Asked by Charlie
An object of mass 3.5x1023kg circles the earth and is attracted to it with a force whose magnitude is given be GMeM/r^2. If the period of rotation is 22 days, what is the distance from the earth to the object? Here G=6.67x10-11 Nm2/kg2, Me=6x1024 kg.
Answers
Answered by
drwls
G Me M/r^2 = M V^2/r
Express V in terms of the period P = 22 days = 1.90*10^6 seconds
2 pi r/P = V
G Me/r^2 = 4 pi^2 r/P^2
4 pi^2/(G*Me) = r^3/P^2
Solve for r.
Express V in terms of the period P = 22 days = 1.90*10^6 seconds
2 pi r/P = V
G Me/r^2 = 4 pi^2 r/P^2
4 pi^2/(G*Me) = r^3/P^2
Solve for r.
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