An asteroid orbits the Sun at an average distance of a= 4 AU. How long does it take to orbit the Sun?Use Kepler’s Third Law(p2= a3) to calculate the answer.

7982 seconds

Explanation:

Parameters given:

Distance of asteroid, a = 4 AU = 5.984 * 10^8 m

Mass of sun, M = 1.9891 * 10^30 kg

Orbital period, P, is given as:

P² = (4π²a³) /(GM)

Where G = 6.6774 * 10^(-11) m³/kgs²

P² = [4 * π² * (5. 984 * 10^8)³]/(6.674 * 10^(-11) * 1.9891 * 10^30)

P² = (8.459 * 10^27)/(1.328 * 10^20)

P² = 6.372 * 10^7

=> P = 7982 seconds

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