ch13-p075 - 75. (a) Using Keplers law of periods, we obtain...

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75. (a) Using Kepler’s law of periods, we obtain 2 34 4 2.15 10 s . Tr GM π §· == × ¨¸ ©¹ (b) The speed is constant (before she fires the thrusters), so v o = 2 π r/T = 1.23 × 10 4 m/s. (c) A two percent reduction in the previous value gives v = 0.98 v o = 1.20 × 10 4 m/s. (d) The kinetic energy is K = ½ mv 2 = 2.17 × 10 11 J. (e) The potential energy is U = GmM/r = 4.53 × 10 11 J. (f) Adding these two results gives E = K + U = 2.35 × 10 11 J.
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