Appendix D questions

Appendix D questions PhL 12.6.16 I have now finished a first draft of 18-page Appendix D on the subject of the center of gravity and the tether. Not surprisingly, trying to answer questions raises more and new questions. So right now here are some of those: 1. I argue that N(rcog) = 0 and N(0) = rcog x F as part of the definition of rcog. The other part is the distance rule which sets magnitude rcog. Question: does this apply only in an inertial frame, or does it apply also in a rotating frame? 2. I compute the torque about tether center point ("origin") and I show "restoring force". What does this torque have to do with the torques of item 1 above? And what about the fact that this "origin" is not the center of gravity or of mass? 3. Does it make sense to say anything about a freeze-frame of the tether since you know it must be in motion? Am I apply statics to a case where there is motion and so statics is wrong? Question 1: I am reviewing section D.1 right now. The cms definitions are OK. I then talk about Fi as being a force "experienced by" mass mi. There is no Newton's law yet, so I can assume this is a real force even if I am in a non-inertial frame. Then the definitions of (D.1.2) are all valid. And so is (D.1.3). So I am arguing that if you are in a non-inertial frame, you should NOT include fictitious forces in Fi or F or any of these torques. So (D.1.4) is valid for cog. So is (D.1.5). Conclusion: Section D.1 is valid in any frame of reference where all F and N are real and do not include fictitious components. Thus in the case of the asteroid, if you hop on the asteroid you are in a non-inertial frame, yes. If you take a freeze frame of it as I have done, you compute rcog as I have done. So I think this section is all OK. Inertial/non-inertial only matters when you apply F = ma or N = Iα. Now section D.2. Again I have a freeze-frame of the tether at angle θ. The forces Fi are only those of gravity (need a comment about tension force -- DONE). So this entire section is a freeze-frame statics analysis. I think it is all valid. Then D.3 and D.4 are just special cases. D.5 on the sphere is another statics analysis. The sphere could be accelerating, but the result is not affected. It is a freeze-frame statics analysis. Section D.6 is more of a concern. ( again, comment on tension forces needed! ). As I do things, the stick tension won't enter the torque calcs since x T = 0. I select an arbitrary origin which is the center of my circle. More static calculations. I think things are OK down through (D.5.8) where I talk about the direction as being restoring. I have no Newton's law yet. The Tether Oscillation section brings up Newton's Law N = I for the first time, and so finally I have to worry about things. Remember that the cog and cms are different and neither is at circle center and we are in a rotating frame, lots of complications at once! Start with fictitious torques which have to enter the above law! I have this most general case N'(c')fict = – (c'- c + b) x ma' + S x [mv' + mω x r' + mS] – 'S' x mv' + (r' - c + b) x F'fict . (11.13) If I pick c = c' = 0 for both frames, things simplify: N'(0)fict = – b x ma' + (r' + b) x F'fict . // c = c' = 0 (11.14) where F'fict = – mS – mω x (ω x r') – 2m ω x v' – m x r' (8.6) (11.12) frame centrifugal Coriolis Euler Note that a' appears, the natural acceleration of particle m in Frame S'. Our two frames are in Special Case #1 as in Fig 4.3. Frame S is Earth center and is assumed fixed and the rotation axis passes through it. It is like the turntable. So let's try to apply some Special 1 casing but I won't assume we are on the surface of the Earth as I do in Section (e). Section 8 (d) does do this general special case. I claim there that S = x b + ω x (ω x b) Special Case #1 (7.12) F'eff = F – mω x (ω x r) – 2m ω x v' – m x r . Special Case #1 (8.12) so therefore (Frame S' is the rotating frame attached to the satellite). F'fict = mω x (ω x r) + 2m ω x v' + m x r // non-swap centrifugal Coriolis Euler Note that r is the long vector here. The first term is -mω2r for a circular orbit whose center is Earth center. I will set any Euler term to 0, so then we have N'(0)fict = – b x ma' + (r' + b) x F'fict . // c = c' = 0 (11.14) Ffict = -mω2r + 2m ω x v' centrifugal Coriolis Nreal(0 for Frame S) = r x Freal Warning: confusion now since r and r have two meanings in the tether analysis! Resolved: I will replace the radius of the "circle" with ρ instead of r. Question: Where is the origin of Frame S' ? It can be wherever I want and I see three candidate locations all different! I think I have to take b = rcog because only then do we have b = constant for the orbit! Remember that you can replace your satellite with a point mass there and it goes in a circle. So for mass m1 the above reads N'(0)fict = – rcog x m1a1' + (r1' + rcog) x F'fict,1 Ffict,1 = -m1ω2r1 1 + 2m1 ω x v1' This is immensely complicated if you really track all the distinctions of radii. We are going to need approximations, but certain things are more delicate than others for said approximations. Plan A. Suppose I take the tether circle center as a reference point for the torque oscillation calculation. It is true that as the masses move, this circle center moves in some strange but generally small manner relative to the rcog point. I really don't care about this motion -- it happens, fine. Suppose I also pick circle center as the origin of Frame S'. Then Frame S' as a whole makes this same small motion of rcog - rcc as time goes by in the orbit. The rcc is tied to rcms really. Would it be better to use rcms as torque origin? Locally you get a cleaner picture since then masses move at different radii relative to the cm. The idea would be this: we just ignore the rcog - rcms small motion after acknowledging that it is present. _________________________________________________________ How big is this difference in practice? The general expressions for rcog is very complicated, so I might restrict my estimate to the two angular special cases. For the aligned θ = 0 case I show that rcog = = h rcms = h (D.3.13) If r/h << 1 as is the case even for low earth orbit, we write rcog = h = h [1 - (r/h)2] [ 1 + (r/h)2]-1/2 ≈ = h [1 - (r/h)2] [ 1 - (1/2) (r/h)2] ≈ h [ 1 - (3/2)(r/h)2 ] = h - (3/2)h (r/h)2 = rcms - (3/2)h (r/h)2 So then rcms - rcog = (3/2)h (r/h)2 = (3/2) r2 / h (rcms - rcog)/h = (3/2)(r/h)2 (rcms - rcog)/(2r) = (3/4) (r/h) where r is the circle radius and h is distance to circle center. Let's evaluate this result several ways. I will do it in Maple and assume low earth orbit. I will assume that RE = 6371 km = 6.371 x 106 m and 2r = stick length = 1 km = 103. Then: r = 500 meters r = 5 meters So this shows that: r = 500 m r = 5 m rcms - rcog = 6 cm rcms - rcog = 6 microns (rcms - rcog ) / h ≈ 10-8 (rcms - rcog ) / h ≈ 10-12 (rcms - rcog ) / (2r) ≈ 6 x 10-5 (rcms - rcog ) / (2r) ≈ 6 x 10-7 So for a 1 km long tether the offset rcms - rcog = 6 cm, while for a 10 m tether we get 6 μ !! At higher earth orbit, h is larger than low earth orbit, (r/h) is smaller and all results are smaller. ________________________________ Conclusion: (1) Let's take rcms as the origin of Frame S' and as our torque calculation origin. (2) We acknowledge that rcog - rcms moves in some complicated but very small manner during an orbit. Now go back to where we were, N'(0)fict = – b x ma' + (r' + b) x F'fict . // c = c' = 0 (11.14) Ffict = -mω2r + 2m ω x v' centrifugal Coriolis In the notation N'(0)fict the (0) means this torque is relative to origin of Frame S' which is at rcms but is also at r' = 0 in the circle frame. I will leave it then with the (0) label. Now we set b = rcms and write the above for mass m1 only r1 N'(0)fict,1 = – rcms x m1a1' + (r1' + rcms) x F'fict,1 Ffict,1 = -m1ω2r1 1 + 2m1 ω x v1' ω x v1' = -ωv'1'1 centrifugal Coriolis Freal,1 = -Gmm1/r12 1 – T '1 N'real,1(0) = r'1 x Freal,1 = -(Gmm1/r12) r'1 x 1 [wrong!] Why? I set c = 0 so this torque is referenced to the Frame S origin and r1 x Freal,1 = 0 so this torque is zero and can be ignored. But then where is the gravitational torque in this problem?? But I don't think you can add torques around different points, so this whole setup needs repair! Needs a clean start. Here ω is the orbital rate of the satellite which is 88 minutes for low earth. Primed quantities are relative to Frame S', unprimed relative to Frame S so r1 is a long vector while r'1 is a short one. r2 N'(0)fict,2 = – rcms x m2a2' + (r2' + rcms) x F'fict,2 Ffict,2 = -m2ω2r2 2 + 2m1 ω x v2' ω x v2' = -ωv'2'2 centrifugal Coriolis Freal,2 = -Gmm2/r22 2 – T '2 N'real,2(0) = r'2 x Freal,2 = -(Gmm2/r22) r'2 x 2 Since the Coriolis force has a simple direction shown, I will restate all the above results N'(0)fict,1 = – rcms x m1a1' + r1 x F'fict,1 Ffict,1 = -m1ω2r1 1 – 2m1ωv'1'1 centrifugal Coriolis Freal,1 = -Gmm1/r12 1 – T '1 N'real,1(0) = r'1 x Freal,1 = -(Gmm1/r12) r'1 x 1 N'(0)fict,2 = – rcms x m2a2' + r2 x F'fict,2 Ffict,2 = -m2ω2r2 2 – 2m2ωv'2'2 centrifugal Coriolis Freal,2 = -Gmm2/r22 2 – T '2 N'real,2(0) = r'2 x Freal,2 = -(Gmm2/r22) r'2 x 2 And now one more time with vigor, N'(0)fict,1 = – rcms x m1a1' + r1 x [ -m1ω2r1 1 – 2m1ωv'1'1] Freal,1 = -Gmm1/r12 1 – T '1 N'real,1(0) = -(Gmm1/r12) r'1 x 1 N'(0)fict,2 = – rcms x m2a2' + r2 x [ -m2ω2r2 2 – 2m2ωv'2'2] Freal,2 = -Gmm2/r22 2 – T '2 N'real,2(0) = -(Gmm2/r22) r'2 x 2 Now one of the cross products drops a term so one more time again N'(0)fict,1 = – rcms x m1a1' – m1ωv'1 r1 x '1 Freal,1 = -Gmm1/r12 1 – T '1 N'real,1(0) = -(Gmm1/r12) r'1 x 1 N'(0)fict,2 = – rcms x m2a2' – m2ωv'2 r2 x '2 Freal,2 = -Gmm2/r22 2 – T '2 N'real,2(0) = -(Gmm2/r22) r'2 x 2 Now I think a1' = -a'1'1 because m1 must go in circular motion. So here we go again N'(0)fict,1 = m1a'1 rcms x '1 – m1ωv'1 r1 x '1 Freal,1 = -Gmm1/r12 1 – T '1 N'real,1(0) = -(Gmm1/r12) r'1 x 1 N'(0)fict,2 = m2a'2 rcms x '2 – m2ωv'2 r2 x '2 Freal,2 = -Gmm2/r22 2 – T '2 N'real,2(0) = -(Gmm2/r22) r'2 x 2 Now again combining the torques N'(0)1 = m1a'1 rcms x '1 – m1ωv'1 r1 x '1 - (Gmm1/r12) r'1 x 1 Freal,1 = -Gmm1/r12 1 – T '1 N'(0)2 = m2a'2 rcms x '2 – m2ωv'2 r2 x '2 -(Gmm2/r22) r'2 x 2 Freal,2 = -Gmm2/r22 2 – T '2 But change slightly, N'(0)1 = m1a'1 rcms x '1 – m1ωv'1/r'1 r1 x r'1 - (Gmm1/r13) r'1 x r1 Freal,1 = -Gmm1/r12 1 – T '1 N'(0)2 = m2a'2 rcms x '2 – m2ωv'2/r'2 r2 x r'2 -(Gmm2/r23) r'2 x r2 Freal,2 = -Gmm2/r22 2 – T '2 Now we can combine some terms to get N'(0)1 = m1a'1 rcms x '1 + [m1ωv'1/r'1 - (Gmm1/r13)] r'1 x r1 Freal,1 = -Gmm1/r12 1 – T '1 N'(0)2 = m2a'2 rcms x '2 + [ m2ωv'2/r'2 -(Gmm2/r23)] r'2 x r2 Freal,2 = -Gmm2/r22 2 – T '2 I am hunting for an equation of motion for m1. Perhaps it is this N'(0)1 = I1 ' ' I1 = m1 r'12 N'(0)2 = I2 ' ' I2 = m2 r'22 These equations better be consistent! Note that both have same ' polar angle accel. Write out m1a'1 rcms x '1 + [m1ωv'1/r'1 - (Gmm1/r13)] r'1 x r1 = ( m1 r'12) ' ' m2a'2 rcms x '2 + [ m2ωv'2/r'2 -(Gmm2/r23)] r'2 x r2 = (m2 r'22) ' ' The constraint is that '2 = - '1 from cms analysis. I can do more now v'1 = r'1' a'1 = r'1' v'2 = r'2' a'2 = r'2' So install these and cancel masses at the same time r'1' rcms x '1 + [ω r'1'/r'1 - (GM/r13)] r'1 x r1 = ( r'12) ' ' r'2' rcms x '2 + [ ωr'2'/r'2 - (GM/r23)] r'2 x r2 = ( r'22) ' ' or r'1' rcms x '1 + [ω ' - (GM/r13)] r'1 x r1 = ( r'12) ' ' r'2' rcms x '2 + [ ω' - (GM/r23)] r'2 x r2 = ( r'22) ' ' We know a lot more of course: m1r'1 + m2r'2 = 0 since cms is center of circle, we are staying r'cms = 0 m1r'1'1 + m2r'2'2 = 0 m1r'1'1 - m2r'2'1 = 0 m1r'1 - m2r'2 = 0 m1r'1 = m2r'2 '1 = - '2 Then still unclear how they will come out being the same, but I continue to eliminate r'2 stuff. First dived things through by r'i ' rcms x '1 + [ω ' - (GM/r13)] '1 x r1 = r'1 ' ' ' rcms x '2 + [ ω' - (GM/r23)] '2 x r2 = r'2 ' ' And now put the masses back in m1' rcms x '1 + m1[ω ' - (GM/r13)] '1 x r1 = m1r'1 ' ' m2' rcms x '2 + m2[ ω' - (GM/r23)] '2 x r2 = m2r'2 ' ' The right sides are now exactly the same, so to be consistent we have to be able to show that m1' rcms x '1 + m1[ω ' - (GM/r13)] '1 x r1 = m2' rcms x '2 + m2[ ω' - (GM/r23)] '2 x r2 or using '2 = - '1 m1' rcms x '1 + m1[ω ' - (GM/r13)] '1 x r1 = - m2' rcms x '1 - m2[ω ' - (GM/r23)] '1 x r2 Picture shows that r2 - r'2 = r1 - r'1 = rcms or r2 + r'2'1 = r1 - r'1'1 or r2 = r1 - (r'1+r'2) '1 Since ' and ' are really independent, we really need to show things which are not going to work! Need m1 = -m2 for example. And then I cannot imagine the 1/r13 terms matching. So I have taken a long trip here and I have ended up in the dump!