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?

Question

Quizplus · Accepted Answer

According to Kepler's third law, the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. If the period of Mariner 10's orbit is twice that of Mercury's orbit, then the ratio of their periods squared (T²) is 4 (since 2² = 4). Therefore, the cube of the semi-major axis of Mariner 10's orbit (a³) must also be 4 times that of Mercury's. Mercury's semi-major axis is approximately 0.39 AU, so to find the semi-major axis of Mariner 10's orbit that is 4 times the cube of Mercury's semi-major axis, we take the cube root of 4 times Mercury's semi-major axis cubed. This calculation gives us a semi-major axis of about 0.61 AU for Mariner 10's orbit.

The Mariner 10 Spacecraft Was Placed in an Elliptical Orbit