Question

Calculate the period of a satellite orbiting the Moon, 96 km  above the Moon's surface. Ignore effects of the Earth. The radius of the Moon is 1740 km..

Answer 1

use Kepler's third law

MP^2 = (4 pi^2/G) d^3

where M= mass of the moon = 7.3x10^22kg

P=period in seconds

G=newtonian grav cst = 6.67x10^-11

d=semi-major axis = distance from center of moon = 96km+1740km = 1836km = 1.836 x 10^6m

solve for P:

P = Sqrt[(4 pi^2/G M)*d^3]

P=Sqrt[(4 pi^2/6.67x10^-22*7.3x10^22kg) x (1.836x10^6m)^3]

P=7079.64s or 1.966hrs