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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