What is the period of revolution of a satellite with mass m that orbits the earth in a circular path of radius 7820 km (about 1450 km above the surface of the earth)?

Answer 1

F = m a
Force of gravity = m v^2/r

G m Me /r^2 = m v^2/r
where G is Newton's gravitational constant and Me is mass of earth
r = 7,820,000 meters

G Me = v^2 r

v^2 = G Me/r
solve for v

(2 pi r)/v = period

Answer 2

6979.13Second