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 don't get what to do afterwards
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