Example:
A satellite is revolving at a height of 200 km above the surface of the Earth. Find its time period and orbital velocity.
Given: mass of the Earth = 6 x 1024 kg and radius of the Earth = 6.4 x 106 m.
Solution:
The height at which the satellite orbits the earth is 200 km or 0.2 x 106 m.
Therefore, the distance of the satellite from the center of the earth is (6.4 + 0.2) x 106 m = 6.6 x 106 m.
Substituting r = 6.6 x 106 and M = 6 x 1024, the time period of the satellite is T = 10621.56 s from equation (G.20) and its orbital velocity is v = 3.04 x 103 m/s = 3.04 km/s according to equation (G.21).
