Asked by jimmyjonas

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?
Answered by charven
i dint know
Answered by tchrwill
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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