' – m1[(ωr'1)ω ' - ω2r'1 '1] – 2m1 ω x v'1 F'fict,2 = + m2ω2b ' + m1[(ωr'1)ω ' - ω2r'1 '1] + 2m1 ω x v'1 or F'fict,1 = + m1ω2b ' – m1(ωr'1)ω ' + m1ω2r'1 '1 – 2m1 ω x v'1 F'fict,2 = + m2ω2b ' + m1(ωr'1)ω ' + m1ω2r'1 '1 + 2m1 ω x v'1 or F'fict,1 = m1ω2r'1 '1 + m1ω2b ' – 2m1 ω x v'1 – m1(ωr'1)ω ' F'fict,2 = m1ω2r'1 '1 + m2ω2b ' + 2m1 ω x v'1+ m1(ωr'1)ω ' Next replace ' = cosφ'sinθ' '1 + cosφ'cosθ' ' - sinφ' ' ' = cosθ' '1 - sinθ' ' . v'1 = v'θ1' + v'φ1 ' Notice that ω x v'1 = ω' x [v'θ1' + v'φ1 '] = ωv'θ1 ' x ' + ω v'φ1 ' x ' = ωv'θ1 [cosφ'sinθ' ' + sinφ' '1 ]+ ω v'φ1 [- cosφ'sinθ' ' + cosφ'cosθ' '1] = ω[ v'θ1sinφ' + v'φ1 cosφ'cosθ'] '1 + ω[- v'φ1cosφ'sinθ'] ' + ω[ v'θ1cosφ'sinθ'] ' Things are now a bit messy. Start with F'fict,1 : F'fict,1 = m1ω2r'1 '1 + m1ω2b ' – 2m1 ω x v'1 – m1(ωr'1)ω ' = m1ω2r'1 '1 + m1ω2b [ cosθ' '1 - sinθ' '] – 2m1ω { [ v'θ1sinφ' + v'φ1 cosφ'cosθ'] '1 + [- v'φ1cosφ'sinθ'] ' + [ v'θ1cosφ'sinθ'] '} – m1(ωr'1) [cosφ'sinθ' '1 + cosφ'cosθ' ' - sinφ' '] = m1ω2r'1 '1 + m1ω2b cosθ' '1 - m1ω2b sinθ' ' [ – 2m1ωv'θ1sinφ' – 2m1ω v'φ1 cosφ'cosθ'] '1 + [ 2m1ωv'φ1cosφ'sinθ'] ' + [ – 2m1ωv'θ1cosφ'sinθ'] ' – m1(ωr'1)cosφ'sinθ' '1 – m1(ωr'1) cosφ'cosθ' ' + m1(ωr'1) sinφ' ' = [ m1ω2r'1 + m1ω2b cosθ'– 2m1ωv'θ1sinφ' – 2m1ω v'φ1 cosφ'cosθ' – m1(ωr'1)cosφ'sinθ'] '1 + [- m1ω2b sinθ' + 2m1ωv'φ1cosφ'sinθ'– m1(ωr'1) cosφ'cosθ'] ' + [– 2m1ωv'θ1cosφ'sinθ'+ m1(ωr'1) sinφ'] ' = m1ω[ ωr'1 + ωb cosθ'– 2v'θ1sinφ' – 2 v'φ1 cosφ'cosθ' –(r'1)cosφ'sinθ'] '1 + m1ω[- ωb sinθ' + 2v'φ1cosφ'sinθ'– (r'1) cosφ'cosθ'] ' + m1ω[– 2v'θ1cosφ'sinθ'+ (r'1) sinφ'] ' Meanwhile (r'1) = r'1(''1) = r'1sinθ'cosφ' So far then we know that F'fict,1 = m1ω[ ωr'1 + ωb cosθ'– 2v'θ1sinφ' – 2 v'φ1 cosφ'cosθ' – r'1sinθ'cosφ'cosφ'sinθ'] '1 + m1ω[- ωb sinθ' + 2v'φ1cosφ'sinθ'– r'1sinθ'cosφ' cosφ'cosθ'] ' + m1ω[– 2v'θ1cosφ'sinθ'+ r'1sinθ'cosφ'sinφ'] ' or F'fict,1 = m1ω[ ωr'1 + ωb cosθ'– 2v'θ1sinφ' – 2 v'φ1 cosθ'cosφ' – r'1sin2θ'cos2φ'] '1 + m1ωsinθ'[- ωb + 2v'φ1cosφ'– r'1cosθ'cos2φ'] ' + m1ωsinθ'cosφ'[– 2v'θ1+ r'1sinφ'] ' Now recall that F1 = [- (GMEm1/r13){1+bcosθ'} - T ] '1 + [ (GMEm1/r13) bsinθ'] ' We combine this with the above then to get Feff,1 = F1 + F'fict,1 = m1ω[ ωr'1 + ωb cosθ'- 2v'θ1sinφ' - 2 v'φ1 cosθ'cosφ' - r'1sin2θ'cos2φ'- (GMEm1/r13){1+bcosθ'} - T] '1 + m1ωsinθ'[- ωb + 2v'φ1cosφ'– r'1cosθ'cos2φ' + (GMEm1/r13) bsinθ'] ' + m1ωsinθ'cosφ'[– 2v'θ1+ r'1sinφ'] ' Now set this to m1a'1 which is this m1a'1 = m1a'r1 '1 + m1a'θ1' + m1a'φ1 ' where a'r1 = - r'1'2 – r'1 '2 sin2θ' a'θ1 = r'1' - r'1 '2 sinθ' cosθ' a'φ1 = 2 r'1 ' ' cosθ' + r'1'sinθ' We then end up with some mind-boggling equations of motion m1ω[ ωr'1 + ωb cosθ'- 2v'θ1sinφ' - 2 v'φ1 cosθ'cosφ' - r'1sin2θ'cos2φ'- (GMEm1/r13){1+bcosθ'} + T] = m1r'1( - '2 – '2 sin2θ') m1ωsinθ'[- ωb + 2v'φ1cosφ'– r'1cosθ'cos2φ' + (GMEm1/r13) bsinθ'] = m1r'1( ' - '2 sinθ' cosθ') m1ωsinθ'cosφ'[– 2v'θ1+ r'1sinφ'] = m1 r'1( 2 ' ' cosθ' + 'sinθ') Question: What determines the value of T ? Put that on hold. Question: How to translate this into the F2 situation? Recall that F1 = [- (GMEm1/r13){1+bcosθ'} - T ] '1 + [ (GMEm1/r13) bsinθ'] ' F2 = [- (GMEm2/r23){1+bcosθ'} + T ] '1 + [ (GMEm2/r23) bsinθ'] ' F'fict,1 = m1ω2r'1 '1 + m1ω2b ' – 2m1 ω x v'1 – m1(ωr'1)ω ' F'fict,2 = m1ω2r'1 '1 + m2ω2b ' + 2m1 ω x v'1+ m1(ωr'1)ω ' With regard to the first two lines change m1/r13 to m2/r23 change T to -T With regard to the last two lines, make these changes In the ω2b terms replace m1 by m2 done Change sign in the Coriolis terms (any terms including velocity) Change sign in the (ωr'1) terms (these are the last term in each component) So backing up, we would then have (changes shown in red) F2 = [- (GMEm2/r23){1+bcosθ'} + T ] '1 + [ (GMEm2/r23) bsinθ'] ' F'fict,2 = m1ω[ ωr'1 - (m2/m1) ωb cosθ' + 2v'θ1sinφ' + 2 v'φ1 cosθ'cosφ' + r'1sin2θ'cos2φ'] '1 + m1ωsinθ'[ + (m2/m1)ωb - 2v'φ1cosφ'+ r'1cosθ'cos2φ'] ' + m1ωsinθ'cosφ'[+ 2v'θ1- r'1sinφ'] ' Here then is how the three equations of motion will look With regard to the first two lines change m1/r13 to m2/r23 DONE change T to -T DONE With regard to the last two lines, make these changes In the ω2b terms replace m1 by m2 DONE Change sign in the Coriolis terms (any terms including velocity) DONE Change sign in the (ωr'1) terms (these are the last term in each component) DONE What are the right sides going to be? Recall that θ'2 = π - θ'1 and φ'2 = φ'1 + π sinθ'2 = + sinθ'1 cosθ'2 = -cosθ'1 so '2 = - '1 and '2= '1 So m1r'1( - '2 – '2 sin2θ') → m2r'2( - '22 – '22 sin2θ'2) = m2r'2( - '2 – '2 sin2θ') m1r'1( ' - '2 sinθ' cosθ') → m2r'2( '2 - '22 sinθ2' cosθ'2) = m2r'2( - ' + '2sinθ'cosθ') m1 r'1( 2 ' ' cosθ' + 'sinθ') → m2 r'2( 2 '2 '2 cosθ'2 + '2sinθ'2) = m2 r'2( + 2 ' ' cosθ' + 'sinθ') Only the second term picks up a minus sign overall on the right side. Here then are the equations of motion from the particle 2 forces equation: m1ω[ ωr'1 - (m2/m1)ωb cosθ'+ 2v'θ1sinφ' + 2 v'φ1 cosθ'cosφ' + r'1sin2θ'cos2φ'- (GMEm2/r23){1+bcosθ'} - T] = m2r'2( - '2 – '2 sin2θ') m1ωsinθ'[+ (m2/m1)ωb - 2v'φ1cosφ'+ r'1cosθ'cos2φ' + (GMEm2/r23) bsinθ'] = m2r'2(-' + '2sinθ'cosθ') m1ωsinθ'cosφ'[+ 2v'θ1- r'1sinφ'] = m2r'2( 2 ' ' cosθ' + 'sinθ') So I end up with SIX equations of motion for the two particles 1. m1ω[ ωr'1 + ωb cosθ'- 2v'θ1sinφ' - 2 v'φ1 cosθ'cosφ' - r'1sin2θ'cos2φ'- (GMEm1/r13){1+bcosθ'} + T] = - m1r'1( '2 + '2 sin2θ') 2. m1ωsinθ'[- ωb + 2v'φ1cosφ'– r'1cosθ'cos2φ' + (GMEm1/r13) bsinθ'] = m1r'1( ' - '2 sinθ' cosθ') 3. m1ωsinθ'cosφ'[– 2v'θ1+ r'1sinφ'] = m1 r'1( 2 ' ' cosθ' + 'sinθ') 4. m1ω[ ωr'1 - (m2/m1)ωb cosθ'+ 2v'θ1sinφ' + 2 v'φ1 cosθ'cosφ' + r'1sin2θ'cos2φ'- (GMEm2/r23){1+bcosθ'} - T] = -m2r'2( '2 + '2 sin2θ') 5. m1ωsinθ'[+ (m2/m1)ωb - 2v'φ1cosφ'+ r'1cosθ'cos2φ' + (GMEm2/r23) bsinθ'] = m2r'2(-' + '2sinθ'cosθ') 6. m1ωsinθ'cosφ'[+ 2v'θ1- r'1sinφ'] = m2 r'2( + 2 ' ' cosθ' + 'sinθ') I also know that v'θ1 = r'1 ' v'φ1 = r'1 ' sinθ' Install these and rewrite the six equations, but at the same time replace m2r'2 = m1r'1 on the right side of the last three equations: 1. m1ω[ ωr'1 + ωb cosθ'- 2 r'1 'sinφ' - 2 r'1 ' sinθ' cosθ'cosφ' - r'1sin2θ'cos2φ'- (GMEm1/r13){1+bcosθ'} + T] = - m1r'1( '2 + '2 sin2θ') 2. m1ωsinθ'[- ωb + 2r'1 ' sinθ'cosφ'– r'1cosθ'cos2φ' + (GMEm1/r13) bsinθ'] = m1r'1( ' - '2 sinθ' cosθ') 3. m1ωsinθ'cosφ'[– 2 r'1 '+ r'1sinφ'] = m1 r'1( 2 ' ' cosθ' + 'sinθ') 4. m1ω[ ωr'1 - (m2/m1)ωb cosθ'+ 2 r'1 'sinφ' + 2r'1 ' sinθ'cosθ'cosφ' + r'1sin2θ'cos2φ'- (GMEm2/r23){1+bcosθ'} - T] = -m1r'1( '2 + '2 sin2θ') 5. m1ωsinθ'[+ (m2/m1)ωb - 2r'1 ' sinθ'cosφ'+ r'1cosθ'cos2φ' + (GMEm2/r23) bsinθ'] = m1r'1(-' + '2sinθ'cosθ') 6. m1ωsinθ'cosφ'[+ 2 r'1 '- r'1sinφ'] = m1 r'1( + 2 ' ' cosθ' + 'sinθ') What are we trying to solve for? r'1(t), θ'(t), φ'(t), T(t) But r'1(t) = r'1 = a constant that I know, so only 3 unknowns θ'(t), φ'(t), T(t) So our F = ma equations result in 6 equations in 3 unknowns, which seems odd. If we subtract 1-4 the right sides cancel, so 1-4: m1ω[ ωr'1 + ωb cosθ'- 2 r'1 'sinφ' - 2 r'1 ' sinθ' cosθ'cosφ' - r'1sin2θ'cos2φ'- (GMEm1/r13){1+bcosθ'} + T] -m1ω[ ωr'1 - (m2/m1)ωb cosθ'+ 2 r'1 'sinφ' + 2r'1 ' sinθ'cosθ'cosφ' + r'1sin2θ'cos2φ'- (GMEm2/r23){1+bcosθ'} - T] = 0 or m1ω[ ωr'1 + ωb cosθ'- 2 r'1 'sinφ' - 2 r'1 ' sinθ' cosθ'cosφ' - r'1sin2θ'cos2φ'- (GMEm1/r13){1+bcosθ'} + T] m1ω[ -ωr'1 + (m2/m1)ωb cosθ'- 2 r'1 'sinφ' - 2r'1 ' sinθ'cosθ'cosφ' - r'1sin2θ'cos2φ'+ (GMEm2/r23){1+bcosθ'} + T] = 0 or m1ω[+ ωb cosθ'- 4 r'1 'sinφ' - 4 r'1 ' sinθ' cosθ'cosφ' - 2r'1sin2θ'cos2φ'- (GMEm1/r13){1+bcosθ'} + 2T + (m2/m1)ωb cosθ' + (GMEm2/r23){1+bcosθ'}] = 0 There is nothing simple about the resulting equation! the 2-4 is similarly ugly. Next, consider 3-6. They both have the same right side, so we get 3 - 6: m1ωsinθ'cosφ'[– 2 r'1 '+ r'1sinφ'] - m1ωsinθ'cosφ'[+ 2 r'1 '- r'1sinφ'] = 0 or [– 2 r'1 '+ r'1sinφ'] - [+ 2 r'1 '- r'1sinφ'] = 0 or – 4 r'1 '+ 2r'1sinφ'= 0 or – 2 r'1 '+ r'1sinφ'= 0 or – 2 '+ sinφ'= 0 or 2 ' = sinφ' How could such a trivial simple equation drop out of the above mess??? More likely I have sign errors. Question: I know there is no ang mom about the r'1 axis because the masses have no extent. Is there some simple math statement of this fact? This fact ought to be true in both frames. In Section D.6 I computed in Frame S' that L'(0) = (m1/m2) M r'12( '' – ' sinθ'') and this verifies that there is no ang mom on the r'1 axis. I go on to compute the fictitious torque and I find that it is also 0 in the r'1 component, so everything is consistent. But this seems to have little bearing on the above mystery. I have shown that in the ' component, F'fict,1 = m1ωsinθ'cosφ'[– 2v'θ1+ r'1sinφ'] ' F'fict,2 = m1ωsinθ'cosφ'[+ 2v'θ1 - r'1sinφ']' so they are equal and opposite. Why should this be true? It then says F'fict, tot = 0 in the ' component Go way back to" F'fict,1 = – m1S – m1ω x (ω x r'1) – 2m1 ω x v'1 – m1 x r'1 F'fict,2 = – m2S – m2ω x (ω x r'2) – 2m2 ω x v'2 – m2 x r'2 and make the trivial replacements so we then have F'fict,1 = – m1S – m1ω x (ω x r'1) – 2m1 ω x v'1 – m1 x r'1 F'fict,2 = – m2S + m1ω x (ω x r'1) + 2m1 ω x v'1 + m1 x r'1 Now add to get F'fict,tot = - MS The centrifugal, Coriolis and Euler terms all cancel out right at the get-go just because of the CMS situation between these two particles. I then show that S = x b - ω2b ' = x ' - ω2b ' Since in GENERAL could be in any direction, I cannot rule out ' component here, but with = 0 I then have S = - ω2b ' = - ω2b[ cosθ' '1 - sinθ' '] which has no ' component. What does that mean? Frame S' is doing circular motion so the origin of Frame S' is accelerating to the Earth which is - ω2b ' which is a down vector in my picture, But ' is always a horizontal vector in the same picture! So to finish, we get F'fict,tot = - MS = M ω2b[ cosθ' '1 - sinθ' '] and yes, this has no ' component. Gravity also has no component horizontally! Therefore, the two masses must have equal and opposite F'fict in the ' direction. But in this direction we also have F' = 0 so then F'eff,tot = 0 in the ' direction Now consider F'eff,1 = m1a'1 F'eff,2 = m2a'2 Add these to learn that 0 = m1a'1+ m2a'2 in the ' direction But the right side is 0 in all directions, so how did I obtain an equation out of this??? Have to stop for now. When resume, find out why equations 3 and 6 have the same right side, when the above equation says they should have opposite sign. Paradox: 1. We know that (leave off the primes here) m2a2 = - m1a2 2. We know that aφ1 = 2 r1 cosθ + r1sinθ aφ2 = 2 r2 2 2 cosθ2 + r22sinθ2 = 2 r2 [2] [2] [cosθ2] + r2[2][sinθ2] = 2 r2 [-] [] [-cosθ] + r2[][sinθ] = 2 r2 cosθ + r2sinθ The component of m2a2 = - m1a1 then reads m2aφ2 = - m1aφ1 or m2(2 r2 cosθ + r2sinθ) = - m1(2 r1 cosθ + r1sinθ) or m2r2(2 cosθ + sinθ) = - m1r1(2 cosθ + sinθ) or m1r1(2 cosθ + sinθ) = - m1r1(2 cosθ + sinθ) or 1 = - 1 and there is your paradox.