F6 in Cartesian coordinates

F.6 Force analysis of the Dumbbell Satellite in Frame S' in Cartesian Coordinates 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 Cartesian 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): m1a1 = m1(ax + ay + az ) (E.3.6) (F.6.9) Term 1: - (GMEm1/r'13)r'1 = - (GMEm1/r'13)(b + r1) // (F.1.8) = - (GMEm1/r'13)( b + r1sinθcosφ + r1sinθsinφ + r1cosθ ) = - (GMEm1/r'13)[ r1sinθcosφ + r1sinθsinφ + (b+r1cosθ)] (F.6.10) Term 2: - T 1 = -T [r1sinθcosφ + r1sinθsinφ + r1cosθ ] (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.6.12) 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 + m1ω2r1[sinθcosφ + sinθsinφ + cosθ] // (E.2.4)) = m1ω2r1[ (sinθcosφ - sinθcosφ) + sinθsinφ + cosθ ] = m1ω2r1[ sinθsinφ + cosθ ] (F.6.13) Term 5: -2m1 ω x v1 = -2m1 [ω] x ( vx + vy + vz) = -2m1ω [ vy - vz] (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 : m1 a1x = - (GMEm1/r'13) r1sinθcosφ - Tr1sinθcosφ 1 2 : m1 a1y = - (GMEm1/r'13) r1sinθsinφ - T r1sinθsinφ + m1ω2r1 sinθsinφ + 2m1ωvz 1 2 4 5 : m1 a1z = - (GMEm1/r'13) (b+r1cosθ) - T (b+r1cosθ) + m1ω2b + m1ω2r1 cosθ - 2m1ω vy 1 2 3 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 : : a1x = - (ω2b3/r'13) r1sinθcosφ - (T/m1)r1sinθcosφ : a1y = - (ω2b3/r'13) r1sinθsinφ - (T/m1) r1sinθsinφ + ω2r1 sinθsinφ + 2ωvz : a1z = - (ω2b3/r'13) (b+r1cosθ) - (T/m1) (b+r1cosθ) + ω2b + ω2r1 cosθ - 2ω vy or : = - (ω2b3/r'13) r1sinθcosφ - (T/m1)r1sinθcosφ : = - (ω2b3/r'13) r1sinθsinφ - (T/m1) r1sinθsinφ + ω2r1 sinθsinφ + 2ω : = - (ω2b3/r'13) (b+r1cosθ) - (T/m1) (b+r1cosθ) + ω2b + ω2r1 cosθ - 2ω Comments: These are "not bad" really, could have been much worse. But now we have the following unknown functions x, y,z,θ,φ. I guess I could have done it all simpler and I now rewrite : = - (ω2b3/r'13) x - (T/m1)x : = - (ω2b3/r'13) y - (T/m1)y + ω2y + 2ω : = - (ω2b3/r'13) (b+z) - (T/m1) (b+z) + ω2(b+z) - 2ω Then r'12 = (r1+b2) = r12 + b2 + 2 r1 b = r12 + b2 + 2b r1 b = r12 + b2 + 2bz Rewrite them again = - [(ω2b3/r'13) + (T/m1)] x r'12 = r12 + b2 + 2bz = - [(ω2b3/r'13) + (T/m1) - ω2]y + 2ω = - [(ω2b3/r'13) + (T/m1) +ω2](b+z) - 2ω x2+y2+z2 = r12 This is a system of 4 equations in 4 unknowns x,y,z,T . Suppose I try to eliminate x. Then with some Maple help, We then end up with 3 equations in three unknowns which are y,z,T. Maybe one should eliminate T early in the game. Rewrite the above y = - [(ω2b3/r'13) + (T/m1)] yx r'12 = r12 + b2 + 2bz x = - [(ω2b3/r'13) + (T/m1) - ω2]yx + 2ωx = - [(ω2b3/r'13) + (T/m1) +ω2](b+z) - 2ω Subtract the first two to get y - x = - ω2yx - 2ωx Go back again (b+z) = - [(ω2b3/r'13) + (T/m1)] x(b+z) r'12 = r12 + b2 + 2bz = - [(ω2b3/r'13) + (T/m1) - ω2]y + 2ω x = - [(ω2b3/r'13) + (T/m1) +ω2](b+z)x - 2ωx Now subtract the third from the first, (b+z) - x = +ω2(b+z)x + 2ωx Now I have three equations in three unknowns which are x,y,z y - x = - ω2yx - 2ωx (b+z) - x = +ω2(b+z)x + 2ωx x2+y2+z2 = r12 Once you solve these, then T is determined by any of the equations. Maybe I can assume b >> z to get y - x = - ω2yx - 2ωx b - x = +ω2bx + 2ωx x2+y2+z2 = r12 I will pause this effort right here and work on the gas valve which just arrived.