dumbbell satellite as rigid body problem v1

Dumbbell Satellite as Rigid Body Problem v1 PhL 3.26.17 do not read below, see summary doc In Appendix F I use swap notation, so Frame S is the body frame. It seems clear that θ,φ are just Euler angles. Before starting, note that for such a satellite GME/r02 = ω2r0 ω2 = GME/r03 ≈ GME/b3 in the far approximation, so ω is determined by outside factors. It is not something you solve for. It is the ω which relates the rotation of the two Frames of reference. What are the moments of a dumbbell? In this rigid body analysis I will assume that the two masses are spheres of radius R1 and R2. In this case, there is inertial about the long axis. It will be I3 = (2/5)m1R12 + (2/5)m2R22 Since the spheres could have any radii we want, I will just maintain I3 as a constant in the problem. The other moments in the body frame are a bit complicated since I would have to integrate over each sphere or I could maybe use the axis-shift theorem which I really might add somewhere *****. So let's just say there is some I1 which I could compute, and I2 = I1 because the rigid object is axisymmetric. For a point-mass dumbbell we would know that I1 = m1r12 + m2r22 with m1r1 = m2r2 so I1 = (1/m2) [ m1m2r12 + m22r22 ] = (1/m2) [ m1m2r12 + m21r12 ] = (m1/m2)r12 [ m2 + m1 ] = (m1/m2)r12M = (m1/μ2)r12 I am going to say the R1 and R2 are very small and then this expression for I1 is valid. Side question: Is the torque on a sphere the same as on a point mass at its center? For me this is a very hard question. I think the answer is NO. Therefore, if I am going to use the results of Appendix F for torque, I will have to assume that the R1,R2 << r1,r2 . Consider that done. What are the equations of motion I start with the equations of motion in the body frame, changed to swap notation: N1 = I1 1 - ω2 ω3 (I2 - I3) N2 = I2 2 - ω3 ω1 (I3 - I1) N3 = I3 3 - ω1 ω2 (I1 - I2) . // Frame S = body frame (I.4.3) and with comments above this becomes N1 = I1 1 - ω2 ω3 (I1 - I3) N2 = I1 2 - ω3 ω1 (I3 - I1) N3 = I3 3 . // Frame S = body frame (I.4.3) Getting pretty simple it would seem. Remember that the vector equation was N = I + ω x (Iω) (I.2.6) and maybe that will be helpful, but ignore for now. I show in (F.3.10) the torque on the satellite (with point masses) to be N'(b) = (GMEm1br1sinθ) [1/r'13 - 1/r'23] and this is about "b" which is the center of Frame S. In the far approximation I write this as N'(b) = - 3GME(m1/μ2) b-3r12sinθcosθ . What are the Cartesian Frame S components of this torque? Use = -sinφ + cosφ (E.2.6) So then N'(b) = - 3GME(m1/μ2) b-3r12sinθcosθ [ -sinφ + cosφ ] all in body Frame S. So I then know that true torque is same in both frames, so N1 = 3GME(m1/μ2) b-3r12sinθcosθ sinφ N2 = -3GME(m1/μ2) b-3r12sinθcosθ cosφ So now I have these three equations of motion in the far approximation 3GME(m1/μ2) b-3r12sinθcosθ sinφ = I1 1 - ω2 ω3 (I1 - I3) -3GME(m1/μ2) b-3r12sinθcosθ cosφ = I1 2 - ω3 ω1 (I3 - I1) 0 = I3 3 So at once we know that ω3 = constant. Now as usual we are worrying about the ω vector and we have to think about what vector that really is. In body frame L = Iω , so L1 = I1ω1 L2 = I1ω2 L3 = I3ω3 Lets set ω3 = K for now. I am now going to assume that I3 << I1 to make these equations more tractable, so rewrite as 3GME(m1/μ2) b-3r12sinθcosθ sinφ = I1 1 - ω2 ω3 I1 -3GME(m1/μ2) b-3r12sinθcosθ cosφ = I1 2 + ω3 ω1I1 and divide by I1 so get 3GME(m1/μ2) I1-1 b-3r12sinθcosθ sinφ = 1 - K ω2 -3GME(m1/μ2) I1-1b-3r12sinθcosθ cosφ = 2 + K ω1 Now define C ≡ 3GME(m1/μ2) I1-1 b-3r12 What do I know about C ? I claim above that C = 3GME(m1/μ2) I1-1 b-3r12 I1 = (m1/μ2)r12 I1-1 = (μ2/m1)r1-2 and then C = 3GME(m1/μ2)(μ2/m1)r1-2b-3r12 = 3GMEb-3 which is at least simpler than it was. Moreover, recall from above that ω2 = GME/b3 so then we can write C =3ω2 Then equations of motion are 3ω2sinθcosθ sinφ = 1 - K ω2 -3ω2sinθcosθ cosφ = 2 + K ω1 Recall that I have assumed that 0 < I3 << I1 and R1,R2 << r1,r2 for the two "balls" . Offhand, does not look too bad. Now would be a good time to roll out ω expressed in body frame Euler angles (I have many weapons now! Removing primes since in swap notation now ω1 = sinθsinψ + cosψ ω2 = sinθcosψ - sinψ ω3 = cosθ + (H.4.7) So then 1 = sinθcosψ + cosθ sinψ + sinθsinψ - sinψ + cosψ 2 = - sinθsinψ + cosθ cosψ - sinθcosψ - cosψ - sinψ Very interesting so far. Now install these into the equations along with the above 3ω2sinθcosθ sinφ = [ sinθcosψ + cosθ sinψ + sinθsinψ - sinψ + cosψ] - K[ sinθcosψ - sinψ] -3ω2sinθcosθ cosφ = [- sinθsinψ + cosθ cosψ - sinθcosψ - cosψ - sinψ] + K [ sinθsinψ + cosψ ] Earlier I wanted to just set ψ ≡ 0, but I have I3 > 0 and there will be some ψ(t) and I do NOT want to try and throw that out. So we are stuck then with the above fairly messy equations for now. Recall now where I want to end up somehow : + sinθcosθ(3ω2 - 2) + ωsinθcosφ (ω cosθ cosφ - 2sinθ ) = 0 // + 2 cotθ – ωcosφ (ωsinφ -2) = 0 // (F.5.7) My rigid body equations of THIS doc do not include the symbol ω, so that is one big difference. Is ω a constant of the motion? Well it is determined by the satellite circular orbit, so it is a constant! So there is another equation I have so far ignored. Write again ω1 = sinθsinψ + cosψ ω2 = sinθcosψ - sinψ ω3 = cosθ + (H.4.7) Then ω12 = 2 sin2θsin2ψ + 2 cos2ψ + 2 sinθsinψ cosψ ω22 = 2 sin2θcos2ψ + 2 sin2ψ - 2 sinθcosψ sinψ ω32 = 2 cos2θ + 2 + 2 cosθ Add the first two to get ω12 + ω22 = 2 sin2θ + 2 Then add the third to get ω2 = ω12 + ω22 + ω32 = 2 sin2θ + 2 + 2 cos2θ + 2 + 2 cosθ = 2 + 2 + 2 + 2 cosθ I also have another equation, ω3 = cosθ + = K Question: How do you interpret the fact that ω12 + ω22 has no dependence on ψ ? (defer) Now if you are enforcing ω as a constant from Earth rotation, we then have lots of equations that I have developed in this appendix K = cosθ + ω2 = 2 + 2 + 2 + 2 cosθ 3ω2sinθcosθ sinφ = - K sinθ (1) -3ω2sinθcosθ cosφ = cosθ - sinθ + K (2) Insert the K results to have then only three equations to deal with ω2 = 2 + 2 + 2 + 2 cosθ 3ω2sinθcosθ sinφ = - ( cosθ + ) sinθ -3ω2sinθcosθ cosφ = cosθ - sinθ + ( cosθ + ) or ω2 = 2 + 2 + 2 + 2 cosθ 3ω2sinθcosθ sinφ = - cosθ sinθ - sinθ -3ω2sinθcosθ cosφ = cosθ - sinθ + cosθ + Somehow you must be able to show that these three equations are the same as + sinθcosθ(3ω2 - 2) + ωsinθcosφ (ω cosθ cosφ - 2sinθ ) = 0 // sinθ + 2 cosθ – ωcosφ sinθ(ωsinφ -2) = 0 // (F.5.7) One problem is that now ψ has entered the problem and it was not present before. Can I somehow eliminate from the set of three equations ? Plan A . Take a look at the first of the three 2 + (2 cosθ) + (ω2-2-2) = 0 Think of this as x2 + bx + c = 0 Then 2x = -b ± Now b2 - 4ac = (2 cosθ)2 - 4 (ω2-2-2) = 4[ 2cos2θ - ω2+ 2 + 2] But recall from above that ω2 = 2 + 2 + 2 + 2 cosθ but this does not make the square root go away. So my quadratic solution does not pay off. Plan B . Treat K as the basic constant. Have K = cosθ + K2 = 2 cos2θ + 2 + 2 cosθ ω2 = 2 + 2 + 2 + 2 cosθ ω2 - K2 = 2 + 2sin2θ = ω2- ω32 = ω2 = constant // agrees with earlier ω2 So now differentiate the above to get 2 + 2sin2θ + 22sinθcosθ = 0 or + sin2θ + 2sinθcosθ = 0. I don't think this fact ever appeared in Appendix F. Meanwhile we have another fact ω2 = 2 + 2 + 2 + 2 cosθ 0 = + + + cosθ - sinθ + cosθ = 0 This seems a useless entanglement of ψ into things when I am trying to remove ψ. So I now have several equations of possible interest cosθ + = K = ω3 2 + 2 + 2 + 2 cosθ = ω2 2 + 2sin2θ = ω2 - ω32 = ω2 + sin2θ + 2sinθcosθ = 0 3ω2sinθcosθ sinφ = - cosθ sinθ - sinθ -3ω2sinθcosθ cosφ = cosθ - sinθ + cosθ + I still don't know how to "eliminate" ψ in my point mass limit of interest. I am trying to get here: + sinθcosθ(3ω2 - 2) + ωsinθcosφ (ω cosθ cosφ - 2sinθ ) = 0 // sinθ + 2 cosθ – ωcosφ sinθ(ωsinφ -2) = 0 // (F.5.7) Plan C: OK, let's just try this to eliminate = ω3 - cosθ Then my two big equations become 3ω2sinθcosθ sinφ = - cosθ sinθ - sinθ ( ω3 - cosθ) -3ω2sinθcosθ cosφ = cosθ - sinθ + cosθ + ( ω3 - cosθ) which compare + sinθcosθ(3ω2 - 2) + ωsinθcosφ (ω cosθ cosφ - 2sinθ ) = 0 // sinθ + 2 cosθ – ωcosφ sinθ(ωsinφ -2) = 0 // (F.5.7) Things are NEVER going to match as long as ω3 is sitting in my equations. Plan D. It is possible that maybe is very small? If that were the case, maybe my rigid equations are 3ω2sinθcosθ sinφ = - cosθ sinθ -3ω2sinθcosθ cosφ = cosθ - sinθ + cosθ But no linear combining of these can give (F.5.7) because nothing can develop a linear ω factor!