# Falling through Earth

Imagine a vertical tunnel passing through the centre of the Earth, providing a direct link between opposite points on the planet’s surface. If you jumped in, how long would it take you to reach the other end of the tunnel?

To work out the equation of motion governing your descent (and subsequent ascent!), we need to know the internal gravitational field of the Earth. An analytic solution can be found by making some simplifying assumptions:

• the Earth is spherical
• the density of the Earth is uniform

The first assertion is reasonable, since the Earth is remarkably spherical for such a vast object. It is only 43 kilometres broader across its equator than between its poles, a distance which constitutes only 0.3% of the Earth’s diameter.

The second assumption requires a greater stretch of the imagination. As you might expect, the Earth’s density varies with depth; the Earth’s core is thought to be about 6 times denser than its crust. This is because deeper layers of the Earth must support, and are therefore compressed by, the weight of the layers above them.

Despite these simplifications, our answer should be the right order of magnitude.

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Let $\bold{g}$ be the internal gravitational field of Earth. It is known that the gravitational field satisfies a form of Gauss’ law, a first-order differential equation which says that

$\nabla\cdot\bold{g}(\bold{r})=-4\pi G\rho(\bold{r})$

where $G$ is the gravitational constant and $\rho$ is the mass density of space. The symbol $\nabla\cdot$ represents the divergence operator, whose form is very simple after we impose the two assertions above.

Since the Earth is assumed to be spherical, the gravitational field must be spherically symmetric. It is therefore not a function of our latitude or longitude, but only our distance from Earth’s centre, which we will call $r$:

$\bold{g}=\bold{g}(r)$

This spherical symmetry also means the gravitational field’s direction must have a component only in the radial direction; that is, directly towards Earth’s centre. These two assumptions reduce the divergence operator to

$\displaystyle \nabla\cdot\equiv\frac{1}{r^2}\frac{d}{dr}\left(r^2 \cdot\right)$

We also impose our dodgy assumption that the density of Earth $\rho$ is constant, reducing Gauss’ law to

$\displaystyle \frac{1}{r^2}\frac{d}{dr}\left(r^2 g\right)=-4\pi G\rho$

Let’s guess a solution of the form

$g=\alpha r$

where $\alpha$ is a real constant. Substituting this ansatz in gives

$\displaystyle \frac{1}{r^2}\frac{d}{dr}\left(\alpha r^3\right)=-4\pi G\rho$

$3\alpha=-4\pi G\rho$

Hence our solution is

$g(r)=g(0)-\frac{4}{3}\pi G\rho r$

Taking the gravitational field at Earth’s centre to be zero without loss of generality,

$g(r)=-\frac{4}{3}\pi G\rho r$

Therefore inside a spherical planet of uniform density, the strength of the gravitational field varies linearly with depth. You will move as if attached to a Hookean spring anchored at Earth’s centre.

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Given the gravitational field, we can work out of the force you experience as you fall:

$F=mg$

Using Newton’s second law, we can form the equation of motion:

$m\ddot{r}=-\frac{4}{3}m\pi G\rho r$

$\ddot{r}=-\frac{4}{3}\pi G\rho r$

Here we can group the constants together by defining a new quantity $\omega_0$, given by

$\omega_0^2=\frac{4}{3}\pi G\rho$,

reducing the differential equation to

$\ddot{r}+\omega_0^2 r=0$

The most general solution to this equation is one of the form

$r(t)=A\cos\omega_0 t+B\sin\omega_0 t$

where $A$ and $B$ are real constants. Stepping into the tunnel from the Earth’s surface sets

$r(0)=r_0$

$\dot{r}(0)=0$

where $r_0$ is Earth’s radius. This sets $A=r_0$ and $B=0$, hence

$r(t)=r_0\cos\omega_0 t$

With this equation we can answer the question: how long does it take to traverse the Earth? The time taken to reach the end of the tunnel $\tau$ is half the time taken to complete one full oscillation, hence

$\displaystyle \tau=\frac{1}{2}\frac{2\pi}{\omega_0}$

$\displaystyle \tau=\frac{1}{2}2\pi\sqrt{\frac{3}{4\pi G\rho}}$

$\displaystyle \tau=\sqrt{\frac{3\pi}{4G\rho}}$

Let’s plug in some numbers. Using the mean density of the Earth, we are given the answer

$\tau=42\,\text{minutes}$

to two significant figures. Pretty speedy! You could get from Spain to New Zealand in less than an hour. Why you need to make such a quick escape is your own business.

Here’s another question we can answer easily with the expression for $r$ above: how fast are you travelling as you pass through the Earth’s centre?

Differentiating $r$ with respect to time gives your velocity:

$\dot{r}=-r_0\omega_0\sin\omega_0 t$

Hence your top speed is

$\displaystyle r_0\omega_0=\sqrt{\frac{4\pi G\rho r_0^2}{3}}\approx 7 920\,\text{ms}^{-1}$

This is slightly faster than the speed at which the International Space Station orbits Earth. But don’t worry! According to our (incorrect) equation, you will pop out the other side of the planet with minimal speed.

Get digging folks!