Drilling a hole through the earth

It’s been way too long since my last posting. A combination of wanting to do some additional research to support the post I was working on, and real life interfering,

So in the meantime I thought I’d do a quick post on a question that pops up here and there on the internet as it’s a fun little calculation: What happens (theoretically) if you fall into a hole that goes all the way through the earth?

As usual in introductory physics problems, we make some simplifying assumptions and ignore some practicalities. We’re not going to worry about the heat that would fry you as you descend to the core, we’re not going to worry about the fact that you would smash into the sides of the hole due to the rotation of the earth, and we’re definitely not going to worry about the fact that the earth is not technically a sphere.

Our idealized earth is non-rotating, a perfect sphere, and has the same density \rho throughout. And doesn’t have any of that pesky volcanic

The calculation is pretty simple once you lay the groundwork on two concepts: Simple Harmonic Motion and Gauss’s Law.

Simple Harmonic Motion

Suppose you have a quantity x(t) which obeys the following differential equation:

\displaystyle \frac{d^2x}{dt^2} = -ax

The general solution to this equation is known to be a sine wave, x = A\sin(\omega t + \phi) where A and \phi are arbitrary constants. Plugging this back into the differential equation:

\displaystyle \begin{aligned} x &= A \sin(\omega t + \phi) \\ \frac{dx}{dt} &= A \omega \cos(\omega t + \phi) \\ \frac{d^2x}{dt^2} &= -A \omega^2 sin(\omega t+ \phi) = \omega^2 x \\ \Rightarrow \omega &= \sqrt{a} \end{aligned}

x(t) oscillates sinusoidally with time. Any motion that has this sinusoidal behavior is called simple harmonic motion. If we define f = \omega/{2\pi} then f is the frequency of the oscillation. For any wave, the period T is the reciprocal of f. From this we can conclude that if we have a variable which obeys a differential equation of the form d^2x/dt^2 = -ax, then x will be simple harmonic motion with period T = 2\pi / \sqrt{a}.

The constants A and \phi are the amplitude and phase, respectively, of the sine wave. The fact that the equation is satisfied no matter what the values of those constants means that the system can have any amplitude and phase.

Example: Mass on a spring

The standard model of a mass m on a spring is that there is a constant k called the spring constant, such that if you displace the spring by an amount x from its equilibrium position, the spring pulls back with a force F = -kx. The negative sign indicates that the force is opposite to the displacement. If you pull the spring down, the force pulls up and vice versa.

That force acts on the mass to accelerate it F = ma = m d^2x/dt^2. And so we see that the mass on the spring follows the equation:

\displaystyle \begin{aligned} m \frac{d^2x}{dt^2} &= -kx \\ \frac{d^2x}{dt^2} &= -\frac{k}{m}x \end{aligned}

and we can see immediately that this motion meets the conditions for simple harmonic motion with a = k/m, and its period is therefore

\displaystyle T = 2\pi / \sqrt{ \frac{k}{m} } = 2\pi \sqrt{\frac{m}{k}}

Example: Pendulum

If you swing a pendulum out of its equilibrium position by an angle \theta, gravity will tend to pull it back down. The component of gravity pulling outward on the string has no effect as we assume the pendulum isn’t free to move in that direction. The component of gravitational force which is perpendicular to the string is $-mg \sin \theta$.

If we set this equal to F = md^2x/dt^2 as usual, the x in that equation refers to distance along the arc, x = L\theta where L is the length of the pendulum. So we have the equation

\displaystyle \begin{aligned} -mg \sin \theta &= m \frac {d^2x}{dt^2} = m L \frac{d^2{\theta}}{dt^2} \\ \frac{d^2\theta}{dt^2} &= -\frac{g}{L}\sin \theta \end{aligned}

Because of the sine, this doesn’t at first glance look like it meets the conditions for simple harmonic motion. But for small angles, \theta \approx \sin \theta with \theta in radians. What does “small” mean? As with any approximation in physics, there’s no firm definition. It depends on what level of accuracy you need. The smaller the angle, the better the approximation. For introductory physics calculations, 20^\circ is often taken as a rule of thumb for “small enough”. 20^\circ is about 0.349 radians, and \sin(0.349) = 0.342, an error of about 2\%.

If we use the small angle approximation then the equation of motion becomes

\displaystyle \frac{d^2\theta}{dt^2} \approx -\frac{g}{L}\theta

and now we see that \theta does (approximately) follow simple harmonic motion, with period T = 2\pi / \sqrt{g/L} = 2\pi \sqrt{L/g}.

Gauss’s Law

The mathematician Carl Friedrich Gauss discovered a very powerful theorem that simplifies many otherwise very difficult calculations. There is a form for the electrostatic force and for the gravitational force, and it arises from certain mathematical properties these two forces share, most especially that they are both inverse square-law forces, proportional to 1/r^2.

The general form for the electric field is this somewhat intimidating notation:

\displaystyle \oint_S \textbf{E} \cdot \textbf{dA} = \frac{Q}{\varepsilon_0}

This relates an integral of the electric field \textbf{E} over a closed surface S, to the total charge Q enclosed inside that surface. It doesn’t matter what charge is outside, only inside. The constant {\varepsilon_0} comes from the electric field of a point charge, E = q/(4\pi\varepsilon_0 r^2).

The analogous result for gravitational fields, where we define the gravitational field of a point mass m to be g = Gm/r^2 is

\displaystyle \oint_S \textbf{g} \cdot \textbf{dA} = -4\pi GM

where M is the total mass enclosed by the surface S. The minus sign expresses the fact that the gravitational field of a mass points inward (masses attract) while the direction of electric field from a positive charge is outward.

Note that “electric field” is defined as “force per unit charge”, while “gravitational field” is analogously defined as “force per unit mass”. But because F = ma, then “force per unit mass” is simply acceleration.

Gravity Inside the Earth

We don’t need to do a complete exploration of Gauss’s Law, we only need to cover the case of a uniform sphere, our model of the earth. To calculate the gravitational field at distance r from the center of the earth, we take as the surface S the sphere of radius r. The field will be the same at every point on the surface of this sphere, which means the left hand side of Gauss’ Law reduces to the field g times the area of the sphere: 4 \pi g r^2.

On the right hand side we have the mass enclosed inside the radius r. We’ll call this M_r. So

\displaystyle \begin {aligned} 4\pi g r^2 &= -4 \pi G M_r \\ g &= -\frac{G M_r}{r^2} \end {aligned}

The gravitational field at radius r looks like the gravitational field of a point whose mass is M_r, all the mass enclosed by the sphere of radius r.

How much mass is that? We are assuming a constant density \rho for the earth, so M_r is \rho times the volume of the sphere, M_r = (4/3) \pi r^3 \rho. Thus

\displaystyle \begin {aligned} g &= -\frac{G M_r}{r^2} \\ &= -\frac{4}{3} \frac{G \pi r^3 \rho}{r^2} \\ &= -\frac{4}{3} \pi G \rho r \end {aligned}

Everything on the right besides r is a constant. So this says that the gravitational field at distance r from the center is proportional to r.

Jumping Into the Hole

As we noted above, the gravitational field is simply the acceleration, d^2r/dt^2. So the equation above says that d^2r/dt^2 is a negative constant times r, which we know means that r will follow simple harmonic motion.

We’re almost done, but it will be convenient to put it in terms of different constants. The average density \rho of the earth is equal to total mass over total volume. Using M_E for the total mass of the earth and R_E for the radius,

\displaystyle \begin {aligned} \rho &= \frac{M_E} {\left ( \frac{4}{3} \pi R_E^3 \right )} \\ g &= -\frac{4}{3} \pi G \rho r \\ &= -\frac{4}{3}\frac {\pi G M_E}{ \left ( \frac{4}{3} \pi R_E^3 \right )} r \\ &= -\frac {GM_E}{R_E^3} r \\ \frac {d^2r}{dt^2} &= -\frac {GM_E}{R_E^3} r \end {aligned}

We know that this describes simple harmonic motion, and from the previous analysis we can immediately conclude that the period is T = 2 \pi \sqrt{R_E^3 / GM_E}. Using values of G = 6.6743 \cdot 10^{-11}\,m^3 kg^{-1}s^{-2}, R_E = 6.357 \cdot 10^6\,m (the polar radius, as I think we should drill the hole pole-to-pole to avoid the rotation), and M_E = 5.9736 \cdot 10^{24} \,kg this gives T \approx 5044 seconds, which means the one way trip is 2522 seconds or almost exactly 42 minutes.

Fans of Douglas Adams (Hitchhiker’s Guide to the Galaxy) will appreciate the cosmic significance of the number 42.

Conclusion: Ignoring rotation, air drag, and the minor annoyance of \textbf{5000}^\circ \textbf{C} core temperatures, if you jump into a hole that goes straight through the earth, you will oscillate from one end of the hole to the other forever, taking 42 minutes to go from one end to the other.


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