Question 1

Multiple choiceHard1 mark

A satellite moves in an elliptical orbit around a planet. Its closest and farthest distances from the planet are 1.0×107 m1.0\times10^7\text{ m} and 7.0×107 m7.0\times10^7\text{ m} A satellite in a circular orbit around the same planet has radius 2.0×107 m2.0\times10^7\text{ m} and period 4.0 d4.0\text{ d} Which pair gives the elliptical orbit’s period, followed by the ratio of its speed at closest approach to its speed at farthest distance?

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Answer: C. For the same planet, Kepler’s third law gives T2/a3=constantT^2/a^3=\text{constant} and Kepler’s second law gives equal swept areas in equal times.

  1. The semi-major axis is a=(1.0+7.0)×107/2=4.0×107 ma=(1.0+7.0)\times10^7/2=4.0\times10^7\text{ m} Thus T=4.0(4.0/2.0)3/2=11.3 dT=4.0(4.0/2.0)^{3/2}=11.3\text{ d} which rounds to 11 d11\text{ d}
  2. At closest and farthest approach, velocity is perpendicular to the radius, so equal area rates give rclosevclose=rfarvfarr_{\rm close}v_{\rm close}=r_{\rm far}v_{\rm far} Hence vclose/vfar=7v_{\rm close}/v_{\rm far}=7 A: omits the cube in Kepler’s third-law relation, using T/T0=a/a0T/T_0=\sqrt{a/a_0} B: reverses the speed ratio. D: uses T/T0=(a/a0)3T/T_0=(a/a_0)^3 omitting the square root. Remember: orbital speed is greater nearer the central body.

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