What is the tangential velocity of a satellite moving in a circular orbit at a height of 5250 km above the surface of Earth? The standard gravitational parameter of Earth is 398,600 km3/s2. The radius of Earth is 6378 km.

Correct Answer: 5.85 km/s
Given, Distance between Earth and satellite (r) = 5250 + 6378 = 11,628 km Standard gravitational parameter of Earth (μ) = 398,600 km3/s2 Since, gravitational attraction between Earth and satellite = centrifugal force of the satellite, we can derive, Tangential velocity of satellite (V) = (μ/r)1/2 = (398,600/11,628)1/2 = 5.85 km/s