Asked by bill
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?
m
above the surface of the earth. What is the magnitude of the acceleration due to gravity at this
distance?
Answers
Answered by
Steve
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
3.59*10^7 = 5.63r, g = 9.8/5.63^2 = 0.31 m/s^2
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