Section F.6 force analysis

This is the Title PhL 1.11.15 Note that page numbering is turned on in this template and view is 125%, located in phil/roaming/microsoft/templates size about 219K. F.6 Force analysis of the Dumbbell Satellite in Frame S' The only true forces on a dumbbell mass are gravity and stick tension T. From ** we then write F'1 = - (GMEm1/r'13) r'1 - T 1 = - (GMEm1/r'13)( b + r1) - T 1 F'2 = - (GMEm2/r'23) r'2 - T 2 = - (GMEm2/r'23)( b + r2) - T 2 . (F.6.1) As noted earlier, the center of gravity is not quite at the center of mass in Fig (F.1.1), but the above equations are exact despite this fact. The force are primed because they are forces in the inertial Frame S'. The reader is reminded that we are using the "swap notation" where prime↔noprime relative to the non-swap notation. In order to use Newton's Law in rotating Frame S, we must include the fictitious forces. We translate the result of (8.1.8) to swap notation to obtain Ffict,1 ≈ – m1S' – m1ω x (ω x r1) – 2m1 ω x v1 – m1 x r1 Ffict,2 ≈ – m2S' – m2ω x (ω x r2) – 2m2 ω x v2 – m2 x r2 (F.6.2) frame centrifugal Coriolis Euler In these equations, the b acceleration is given by the swap version of (7.13) which then states S' = x b + ω x (ω x b) . // Special Case #1 (F.6.3) and the vectors b and ω are given by ω = ω b = b . (F.6.4) Finally we may state Newton's Law for each mass, Feff,1 = m1 a1 (F.6.5) ≈ - (GMEm1/r'13)r'1 - T 1 – m1S' – m1ω x (ω x r1) – 2m1 ω x v1 – m1 x r1 Feff,2 = m2a2 (F.6.6) ≈ - (GMEm2/r'23)r'2 - T 2 – m2S' – m2ω x (ω x r2) – 2m2 ω x v2 – m2 x r2 where we have now a set of six scalar equations. Using (F.1.9) through (F.1.11), (F.6.6) can be rewritten, Feff,2 = - m1a1 (F.6.7) ≈ - (GMEm2/r'23)r'2 + T 1 – m2S' + m1ω x (ω x r1) + 2m1 ω x v1 – m1 x r1 Adding (F.6.5) and (F.6.7) gives 0 = - (GMEm1/r'13) r'1+ - (GMEm2/r'23)r'2 - (m1+m2)S' (F.6.8) This equation is just F = ma in inertial Frame S' for the total satellite where b is the center of mass. Ignoring the small offset between center of mass and center of gravity, the three equations (F.6.8) describe the circular orbit of the satellite around the Earth. We may then regard the equation (F.6.5) as a set of three scalar equations for the three unknowns θ,φ and T where recall r1 = (r1,θ,φ) in the spherical coordinates of Fig (F.1.1). Our next task is to write vector equation (F.6.5) in spherical coordinates to obtain the three equations of motion. After expanding the left side, we then consider the right side of (F.6.5) one term at a time: Left side of (F.6.5): m1 a1 = m1(ar + aθ + aφ ) (E.3.6) = m1r1[( - 2 - 2 sin2θ) 1 + ( - 2 sinθ cosθ) + (2 cosθ + sinθ) ] (F.6.9) Term 1: - (GMEm1/r'13)r'1 = - (GMEm1/r'13)(b + r1) // (F.1.8) = - (GMEm1/r'13)(b + r1) // (F.6.4) = - (GMEm1/r'13)(bcosθ 1 - b sinθ + r1) // (E.2.7) = - (GMEm1/r'13)[ (bcosθ +r1) 1 - b sinθ ] (F.6.10) Term 2: - T 1 as is (F.6.11) Term 3: – m1S' = – m1 x b - m1 ω x (ω x b) // (F.6.3) = - m1ω x (ω x b) // satellite in circular orbit, = 0 = - m1(ωb)ω + m1ω2b //- A x (A x C) = -(AC)A + A2C = m1ω2b = m1ω2b // (F.6.4) = m1ω2b [ cosθ 1 - sinθ ] // (E.2.7) (F.6.12) Term 4: – m1ω x (ω x r1) = -m1(ωr1)ω + m1ω2r1 // identity shown above = -m1ω2r1(1) + m1ω2r1 1 // (F.6.4) = -m1ω2r1sinθcosφ + m1ω2r1 1 // (E.2.4)) = -m1ω2r1sinθcosφ[sinθcosφ 1 + cosθcosφ - sinφ ] + m1ω2r1 1 // (E.2.7)) = - m1ω2r1 [ (sin2θcos2φ - 1) 1 + (sinθcosθcos2φ) + (- sinθcosφsinφ) (F.6.13) Term 5: -2m1 ω x v1 = -2m1 [ω] x ( vθ + vφ) // (E.3.5) = -2m1ω [vθ x + vφ x ] = -2m1ω [vθ (sinθcosφ + sinφ 1)+ vφ(-sinθ cosφ + cosθcosφ 1)] // (E.2.13) = -2m1ω [ (vθsinφ + vφcosθcosφ)1 + (-vφsinθ cosφ) + (vθsinθcosφ) ] = -2m1ωr1 [ ( sinφ + sinθ cosθcosφ)1 + (- sin2θ cosφ) + ( sinθcosφ) ] // (E.3.2) (F.6.14) Term 6: – m1 x r1 = 0 because we assume = 0 (F.6.15) Having all the bits and pieces, we now assemble the three component equations of (F.6.5). The numbers show the Term above associated with each piece: Feff,1 = m1 a1 (F.6.5) ≈ - (GMEm1/r'13)r'1 - T 1 – m1S' – m1ω x (ω x r1) – 2m1 ω x v1 – m1 x r1 1 2 3 4 5 6 1: m1r1( - 2 - 2 sin2θ) = - (GMEm1/r'13)(bcosθ +r1) - T + m1ω2b cosθ - m1ω2r1(sin2θcos2φ - 1) 1 2 3 4 - 2m1ωr1 ( sinφ + sinθ cosθcosφ) (F.6.16)ok 5 : m1r1 ( - 2 sinθ cosθ) = + (GMEm1/r'13) b sinθ - m1ω2bsinθ 1 3 - m1ω2r1 sinθcosθcos2φ + 2m1ωr1 sin2θcosφ (F.6.17)ok 4 5 : m1r1(2 cosθ + sinθ) = + m1ω2r1sinθcosφsinφ - 2m1ωr1( sinθcosφ) (F.6.18)ok 4 5 We now rewrite the three equations dividing by m1 and using (F.5.1) that GME = ω2b3 : 1: r1( - 2 - 2 sin2θ) = - ( ω2b3/r'13)(bcosθ +r1) - T/m1 + ω2b cosθ - ω2r1(sin2θcos2φ - 1) - 2ωr1 ( sinφ + sinθ cosθcosφ) (F.6.19)ok : r1 ( - 2 sinθ cosθ) = + (ω2b3/r'13) b sinθ - ω2bsinθ - ω2r1 sinθcosθcos2φ + 2ωr1 sin2θcosφ (F.6.20)ok : (2 cosθ + sinθ) = + ω2sinθcosφsinφ - 2ω( sinθcosφ) (F.6.21)ok If one uses (F.1.4) that r'12 = b2 + r12 + 2br1cosθ in (F.6.20), the pair of equations (F.6.20) and (F.6.21) can in theory be solved for θ(t) and φ(t), given appropriate initial conditions. The solutions can then be inserted into (F.6.19) to obtain a result for the stick tension T(t). We can rewrite(F.6.21) as : sinθ + 2 cosθ - ωsinθcosφ (ωsinφ - 2) (F.6.23)ok which exactly matches the torque equation (F.5.7), sinθ + 2 cosθ – ωsinθcosφ (ωsinφ -2) = 0 // (F.5.7) Next, the equation (F.6.20) may be rewritten r1 ( - 2 sinθ cosθ) = + (ω2b3/r'13) b sinθ - ω2bsinθ - ω2r1 sinθcosθcos2φ + 2ωr1 sin2θcosφ or r1 ( - 2 sinθ cosθ) = +ω2bsinθ [(b/r'1)3 - 1] - ω2r1 sinθcosθcos2φ + 2ωr1 sin2θcosφ or - 2 sinθ cosθ = +ω2bsinθ [(b/r'1)3 - 1]/r1 - ω2sinθcosθcos2φ + 2ω sin2θcosφ . (F.6.24) Now for the first time we assume the far approximation where r'1, b >> r. Recall that r'12 = b2 + r12 + 2br1cosθ (r'1/b)2 = 1 + (r1/b)2 + 2(r1/b)cosθ (r'1/b)3 = [ 1 + (r1/b)2 + 2(r1/b)cosθ ]3/2 (b/r'1)3 = [ 1 + (r1/b)2 + 2(r1/b)cosθ ]-3/2 ≈ 1 + (-3/2) [(r1/b)2 + 2(r1/b)cosθ ] ≈ 1 + (-3/2) 2(r1/b)cosθ = 1 - 3(r1/b)cosθ so [(b/r'1)3 - 1] ≈ - 3(r1/b)cosθ . (F.6.25) Then (F.6.24) becomes - 2 sinθ cosθ = +ω2bsinθ [- 3(r1/b)cosθ]/r1 - ω2sinθcosθcos2φ + 2ω sin2θcosφ or - 2 sinθ cosθ = -3ω2sinθcosθ - ω2sinθcosθcos2φ + 2ω sin2θcosφ or - 2 sinθ cosθ + 3ω2sinθcosθ + ω2sinθcosθcos2φ -2ω sin2θcosφ = 0 or + 3ω2sinθcosθ - 2 sinθ cosθ + ωsinθcosφ(ωcosθcosφ - 2sinθ) = 0 (F.6.26) which exactly matches the torque equation (F.5.7), + 3ω2sinθcosθ - 2 sinθ cosθ + ωsinθcosφ (ω cosθ cosφ - 2sinθ ) = 0 // (F.5.7) Finally we come to the 1 equation which we divide by b: 1: [r1/b]( - 2 - 2 sin2θ) = - ( ω2b3/r'13)(cosθ +[r1/b]) - T/(m1b) + ω2cosθ - ω2[r1/b](sin2θcos2φ - 1) - 2ω[r1/b] ( sinφ + sinθ cosθcosφ) (F.6.27) If we now drop terms over order [r1/b] in the far approximation, we then have 1: 0 ≈ - ( ω2b3/r'13)(cosθ) - T/(m1b) + ω2cosθ (F.6.28) From (F.6.25) one has (b/r'1)3 ≈ 1 - 3(r1/b)cosθ so (F.6.28) becomes 0 ≈ - ( ω2[1 - 3(r1/b)cosθ])(cosθ) - T/(m1b) + ω2cosθ = -ω2cosθ + 3 ω2(r1/b)cos2θ - T/(m1b) + ω2cosθ = 3 ω2(r1/b)cos2θ - T/(m1b) . Thus, in the far approximation, the tension in the stick is given by T = 3m1 ω2r1cos2θ (F.6.29) Since this is non-negative for all angles θ, we may conclude that the stick may be replaced with a tether in the dumbbell satellite and the tether will never be under compression.