Question

In an experiment performed at the bottom of a very deep vertical mine shaft, a ball is tossed vertically in the air with a known initial velocity of 10.0 m/s, and the maximum height the ball reaches (measured from its launch point) is determined to be 5.101 m. Knowing the radius of the Earth, RE = 6370 km, and the gravitational acceleration at the surface of the Earth, g(0) = 9.81 m/s2, calculate the depth of the shaft.

Answer 1

For determining the gravity inside the (spherical) earth, at a distance r<R from the center, only the mass of the earth within the sphere of radius r matters. This is Newton's " shell theorem" which itself simply follows from the 1/r^2 character of gravitational force. Let's call this mass enclosed in the sphere of radius r: M(r), so M(R) = M.
For r<R, F = GM(r)m/r^2 . Since (assuming constant mass density and the fact that the volume of a sphere increases as r^3) M/R^3 = M(r)/r^3, we have F = (GMm/R^3 ) r. In other words we can say that for r<R, g depends on r as g(r) = g(0) r/R.

Now for the motion: The potential energy increase equals integral mg(r) dr = mg(0)/(2 R) (r^2- r0^2). This equals the decrease in kinetic energy : 1/2 m v0^2.
Hence r^2 - r0^2 = R/g(0) * v0^2 = 64,933,741.1 m^2. Now we are given r-r0 = 5.101 m and we derived r^2 - r0^2 = (r - r0)(r + r0) = 64,933,741.1 m^2, so we can solve for r and r0.
r+r0 = 64,933,741.1 m^2 / 5.101 m = 12,729610.1 m

Hence
r + r0 = 12,729610.1 m
r - r0 = 5.101 m

2 r0 = 12,729605 m
r0 = 6364802.5 m
So the depth is: R - r0 = 6370 km - 6364.8 km = 5.19 km