Appendix F v0

Appendix F v0 PhL 1.11.15 This was an early attempt at doing the torque, and I get the result ωt ≈ ω. Still using the planar picture and Earth at the top. D.6 Gravitational torque on the tethered satellite As noted in the main text, gravity exerts a torque on a tethered satellite which acts to restore it to the nearest vertically aligned position. Here we derive an expression for that torque. We start by recalling Fig (D.2.1), where points toward the viewer, (D.5.1) We shall compute the torque about the origin of this figure, but this is the same as the torque about the center of gravity because I think you have to include the tether tension in this discussion. Maybe it should be computed as a first step. I cannot just say the above picture is "static" because I know it is not static, it is a snapshot of a position while the thing is librating say. The position shown is not stable. The forces F1 and F2 were found in (D.2.6) to be (these forces act on m1 and m2) , F1 = (Gmm1/r31) rsinθ + (Gmm1/r31) (h + rcosθ) F2 = -(Gmm2/r32) rsinθ + (Gmm2/r32) (h - rcosθ) (D.2.6) where from (D.2.3), r12 = (rsinθ)2 + (h+rcosθ)2 r22 = (rsinθ)2 + (h-rcosθ)2 . (D.2.3) (D.5.2) Mass m2 is at position r which is r = y + z == rsinθ + rcosθ . (D.5.3) With respect to the origin of the drawing we compute these torques on m1 and m2 N1 = (-r) x F1 = -(Gmm1/r13)[ rsinθ + rcosθ] x [ rsinθ + (h + rcosθ)] = -(Gmm1/r13){ rsinθ (h + rcosθ) + rcosθ rsinθ ( - ) } = -(Gmm1r/r13){ sinθ (h + rcosθ) - cosθ rsinθ } = -(Gmm1r/r13) h sinθ (D.5.4) N2 = r x F2 = (Gmm2/r32)[ rsinθ + rcosθ] x [ - rsinθ + (h - rcosθ) ] = (Gmm2/r32) { rsinθ (h - rcosθ) - rcosθ rsinθ( - ) } = (Gmm2r/r32) { sinθ (h - rcosθ) + cosθ rsinθ } = (Gmm2r/r32) h sinθ (D.5.5) The total torque on the satellite about the origin is then N = N1+ N2 = (Gmr) ( m2/r32 - m1/r13) h sinθ dim ok (D.5.6) which is in or out of the plane of paper depending on parameter sizes. In Fig (D.2.1) we take r,θ to be polar coordinates where -π < θ ≤ π. When m2 is on the right side of the drawing, sinθ > 0, and when it is on the left, sinθ < 0. The torque direction is given by, = sign( m2/r32 - m1/r13) sign(sinθ) (D.5.7) or = sign( m2/r32 - m1/r13) for m2 on the right = - sign( m2/r32 - m1/r13) for m2 on the left (D.5.8) Suppose m2 is a large mass so that (m2/r32 - m1/r13) > 0 and m2 is on the right. In this case, = and the torque acts to move m2 to the top of the circle. If m2 were on the left, one gets instead = - and again the torque acts to move m2 to the top of the circle. In this case, the stable position of the satellite has m2 on the top of the circle, which means it is closest to the Earth. Suppose m2 is a small mass so that (m2/r32 - m1/r13) < 0. In this case the torque acts to move m2 to the bottom of the circle, farthest from Earth. If the masses are equal, for m2 on the upper half of the circle one has r1 > r2 so then = sign( 1/r32 - 1/r13) = + for m2 on the right = - sign( 1/r32 - 1/r13) = - for m2 on the left (D.5.8) and this is the situation of a restoring torque, so either mass is stable in the vertically aligned case. Tether Oscillation Suppose m2 is a large enough mass so m2 is restored by the torque to the top of the circle. If θ is a small angle, the satellite oscillates in a harmonic fashion about this position. Similar to F = m for linear motion, one has N = I for angular motion where I = (m1+m2)r2 is the moment of inertia about the origin. If θ is small then sinθ ≈ θ and we then have N = I N = I (-) // because θ is defined clockwise in Fig (D.5.1), the "wrong way" N = -I (Gmr) ( m2/r32 - m1/r13) h θ + (m1+m2)r2 = 0 [ (Gmh/r) ( μ2/r32 - μ1/r13)] θ + = 0 dim ok + ωt2θ = 0. (D.5.9) The tethered satellite oscillates with motion θ = sin(ωtt + C) where ωt = (D.5.10) But for small θ in this position we have r1 ≈ h+r and r2≈ h-r so ωt = (D.5.11) What happens in the limit that m1 → 0? Then have ωt ≈ ≈ = (1/h) ωt ≈ (1/RE) = = = ω This says as r→ 0, ωt → ∞. How do I explain that? Just have a single point mass m2 in orbit, why should there be a libration going on in the first place? RE = 6371 km ≈ 6.4x106 m r ≈ 10m = ≈ 2500 Tt = 88 min / 2500 = 88 x 60 sec/2500 = 2 second For h >> r (typically the case!) one has 1/(h-r)3 = (h-r)-3 = h-3(1- r/h)-3 ≈ h-3 [1 + (-3)(-r/h)] // (1+a)n ≈ 1 + nb = h-3 [1 + 3(r/h)] 1/(h+r)3 = h-3 [1 - 3(r/h)] . (D.5.12) Then ( μ2/r32 - μ1/r13) ≈ μ2/(h-r)3 - μ1/(h+r)3 ≈ μ2 h-3 [1 + 3(r/h)] - μ1 h-3 [1 - 3(r/h)] = h-3 [ (μ2-μ1) + 3(r/h)(μ2+μ1)] = h-3[ (μ2-μ1) + 3(r/h) ] (D.5.13) and therefore ωt ≈ = (1/h) dim ok (D.5.14) If the masses are exactly equal so μ1=μ2 one gets ωt ≈ (1/h) = (1/h) = Recall from (8.22) that GME/r30 = ω2 . (8.22) where ω is the orbital frequency of a satellite and r0 is the radius of its center of gravity. For h>> r we have r0 ≈ h and then ω2 ≈ GME/h3 and this the natural tether frequency can be written ωt ≈ ω or Tt = T/ A low earth orbit satellite has a period of about 88 minutes so this tether oscillation period would be about 51 minutes, a little less than an hour, so the oscillation is relatively "slow". Result *** appears on page 126 of Ref [] . Suppose the masses are slightly unequal such that (μ2-μ1) >> 3(r/h) . Then we get ωt ≈ (1/h) dim ok and then ωt = ω ******************** Question: In (D.1.4) I claim that N(r) = 0 . I also say that N(0) = rcog x F where F is the total force on the system. In finding rcog I assume it is along F so then N(0) = 0 as well. So what the heck is the non-zero torque I am talking about above???