Question

A satellite revolves around a planet at an altitude equal to the radius of the planet. the force of gravitational interaction between the satellite and the planet is f0. then the satellite moves to a different orbit, so that its altitude is tripled. find the new force of gravitational interaction f2.

Answer 1

f2 = f0/4 The gravity from the planet can be modeled as a point source at the center of the planet with all of the planet's mass concentrated at that point. So the initial condition for f0 has the satellite at a distance of 2r, where r equals the planet's radius. The expression for the force of gravity is F = G*m1*m2/r^2 where F = Force G = Gravitational constant m1,m2 = masses involved r = distance between center of masses. Now for f2, the satellite has an altitude of 3r and when you add in the planet's radius, the distance from the center of the planet is now 4r. When you compare that to the original distance of 2r, that will show you that the satellite is now twice as far from the center of the planet as it was when it started. So let's compare the gravitational attraction, before and after. f0 = G*m1*m2/r^2 f2 = G*m1*m2/(2r)^2 f2/f0 = (G*m1*m2/(2r)^2) / (G*m1*m2/r^2) The Gm m1, and m2 terms cancel, so f2/f0 = (1/(2r)^2) / (1/r^2) f2/f0 = (1/4r^2) / (1/r^2) And the r^2 terms cancel, so f2/f0 = (1/4) / (1/1) f2/f0 = (1/4) / 1 f2/f0 = 1/4 f2 = f0*1/4 f2 = f0/4 So the gravitational force on the satellite after tripling it's altitude is one fourth the original force.