Posted by jimmyjonas on Tuesday, February 22, 2011 at 8:50am.
2.A spy satellite is in circular orbit around Earth. It makes one revolution in 6.02 hours.
(a) How high above Earth's surface is the satellite?
(b) What is the satellite's acceleration?

physics  charven, Tuesday, February 22, 2011 at 10:17am
i dint know

physics  tchrwill, Tuesday, February 22, 2011 at 10:43am
The period of an orbitig satellite is given by T = 2(Pi)sqrt[r^3/µ] where T = the orbital period in seconds, r = the orbit radius and µ = the gravitational constant of the Earth.
Therefore, with T = 6.02(3600) = 21,672 seconds
21,672 = 2(3.14)sqrt[r^3/1.408x10^16]
from which r = 10,440 miles.
Subtracting the Earth's radius of 3963 miles results in an orbit altitude of 6477 miles.
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