Posted by **Max** on Tuesday, March 2, 2010 at 7:51pm.

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?

- Physics -
**bobpursley**, Tuesday, March 2, 2010 at 8:09pm
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

- Physics -
**Max**, Tuesday, March 2, 2010 at 8:37pm
so W = (2PI(6.37x10^6))/21,744sec

6.04hr = 21,744 sec

and then for r = 117,367

I dont get what to do afterwards

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