Question

Assume a planet is a uniform sphere of radius R that (somehow) has a narrow radial tunnel through its center. Also assume we can position an apple anywhere along the tunnel or outside the sphere. Let F_R be the magnitude of the gravitational force on the apple when it is located at the planet's surface. How far from the surface (what multiple of R) is there a point where the magnitude of the gravitational force on the apple is 0.7 F_R if we move the apple (a) away from the planet and (b) into the tunnel?

Answer 1

The force of gravity one experiences is due to the mass which is at a distance from the center of mass that the person is. So if you are above a planet the force of gravity is due to the entire mass of the planet, but if you are under the surface, only the mass that is contained in the sphere of radius equal to the distance you are from the center gives rise to the force of gravity.
a) F = GMm/r^2
FR = GMm/R^2
0.7FR = GMm/r^2
0.7(GMm/R^2) = GMm/r^2
0.7/R^2 = 1/r^2
r^2 = R^2/0.7
r = R/sqrt(0.7)
r = 1.2R
b)GM'm/r^2 = 0.7GMm/R^2
We need to find M' (the mass of the sphere with radius r)
M = p(4/3)piR^3 -------- p(4/3)pi = M/R^2
M' = p(4/3)pi(r^3)
M' = Mr^3/R^3
GMmr^3/R^3(r^2) = 0.6GMm/R^2
r/R^3 = 0.7/R^2
r = 0.7R

Hope this helps.

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