One of the moons of a planet orbits at a distance of 9405 km from the center of the planet. If the mass of the planet is 6.42 ✕ 1023 kg, what is the period of the moon?

Answer 1

From setting centripetal force = gravitational force, v² = GM/r. The orbital speed is also circumference divided by period T, so v² = (2π r / T)² = 4π² r² / T². Note that r = 9.405 x 10³ km or 9.405 x 10⁶ m. Then T² = 4π² r³ /GM = 4π² (9.405 x 10⁶ m)³ / [(6.67 x 10^-11)(6.42 x 10²³)] = 7.67 x 10⁸ and T = 2.77 x 10⁴ seconds = 462 minutes.