posted by jimmyjonas on .
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?
i dint know
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.