scraps

This is the Title PhL 1.11.15 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) = (ω2b4m1rsinθ) (1/r'3 - 1/r'23) N(0)fict = - (m1/m2)Mωr2 sinθcosφ [ (ωsinφ -2) + (ω cosθ cosφ - 2sinθ ) ] 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 ************************* D.10 Force analysis of the Dumbbell Satellite in Frame S' A. COMPUTE TRUE FORCES The only forces on a dumbbell mass are gravity and stick tension T. From ** we then write F1 = - (GMEm1/r'13) r'1 - T 1 = - (GMEm1/r'13)( b + r1) - T 1 F2 = - (GMEm2/r'23) r'2 - T 2 = - (GMEm2/r'23)( b + r2) - T 2 Given that b is the distance to CMS, the above equations are exact no matter where COG lies. B. COMPUTE FICTITIOUS FORCES We must now compute the fictitious force on each mass seen in Frame S: Ffict,1 ≈ – m1S' – m1ω x (ω x r1) – 2m1 ω x 1 – m1 x r1 Ffict,2 ≈ – m2S' – m2ω x (ω x r2) – 2m2 ω x 2 – m2 x r2 What is the meaning of ω here? I have assumed that the CMS is in orbit! So the above equations are not exactly true. C. TOTAL FORCES and EQUATIONS OF MOTION We then know that, Feff,1 = m1 1 ≈ - (GMEm1/r'13)(b + r1) - T 1 – m1S' – m1ω x (ω x r1) – 2m1 ω x 1 – m1 x r1 Feff,2 = m22 = - m11 ≈ - (GMEm2/r'23)(b + r2) - T 2 – m2S' – m2ω x (ω x r2) – 2m2 ω x 2 – m2 x r2 We also know that b = b = b = b' // = bcosθ 1 - bsinθ r'12 = b2 + r12 + 2br1cosθ r2 = -(m1/m2)r1 r'22 = b2 + r22 - 2br2cos θ S = x b - ω2b ' GME = ω2b3 ok to here 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 ' The following items are givens: ω, , m1, m2, b, GME The following items are the unknowns: r'1(t), T = 4 unknowns So we seem to have a system of 6 2nd-order differential equations in 4 unknowns, so the ODE's must not all be linearly independent. The Sum Equation : What happens if you add the two ODE's (so the right sides cancel out): - (GMEm1/r13)(b + r'1) - T '1 – m1S – m1ω x (ω x r'1) – 2m1 ω x '1 – m1 x r'1 - (GMEm2/r23)(b + r'2) - T '2 – m2S – m2ω x (ω x r'2) – 2m2 ω x '2 – m2 x r'2 ≈ 0 In the second line make the obvious replacements to get m2r'2 = - m1r'1 - (GMEm1/r13)(b + r'1) - T '1 – m1S – m1ω x (ω x r'1) – 2m1 ω x '1 – m1 x r'1 - (GMEm2/r23)(b + r'2) + T '1 – m2S + m1ω x (ω x r'1) + 2m1 ω x '1 + m1 x r'1 ≈ 0 or - (GMEm1/r13)(b + r'1) – m1S - (GMEm2/r23)(b + r'2) – m2S ≈ 0 or - (ω2b3m1/r13)(b + r'1) - (ω2b3m2/r23)(b + r'2) ≈ (m1+m2)S or - (m1/r13)(b + r'1) - (m2/r23)(b + r'2) ≈ (m1+m2)S /(ω2b3) or - (1/r13)(m1b + m1r'1) - (1/r23)(m2b +m2 r'2) ≈ (m1+m2)S /(ω2b3) or (1/r13)(m1b + m1r'1) + (1/r23)(m2b - m1r'1) ≈ -(m1+m2)S /(ω2b3) or b[ (m1/r13) + (m2/r23)] + m1r'1 [ (1/r13) - (1/r23)] ≈ -(m1+m2)S /(ω2b3) where S = x b - ω2b ' b = b' Suppose = 0. Then our sum equation becomes b'[ (m1/r13) + (m2/r23)] + m1r'1 [ (1/r13) - (1/r23)] ≈ (m1+m2)ω2b ' /(ω2b3) or [ b(m1/r13) + b(m2/r23)]' + m1r'1 [ (1/r13) - (1/r23)]'1 ≈ M ' /b2 or [ b(m1/r13) + b(m2/r23) - (M/b2)]' + m1r'1 [ (1/r13) - (1/r23)]'1 ≈ 0 But ' and '1 can be regarded as two linearly independent vectors (if θ' ≠ 0) so then b(m1/r13) + b(m2/r23) ≈ (M/b2) (1/r13) ≈ (1/r23) Idea: I have assumed that CMS is COG and maybe this assumption is forcing these results? I think that is in fact the case. Putting r1 = r2 = b into the first equation gives an identity. So does this mean that this whole method is dead in the water? All it really says is that we don't get a contradiction from the sum equation, given that it is the sum of approximate equations. So I think it is OK to move on. The sum equation is approximately valid, so maybe that means that only one of the two vector equations of motion is linearly independent so there are then 3 equations. Let's restate the equations with b = b' and S = - ω2b both installed : (no Euler term) F'eff,1 = m1 '1 ≈ - (GMEm1/r13)(b' + r'1) - T '1 + m1 ω2b – m1ω x (ω x r'1) – 2m1 ω x '1 F'eff,2 = m2'2 = - m1'1 ≈ - (GMEm2/r23)(b' + r'2) - T '2 +m2ω2b – m2ω x (ω x r'2) – 2m2 ω x '2 So we only need to study the first equation: F'eff,1 = m1 '1 ≈ { - (GMEm1/r13)(b' + r'1) - T '1 + m1 ω2b} – m1ω x (ω x r'1) – 2m1 ω x '1 The following items are givens: ω, b, m1, m2, r'1 The following items are the unknowns: '1(t), T = 3 unknowns ( so 3 in 3, good). Consider its first three terms which are in the {...} brackets, - (GMEm1/r13)(b' + r'1) - T '1 + m1 ω2b - (ω2b3m1/r13)(b' + r'1) - T '1 + m1 ω2b - (ω2b3m1/r13)(b') - (ω2b3m1/r13) r'1 - T '1 + m1 ω2b - (ω2b3m1/r13)(b') - (ω2b3r'1m1/r13) '1 - T '1 + m1 ω2b [ - (ω2b3m1/r13)b + m1 ω2b ] + [ - (ω2b3r'1m1/r13) - T ] '1 ω2bm1[ - (b3/r13) + 1 ] - [ (ω2b3r'1m1/r13) + T ] '1 But we know that ' = cosθ' '1 - sinθ' ' so the first three terms are ω2bm1[ - (b3/r13) + 1 ] [ cosθ' '1 - sinθ' '] - [ (ω2b3r'1m1/r13) + T ] '1 = { ω2bm1 ( - (b3/r13) + 1) cosθ' - [ (ω2b3r'1m1/r13) + T ] } '1 + [- ω2bm1 ( -(b3/r13) + 1) sinθ'] ' I am wondering whether you should just say b ≈ r1 to clear this first term, to our level of approx? Well consider 1 - (b3/r13) = (r13 - b3) / r13 = (r1- b)(r12 +r1b+ b2)/r13 ≈ (r1- b)(3b2/b3) = 3 (r1- b)/b So this factor is linear in (r1-b). Now since r12 = b2 + r'12 + 2br'1cosθ' I have shown on scratch that r1-b ≈ r'1cosθ' So then I claim that 1 - (b3/r13) ≈ 3 r'1cosθ'/b So rather than dropping the first term, I will use the above approximation. Let's next to the centrifugal term: ω x (ω x r'1) = (ωr'1)ω - ω2 r'1 = (ω 'r'1)ω ' - ω2r'1 '1 = ω2r'1(''1) ' - ω2r'1 '1 = ω2r'1sinθ'cosφ' ' - ω2r'1 '1 = ω2r'1sinθ'cosφ'[ cosφ'sinθ' '1 + cosφ'cosθ' ' - sinφ' '] - ω2r'1 '1 = ω2r'1sinθ'cosφ'cosφ'sinθ' '1 + ω2r'1sinθ'cosφ'cosφ'cosθ' ' -ω2r'1sinθ'cosφ' sinφ' '- ω2r'1 '1 = ω2r'1sin2θ'cos2φ' '1 + ω2r'1sinθ'cosθ'cos2φ' ' -ωr'12sinθ'cosφ' sinφ' '- ω2r'1 '1 = [ω2r'1sin2θ'cos2φ' - ω2r'1] '1 + [ω2r'1sinθ'cosθ'cos2φ'] ' + [- ω2r'1sinθ'cosφ' sinφ'] ' = ω2r'1 { [sin2θ'cos2φ' - 1] '1 + [sinθ'cosθ'cos2φ'] ' + [- sinθ'cosφ' sinφ'] '} This certainly is an ugly result. Now move on the Coriolis term factor which is 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'] ' where v'θ1 = ' v'φ1 = r'1 ' sinθ' so = ωr'1[ 'sinφ' + ' sinθ' cosφ'cosθ'] '1 + ωr'1[- ' sin2θ'cosφ'] ' + ω r'1[ 'cosφ'sinθ'] ' = ωr'1{ [ 'sinφ' + ' sinθ' cosφ'cosθ'] '1 + [- ' sin2θ'cosφ'] ' + [ 'cosφ'sinθ'] ' = ωr'1{ [ 'sinφ' + ' sinθ' cosθ' cosφ'] '1 + [- ' sin2θ'cosφ'] ' + [ 'sinθ'cosφ'] ' Here are the conclusions so far ω x (ω x r'1) = = ωr'1 { [ωsin2θ'cos2φ' - ω] '1 + [ωsinθ'cosθ'cos2φ'] ' + [- ωsinθ'cosφ' sinφ'] '} ω x v'1 = = ωr'1{ [ 'sinφ' + ' sinθ' cosθ' cosφ'] '1 + [- ' sin2θ'cosφ'] ' + [ 'sinθ'cosφ'] ' Then -m1ω x (ω x r'1) = -m1ωr'1 { [ωsin2θ'cos2φ' - ω] '1 + [ωsinθ'cosθ'cos2φ'] ' + [- ωsinθ'cosφ' sinφ'] '} -2m1ω x v'1 = = -2m1ωr'1{ [ 'sinφ' + ' sinθ' cosθ' cosφ'] '1 + [- ' sin2θ'cosφ'] ' + [ 'sinθ'cosφ'] '} The sum of these terms will be sum = -m1ωr'1Q where Q = [ωsin2θ'cos2φ' - ω] '1 + [ωsinθ'cosθ'cos2φ'] ' + [- ωsinθ'cosφ' sinφ'] ' + [ 2'sinφ' + 2' sinθ' cosθ' cosφ'] '1 + [- 2 ' sin2θ'cosφ'] ' + [ 2'sinθ'cosφ'] ' = [ ωsin2θ'cos2φ' - ω + 2'sinφ' + 2' sinθ' cosθ' cosφ' ] '1 + [ ωsinθ'cosθ'cos2φ' - 2 ' sin2θ'cosφ' ] ' + [ - ωsinθ'cosφ' sinφ' + 2'sinθ'cosφ' ] ' = [ ωsin2θ'cos2φ' - ω + 2'sinφ' + 2' sinθ' cosθ' cosφ' ] '1 + sinθ'cosφ'[ ωcosθ'cosφ' - 2 ' sinθ' ] ' + sinθ'cosφ' [ - ωsinφ' + 2' ] ' Finally we can assemble all the terms: F'eff,1 = m1 '1 = { ω2bm1 ( - (b3/r13) + 1) cosθ' - [ (ω2b3r'1m1/r13) + T ] } '1 + [- ω2bm1 ( -(b3/r13) + 1) sinθ'] ' -m1ωr'1Q Take it one component at a time. First the '1 component: { ω2bm1 ( - (b3/r13) + 1) cosθ' - [ (ω2b3r'1m1/r13) + T ] } -m1ωr'1 {[ ωsin2θ'cos2φ' - ω + 2'sinφ' + 2' sinθ' cosθ' cosφ' ]} ≈ { ω2bm1 (3 r'1cosθ'/b) cosθ' - [ (ω2b3r'1m1/r13) + T ] } // approx from above -m1ωr'1 {[ ωsin2θ'cos2φ' - ω + 2'sinφ' + 2' sinθ' cosθ' cosφ' ]} ≈ { ω2m1 3 r'1cos2θ' - [ (ω2b3r'1m1/r13) + T ] } -m1ωr'1 {[ ωsin2θ'cos2φ' - ω + 2'sinφ' + 2' sinθ' cosθ' cosφ' ]} ≈ { ω2m1 3 r'1cos2θ' - [ (ω2r'1m1) + T ] } // set r1 ≈ b in ratio -m1ωr'1 {[ ωsin2θ'cos2φ' - ω + 2'sinφ' + 2' sinθ' cosθ' cosφ' ]} ≈ { ω2m1 3 r'1cos2θ' - ω2r'1m1 - T ] } // set r1 ≈ b in ratio -m1ωr'1 {[ ωsin2θ'cos2φ' - ω + 2'sinφ' + 2' sinθ' cosθ' cosφ' ]} ≈ { ω2m1 r'1(3cos2θ'-1) - T ] } // set r1 ≈ b in ratio -m1ωr'1 {[ ωsin2θ'cos2φ' - ω + 2'sinφ' + 2' sinθ' cosθ' cosφ' ]} ≈ ω2m1 r'1(3cos2θ'-1) - T -m1ωr'1 {[ ωsin2θ'cos2φ' - ω + 2'sinφ' + 2' sinθ' cosθ' cosφ' ]} This is going to get set equal to m1a'1 which has this '1 component m1r'1[ - '2 – '2 sin2θ' ] So here is the equation of motion in the '1 component: - [ ω2(3cos2θ'-1) + T/(m1r'1) ] - ω {[ ωsin2θ'cos2φ' - ω + 2'sinφ' + 2' sinθ' cosθ' cosφ' ]} = - [ '2 + '2 sin2θ' ] or ω2(3cos2θ'-1) + T/(m1r'1) + ω [ωsin2θ'cos2φ' - ω + 2'sinφ' + 2' sinθ' cosθ' cosφ' ] - '2 - '2 sin2θ' = 0 ω2(3cos2θ'-2) + T/(m1r'1) + ω [ωsin2θ'cos2φ' + 2'sinφ' + 2' sinθ' cosθ' cosφ' ] - '2 - '2 sin2θ' = 0 The unknowns here are: θ'(t), φ'(t), T(t) This ODE has the θ' and φ' stuff cross coupled, and it is non-linear because it has '2 . Let's get the other two equations at least written down. First here is the ' component: [- ω2bm1 ( -(b3/r13) + 1) sinθ'] -m1ωr'1 sinθ'cosφ'[ ωcosθ'cosφ' - 2 ' sinθ' ] = m1 r'1 [ ' - '2 sinθ' cosθ' ] I will approx the first term as before, leaving then - ω2m13 r'1cosθ' sinθ' -m1ωr'1 sinθ'cosφ'[ ωcosθ'cosφ' - 2 ' sinθ' ] = m1 r'1 [ ' - '2 sinθ' cosθ' ] - ω23 cosθ' sinθ' -ω sinθ'cosφ'[ ωcosθ'cosφ' - 2 ' sinθ' ] = ' - '2 sinθ' cosθ' What happens here if θ' is very small? -3ω2 θ' - ωθ'cosφ' [ ωcosφ' - 2 ' θ'] = ' - '2 θ' -3ω2 θ' - [ ωθ'cosφ' ωcosφ' - ωθ'cosφ' 2 ' θ'] = ' - '2 θ' -3ω2 θ' - [ ω2θ'cos2φ' -order(θ')2] = ' - '2 θ' -3ω2 θ' - ω2θ'cos2φ' = ' - '2 θ' 0 = ' - '2 θ' + 3ω2 θ' + ω2θ'cos2φ' ' + [ 3ω2 - '2 + ω2cos2φ'] θ' = 0 Now make the ansatz that φ' = π/2 so we have a fixed planar situation in paper. Then cosφ' = 0 so ' + [ 3ω2] θ' = 0 and this agrees with what I got in the torque analysis! But there I had to assume ω large I think. But this basic resulting is still lurking in the works ! What about the ' equation just to fill this out. Only Q has such a component and we get F'eff,1 ' = -m1ωr'1Q ' = -m1ωr'1 sinθ'cosφ' [ - ωsinφ' 2' ] and the right side is m1a'φ1 so we get -m1ωr'1 sinθ'cosφ' [ - ωsinφ' + 2' ] = m1r'1 [ 2 ' ' cosθ' + 'sinθ'] -ωsinθ'cosφ' [ - ωsinφ' + 2' ] = [ 2 ' ' cosθ' + 'sinθ'] - [ - ωsinθ'cosφ'ωsinφ' + ωsinθ'cosφ'2' ] = [ 2 ' ' cosθ' + 'sinθ'] ω2sinθ'cosφ'sinφ' - ωsinθ'cosφ'2' = 2 ' ' cosθ' + 'sinθ' 'sinθ' + 2 ' ' cosθ' + ωsinθ'cosφ'2' - ω2sinθ'cosφ'sinφ' = 0 What does this one do for small θ' ? 'θ' + 2 ' '+ ωθ'cosφ'2' - ω2θ'cosφ'sinφ' = 0 ' + 2 ('/θ') '+ ωcosφ'2' - ω2cosφ'sinφ' = 0 Suppose again that φ' = π/2 , then it says ' + 2 ('/θ') ' = 0 and φ' = constant is a fine solution to this equation. So we have a self-consistent solution. Now recall θ' = Asin(ω't + α) ' = Aωcos(ω't+α) ('/θ') = ωcos(ω't+α)/ sin(ω't + α) Then sin(ω't + α) ' + 2 ωcos(ω't+α) ' = 0 which is some strange but well defined first order ODE for ' . Summary of resulting Equations of Motion (the order is r,θ,φ) ω2(3cos2θ'-1) + T/(m1r'1) +ω [ωsin2θ'cos2φ' - ω+ 2'sinφ' +2' sinθ' cosθ' cosφ' ] = '2 + '2 sin2θ' - ω23 cosθ' sinθ' -ω sinθ'cosφ'[ ωcosθ'cosφ' - 2 ' sinθ' ] = ' - '2 sinθ' cosθ' ω2sinθ'cosφ'sinφ' - ωsinθ'cosφ'2' = 2 ' ' cosθ' + 'sinθ' What does the first equation say for small θ' ? ω2(3-2) + T/(m1r'1) +ω [ωθ'2cos2φ'+ 2'sinφ' +2'θ' cosφ' ] = '2 + '2θ' ω2 + T/(m1r'1) +ω [2'sinφ' +2'θ' cosφ' ] = '2 + '2 θ' 2ω2 + T/(m1r'1) + 2ω'sinφ' +2ω'θ' cosφ' = '2 + '2 θ' ω2 + T/(m1r'1) + 2ω'sinφ' = '2 + ['2 -2ω' cosφ'] θ' Suppose it is static. Then equation says ω2 + T/(m1r'1) = 0 T = -m1r'1ω2 what I get here This is the first time I have had any estimate for T. But what happened to T = 3 (mMG/r03)(Δr) . (8.6.9) If I set Δr = r'1 for mass m1 and r0 = b, this says T = 3 (m1r'1MEG/b3) . But as usual GME = ω2b3 so my main text says T = 3 (m1r'1 ω2) what I get in the main text and things are not right, though dimensions are correct. Start again with the r equation ω2(3cos2θ'-1) + T/(m1r'1) +ω [ωsin2θ'cos2φ' - ω+ 2'sinφ' +2' sinθ' cosθ' cosφ' ] = '2 + '2 sin2θ' Set θ' = 0 and ' = 0 to get 2ω2 + T/(m1r'1) +ω [ - ω ] = 0 ω2 + T/(m1r'1) = 0 Same bad result. I think T has to be positive, so I have a bug in the r equation which of course is the messiest. I will review this tomorrow and hope it repairs itself.