The Mariner 10 spacecraft was placed in an elliptical orbit around the Sun,with the period of its orbit twice that of the period of Mercury's orbit.What is the length of the semimajor axis of Mariner 10's orbit? (Be careful with units,and review Kepler's laws,Section 2-5,Comins,Discovering the Essential Universe,6th ed.)

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According to Kepler's third law, (T^2 /r^3) is the same for all objects orbiting the Sun. Here, Mariner 10's period is twice that of Mercury's period, so T(Mariner 10) = 2T(Mercury). Plugging this into Kepler's third law and solving for r, we get r(Mariner 10) = 2^(2/3) r(Mercury) = 1.122 r(Mercury). Mercury's average distance from the Sun is about 0.39 AU, so we get r(Mariner 10) = 0.39 x 1.122 = 0.437 AU. However, this is only the semi-minor axis of Mariner 10's elliptical orbit. Since we know that its orbit is twice as long as Mercury's, the semi-major axis of Mariner 10's orbit is simply twice its semi-minor axis, or 2 x 0.437 AU = 0.874 AU. Therefore, the length of the semi-major axis of Mariner 10's orbit is 0.874 AU, which is closest to option B) 0.61 AU.

The Mariner 10 Spacecraft Was Placed in an Elliptical Orbit