Given: G = 6.67259 × 10−11 Nm2/kg2

Question

The acceleration of gravity on the surface of a planet of radius R = 3430 km is 4.43 m/s2.What is the period T of a satellite in circular orbit h = 7683.2 km above the surface?
Answer in units of s

Answer 1

The satellite is
R = 3430 + 7683 = 11,113 km from the center of the planet.

The value of the acceleration of gravity at the orbit distance R is

g' = 4.43 m/s^2 * (3430/11,113)^2 = 0.4220*10^-2 m/s^2

Set g' = V^2/R to get the satellite velocity V
Make sure R is in meters and V is in m/s

Then use V*T = 2 pi R to get the period, T.