A satellite is to be launched into a circular orbit around t

A satellite is to be launched into a circular orbit around the Earth so that it orbits the planet once every T seconds. Treating the Earth as a perfect sphere with its center of mass in the middle, show that the altitude h above the Earth\'s surface that the satellite must have is h = (GMT^2/4 pi^2)^1/3 - R, where G = 6.67 x 10^-11 m^3 kg^-1 s^-2 is Newton\'s gravitational constant, M = 5.97 x 10^24 kg is the mass of the Earth, and R = 6371 km is its radius.

Solution

Gravitation force = GM m / (R + h)^2

as satellite moves in circular orbit,

so F = m * centripetal acc. = mw^2d

w = 2pi/ T

GM m / (R + h)^2 = m (2pi / T)^2 ( R+h)

(R + h)^3 = GMT^2 / 4 pi^2

R + h = (GMT^2 / 4 pi^2 )^1/3

h = (GMT^2 / 4 pi^2 )^1/3 - R