free fall - Gravity and bottomless pits

Assuming that someone is theoretically able to make a hole through the center of a large planet, and then jumps down the hole, what will happen? Given my understanding of gravity and energy, my estimate is that in the absence of any resistive forces, the person would fall right through to the other side, with at least part of their body coming through, and would continue oscillating like this. If there were resistive forces, such as air resistance, the person would oscillate similar to a spring winding down, eventually stopping in the middle of the planet. Is this what would happen, or are my calculations off somewhere?

Answer

If you use Gauss's theorem for the gravitational field of the Earth at radius $R$ and assume it has uniform density $\rho$.

$$g(R) = -GM(R)/R^2 = -4\pi R G \rho /3,$$ where $M(R)$ is the mass inside radius $R$, and is given by $4\pi R^3 \rho/3$.

Thus the force the body feels towards the centre of the Earth (neglecting air resistance - see below) is proportional to $R$ and thus

$$\frac{d^2 R}{dt^2} = -\frac{4\pi G \rho}{3} R,$$ which is a simple harmonic motion equation with an oscillation frequency of $(4\pi G\rho/3)^{1/2}$.

If I assume an average $\rho =5000\ kg/m^3$, this means an oscillation period of 88 minutes.

If you really did hollow out such a tunnel, it would quickly fill with air. You would reach a terminal velocity of order 200 km/h at best. Travelling at this sort of speed as you went through the centre of the Earth would mean you would only rise by a height on the other side given approximately by

$$\frac{1}{2} m v^2 = m \int_0^{H} g(R)\ dR = m \frac{2\pi G \rho H^2}{3}$$ $$H = \left(\frac{3}{4\pi G \rho}\right)^{1/2} v$$ Putting in some numbers then gives $H \simeq 50\ km$. So nowhere near emerging at the antipodes.