Body falling through Earth

Question

Imagine a body with mass $m$ moves through a pipe through the centre of earth. The gravitational force is given by $\vec{F}=-C \vec{r}$ where $C$ is constant. I want to determine the 1D equation of motion of the body and solve it for the case that the body starts at the surface of the earth with initial velocity $v_0=0$.

My attempt: $F=m\ddot{r}=-Cr$ so $r=Ae^{x\sqrt{-C/m}}+Be^{-x\sqrt{-C/m}}$. But I think this isn't right, shouldn't the body stay at the centre of the earth? Because I think the attraction is largest there.

Answer 1

The full mass of the earth doesn't contribute to the gravitational field. You can use Gauss' Law for the gravitational field to get the right dependence.

Here $\vec{a}$ is the acceleration due to gravity.

$\nabla\cdot\vec{a}=-4\pi G\rho$

$a4\pi r^2=-4\pi G (\frac{3M}{4\pi R^3})(\frac{4\pi r^3}{3})=\frac{-4\pi GMr^3}{R^3}$

$\vec{a}=\frac{-GMr}{R^3}\hat{r}$

$m\frac{d^2\vec{r}}{dt^2}=\frac{-GMm}{R^3}\vec{r}$

So you have simple harmonic motion.

$\vec{r}=R\cos(\omega t)$ where $\omega^2=\frac{GM}{R^3}$.

Note the angular velocity to the cube of the radius follows Kepler's Third Law despite the absence of an inverse square force law. That's a good mnemonic for remembering it. It can be shown the straight line back and forth time round trip time is equal to a full orbit about the earth just above its surface, approximately 84 minutes, so just 42 minutes to the opposite side of the Earth.