Question

A spacecraft is in orbit around a planet. The radius of the orbit is 2.9 times the radius of the planet (which is R = 71451 km). The gravitational field at the surface of the planet is 21 N/kg. What is the period of the spacecraft's orbit?

Answer 1

Kepler's third law tells us that

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

M is the mass of the planet (in kg)
P the period (in secs)
G is the newtonian grav cst = 6.67 x10^-11 in MKS units
d = orbital radius in m

we can find M knowing that the

surface gravity g = 21N/kg from

g = GM/R^2

M = g R^2/G = 21N/kg x (71451*1000m)² / 6.67x10^-11 = 1.60735x10²⁷ kg

we know that d = 2.9R =207207900 m

so that
P^2 = (4 pi^2/G)(207207900 m)^3/1.60735x10²⁷kg

P^2 = 3.275 x 10⁹s

P = 57236.3 s