center of gravity of satellite v2

Center of Gravity for Satellite of two masses on massless stick PhL 1.11.15 The Earth is treated as a point particle of mass M located at its center. Here is the picture: The "satellite" consists of masses m1 and m2 at the ends of a massless rigid stick. Mass m2 is located on the z axis at z = h, while mass m1 is located relative to mass m2 by the stick vector s. By this means we can have the two masses be in an arbitrary orientation for this problem. Now in spherical coordinates we know that = cosφ sinθ + sinφ sinθ + cosθ Therefore for our little stick unit vector we can write = cosφ sinψ + sinφ sinψ + cosψ I think of the Earth as the test mass mE, a point mass located at its center. The force on this test mass due to the two masses is this F = GmE [ (m1/r2) + (m2/h2) ] = GmE [ (m1/r3) r + (m2/h3) h ] = F This vector certainly lies in the plane φ. If m1 = 0 it lies in the h direction and if m2 = 0 it lies in the r direction, so for general masses it lies in the φ plane between the vectors r and h. Let R be the solution center of gravity vector (point) which lies in this plane and which has length R so R = R Then the main act is this GmEM/R2 = F M = m1+ m2 We know something about θ from the law of cosines s2 = r2+h2 - 2rhcosθ cosθ = (r2+h2-s2)/(2rh) Now = cosθ = (r2+h2-s2)/(2rh) We know that F2 = (GmE)2[ (m1/r2)2 + (m2/h2)2 + 2(m1/r2)(m2/h2) * (r2+h2-s2)/(2rh) ] = (GmE)2[ m12/r4 + m22/h4 + m1m2(r2+h2-s2) /(r3h3) ] = (GmE/r2h2)2 [ m12h4 + m22r4 + m1m2(r2+h2-s2) rh ] F = (GmE/r2h2) = (GmE/r12r22) So then the distance to the COG point would be R2 = GmE(m1+m2)/F It seems unlikely that this COG point lies on the stick! Special case. Suppose ψ = 0 in which case we have h + s = r. The stick is then vertical and on the z axis. Then F = GmE [ (m1/r3) r + (m2/h3) h ] = GmE [ (m1/r2) + (m2/h2) ] = GmE[(m1/r2) + (m2/h2)] F = GmE[(m1/r2) + (m2/h2)] Rcog2 = GmE(m1+m2)/F = = The center of mass is this Rcms = (r1m1 + r2m2)/(m1+m2) = If the masses are equal we find Rcog2 = = and Rcog = r1r2/ Rcms = (r1+r2)/2 Rcog = Here then are the results for the Special Case (tethered satellite): general mass equal mass Rcms = Rcms = Rcog = Rcog = I know these both lie in the range (r2, r1). I would guess that Rcog < Rcms since gravity is stronger on the closer mass. Let's check Rcog2 < Rcms2 ? < ( )2 ? < (r1m1 + r2m2)2 ? < ( )2 ? μi = mi/(m1+m2) < ( r1μ1 + r2μ2)2 ? ( r1μ1 + r2μ2)2 [ (μ1/r12) + (μ2/r22)] > 1 ? ( r1μ1 + r2μ2)2 ( μ1r22 + μ2r12) > r12r22 ? ( r1μ1 + r2μ2) ( r1μ1 + r2μ2)( μ1r22 + μ2r12) > r12r22 ? Let r2/r1 = x. Then ( μ1 + xμ2) (μ1 +xμ2)( μ1 + μ2x-2) > 1 ? This is just not very obvious! ( μ1 + x[1-μ1]) (μ1 +x[1-μ1])( μ1 + [1-μ1]x-2) > 1 ? Now only two variables μ1 in (0,1) and x in (0,∞). Write μ1 = y to simplify ( y + x[1-y]) (y +x[1-y])(y + [1-y]x-2) > 1 ? I bet it is not true! ( y + x[1-y])2(y + [1-y]x-2) > 1 ? Plot this in Maple. Maple shows that it is really true! Wow. ( y + x[1-y])2(y + [1-y]x-2) > 1 ? ( y + x[1-y])2(x2y + [1-y]) > x2 ? ( y + x[1-y])2(x2y + [1-y]) - x2 > 0 ? Maple can expand then factor this thing y(x-1)2(y-1)(x2y-y-x2-2x) > 0 ? 0 < y < 1 y(x-1)2(1-y)(-x2y+y+x2+2x) > 0 ? 0 < y < 1 + + + ? I would be done if I could show that -x2y+y+x2+2x > 0 x2(y-1) +x2+2x > 0? The smallest this can be occurs with y = 0. At that point the question is -x2+x2+2x > 0? 2x > 0 ? Yes, since 0 < x < ∞, this is always true. Thus I have shown that ( μ1 + x[1-μ1]) (μ1 +x[1-μ1])( μ1 + [1-μ1]x-2) > 1 and therefore < ( )2 and therefore Rcog < Rcms as I hoped would be the case. This confirms that tether picture. Question: How do you interpret the COG point in terms of an orbit? You have a 1-mass M system orbiting at radius r in circular orbit. You replace it with a 2-mass system of the same total mass, and you do it in such a way that the 2-mass system is swapped in so its COG point lies on the same orbit of radius r. This system then has the same total force toward Earth center that the single mass had, so it has the same acceleration a, so it has the same orbit. Both masses of the 2-mass system have this same acceleration a toward earth center. There is tension in the stick. Exercise. Suppose masses are equal. Then write r1 = r0 + ε r2 = r0 – ε r0 = rcms If you use these in the rcog equation you get rcog = = = = (r0- ε)(r0+ε) = (r02 - ε2) = (r02- ε2) (r02+ε2)-1/2 = r02 (1 - (ε/r0)2) (1/r0) (1 +(ε/r0)2)-1/2 = r0 (1 - (ε/r0)2) (1 +(ε/r0)2)-1/2 (1+a)n = 1 + na + order(a2) ≈ r0 (1 - (ε/r0)2) (1 -1/2(ε/r0)2 ≈ rcms [ 1 - (3/2) (ε/r0)2 + order (ε/r0)4 ] In the case that ε ~ r0/10 say, you then have an estimate for the difference rcog ≈ rcms [ 1 - (3/2) (ε/rcms)2 ] which is accurate to an error on the order of 10-4 inside the bracket. Appendix A r = h + s = h + s = h + scosφ sinψ + ssinφ sinψ + scosψ = scosφ sinψ + ssinφ sinψ + (scosψ+h) Then r2 = [scosφ sinψ + ssinφ sinψ + (scosψ+h) ]2 = s2cos2φ sin2ψ + s2sin2φ sin2ψ + (scosψ+h)2 + 2s2sinψ sinφ cosφ + 2ssinψcosφ(scosψ+h) + 2ssinψsinφ(scosψ+h) = s2sin2ψ + s2cos2ψ + 2hscosψ + h2 + 2s2sinψ sinφ cosφ + 2s2sinψcosφcosψ + 2s2sinψsinφcosψ + 2shsinψcosφ + 2shsinψsinφ = s2 + 2hscosψ + h2 + 2s2 sinψ (sinφ cosφ + cosφcosψ +sinφcosψ) + 2sh sinψ(cosφ + sinφ) It is a real mess. But for specific numbers, I can write an expression for F and for . Assume then that we have these two quantities. We then want to do this: