g at r=6.37 * 10^6 m is 9.8
3.59*10^7 = 5.63r, g = 9.8/5.63^2 = 0.31 m/s^2
Geosynchronous communications satellites are placed in a circular orbit that is 3.59 107
m
above the surface of the earth. What is the magnitude of the acceleration due to gravity at this
distance?
1 answer