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 of 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 is there a point where the magnitude is $\frac{1}{2}{F}_R$ if we move the apple (a) away from the planet and (b) into the tunnel?

Answer 1

we set $GmM/r^2$ equal to $\frac{1}{2}GmM/R^2$ , and we find $r=R\sqrt{2}$. Thus,
the distance from the surface is $(\sqrt{2}-1)$ R= 0.414 R.
(b) Setting the density p equal to M/V where v=$\frac{4}{3}\pi R^3$, we use
$F=\frac{4\pi Gmrp}{3}=\frac{4\pi Gmr}{3}\left(\frac{M}{4\pi R^3/3}\right)=\frac{GMmr}{R^3}=\frac{1}{2}\frac{GMm}{R^2}\Rightarrow r=R/2.$