An Earth satellite moves in a circular orbit 824 km above Earth's surface with a period of 101.2 min. What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?

Answer 1

With a mean earth radius of 6378km, the orbital radius becomes 6378 + 824= 7202km.

The orbital circumference is then C = 2(Pi)7202 = 45,251km.

The orbital velocity is then Vc = 45,251km/6072sec. = 7.452km/sec = 7452m/s.

The centripetal acceleration is a = V^2/r.