Question

A spy satellite is in circular orbit around Earth. It makes one revolution in 5.95 hours. (Radius of the Earth=6.371 106 m)

(a) How high above Earth's surface is the satellite?
(b) What is the satellite's acceleration?

Answer 1

velocity=2PI*(re+h)/period

but velocity can be found..

9.8(re/(re+h))^2=v^2/(re+h)

or v=sqrt (9.8 (re^2/(re+h))

then put that v into the first equation, and solve for h.

This is a form of Keplers' law.