dumbbell satellite as rigid body problem

Dumbbell Satellite as Rigid Body Problem 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 and that ψ probably plays no role since it's axis has no inertia. 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) For my particular problem, I3 = 0 since just two point masses, so then have N1 = I1 1 - ω2 ω3 I2 N2 = I2 2 + ω3 ω1 I1 N3 = I3 3 - ω1 ω2 (I1 - I2) . // Frame S = body frame (I.4.3) If I take the rigid body lined up on z' say, then clearly I1 = I2 so then have N1 = I1 1 - ω2 ω3 I1 N2 = I2 2 + ω3 ω1 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. For the dumbbell we 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 show in (F.3.10) the torque on the satellite 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 -3GME(m1/μ2) b-3r12sinθcosθ cosφ = I1 2 + ω3 ω1 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 . On scratch I write that out as a matrix equation and it tells me this L1 = I1ω1 L2 = I1ω2 L3 = 0 Does this make sense for the dumbbell? Why is L3 = 0, that just seems wrong. Ah, but this is all in the body frame remember, so it does make sense. So we then have a very simple connection in the body frame between L and ω . Lets set ω3 = K for now and divide by I and rewrite our equations 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. Then equations of motion are Csinθcosθ sinφ = 1 - K ω2 -Csinθcosθ cosφ = 2 + K ω1 C = 3GMEb-3 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) Can I remove ψ completely from the problem? Let's be safe and maintain it, and later with larger sphere masses I will want to maintain it anyway. 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 Csinθcosθ sinφ = [ sinθcosψ + cosθ sinψ + sinθsinψ - sinψ + cosψ] - K[ sinθcosψ - sinψ] -Csinθcosθ cosφ = [- sinθsinψ + cosθ cosψ - sinθcosψ - cosψ - sinψ] + K [ sinθsinψ + cosψ ] NOW let's get rid of ψ somehow. Assume ψ ≡ 0 at this point, so sinψ = 0 and cosψ = 1, Csinθcosθ sinφ = [] - K[ sinθ] -Csinθcosθ cosφ = [ + cosθ - sinθ] + K [ + ] which rewrite as Csinθcosθ sinφ = - K sinθ -Csinθcosθ cosφ = cosθ - sinθ + K or - K sinθ - Csinθcosθ sinφ = 0 sinθ - K - cosθ - Csinθcosθ sinφ = 0 C = 3GMEb-3 How do these EOM compare with my results in Appendix F ?? + sinθcosθ(3ω2 - 2) + ωsinθcosφ (ω cosθ cosφ - 2sinθ ) = 0 // + 2 cotθ – ωcosφ (ωsinφ -2) = 0 // (F.5.7) My rigid body equations 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 And now set = 0 to get ω2 = 2 + 2 simpler than I thought it would be. I also have another equation, ω3 = cosθ + = K which then says K = cosθ for our simple case. 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 = constant which implies the next line + = 0 // I have never seen this equation before! Csinθcosθ sinφ = - K sinθ -Csinθcosθ cosφ = cosθ - sinθ + K C = 3GMEb-3 But this still looks a far cry from my Appendix F equations of motion which are these + sinθcosθ(3ω2 - 2) + ωsinθcosφ (ω cosθ cosφ - 2sinθ ) = 0 // + 2 cotθ – ωcosφ (ωsinφ -2) = 0 // (F.5.7) I am now worried about the terms in (F.5.7) that are linear in ω since ω = . So here are our four rigid body equations again K = cosθ K = ω3 ω2 = 2 + 2 = constant 3GMEb-3sinθcosθ sinφ = - K sinθ (1) - 3GMEb-3sinθcosθ cosφ = cosθ - sinθ + K (2) which again I am comparing to + sinθcosθ(3ω2 - 2) + ωsinθcosφ (ω cosθ cosφ - 2sinθ ) = 0 // + 2 cotθ – ωcosφ (ωsinφ -2) = 0 // (F.5.7) We still seem to be in separate worlds! Let's work on (1) and (2) this way, 3GMEb-3sinθcosθ sinφcosφ = cosφ - K sinθcosφ (1)cosφ - 3GMEb-3sinθcosθ cosφsinφ = cosθsinφ - sinθsinφ + Ksinφ (2)sinφ Add these to get cosφ - K sinθcosφ + cosθsinφ - sinθsinφ + Ksinφ = 0 Now insert K = cosθ to get cosφ - cosθ sinθcosφ + cosθsinφ - sinθsinφ + cosθsinφ = 0 or cosφ - 2 cosθ sinθcosφ + cosθsinφ - sinθsinφ + cosθsinφ = 0 or cosφ - 2 cosθ sinθcosφ - sinθsinφ = 0 We are still operating in different worlds it seems to me. I need to assert the constraint that ω is constant somehow ω2 = 2 + 2 and + = 0