Physics
posted by Max on .
A spy satellite is in circular orbit around Earth. It makes one revolution in 6.04 hours.
(a) How high above Earth's surface is the satellite?
(b) What is the satellite's acceleration?

YOu start with centripetal acceleration= gravitational attraction.
w^2 * r= 9.8 (re^2/(r^2)) where re is the radius of the Earth, and r is the distance from the center of the Earth to the satellite.
w= 2PIr/period work in m/s, m 
so W = (2PI(6.37x10^6))/21,744sec
6.04hr = 21,744 sec
and then for r = 117,367
I don't get what to do afterwards