**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 10^{24} kg and radius of the Earth = 6.4 x 10^{6} m. **Solution:** The height at which the satellite orbits the earth is 200 km or 0.2 x 10^{6} m. Therefore, the distance of the satellite from the center of the earth is (6.4 + 0.2) x 10^{6} m = 6.6 x 10^{6} m. Substituting *r* = 6.6 x 10^{6} and *M* = 6 x 10^{24}, the time period of the satellite is *T* = 10621.56 s from equation (G.20) and its orbital velocity is *v* = 3.04 x 10^{3} m/s = 3.04 km/s according to equation (G.21). |