Assume a planet is a uniform sphere of radius R that (somehow) has a narrow radi

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 FR 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 0.6 FR if we move the apple (a) away from the planet and (b) into the tunnel?

Please provide units as well thanks

Explanation / Answer

on the surface of the planet , FR = GMm/R^2

m = mass of the apple

if we move away from the planet to a distance h from the surface

FR' = G*M*m/(R+h)^2

FR' = 0.6FR

G*M*m/(R+h)^2 = 0.6*G*M*m/R^2

R^2 = (R+h)^2*0.6

R = (R+h)*0.77

R *(1-0.77) = 0.77h

h = 0.29 R <------answer

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on the surface of the planet , FR = GMm/R^2

volume of the planet = (4/3)*pi*R^3

density of the planet = mass/volume = M/(4/3)*pi*R^3

if we move into the planet a distance h from the surface,the effective volume of the planet

dV = (4/3)*pi*r^3

r = (R-h) = distnce of the point from the center

mass = density * volume

M' = M/(4/3)*pi*R^3 *(4/3)*pi*r^3

M' = M*r^3/R^3

gravitaitonal force FR' = GM'*m/r^2

FR' = G*M*m*r^3/(R^3*r^2)

FR' = G*M*m*r/R^3

given FR' = 0.6FR

G*M*m*r/R^3 = 0.6*G*M*m/R^2

r/R = 0.6

r = 0.6R

R - h = 0.6R

h = R - 0.6R = 0.4 R <----answer

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