# Falling Through The Earth

If the alien cyborgs have constructed this miraculous planet-coring device with the precision I would expect of them, I predict we shall plunge entirely through the center and out to the other side.

There’s an old puzzle in physics, to work out how long it would take a person to fall right through the centre of the Earth to the other side, along a tunnel constructed through its core. Gregory Benford is the only science fiction writer I’ve ever seen attempt to incorporate that scenario into a story—a particularly striking bit of audacity that made it on to the cover of some editions of his novel.

For simplicity, the puzzle stipulates that the tunnel contains no air to slow the faller, and usually also specifies that it has a frictionless lining—any trajectory that doesn’t follow the rotation axis will result in the faller being pressed against the side of the tunnel by the Coriolis effect. (Benford had his protagonist Killeen fall through an evacuated tunnel from pole to pole, thereby avoiding Coriolis.)

So our unfortunate faller drops through a hole in the surface of the Earth, accelerates all the way to the centre, and then decelerates as she rises towards the antipodal point, coming to a momentary halt just as she pops out of the hole on the far side of the planet—hopefully not too flustered to remember to grab a hold of something at that point, so as to avoid embarking on the return journey.

We can set a lower bound to the duration of that journey by working out how long it would take to fall from the surface of the Earth to the centre, assuming all the Earth’s mass is concentrated at that point. Because the journey is one that involves symmetrical acceleration and deceleration, doubling this number will give us the duration of the complete traverse of the tunnel.

This means our faller moves under the influence of an inverse-square gravitational field throughout her fall. The acceleration at any given distance from the centre of such a field is given by:

\Large
a=\frac{GM}{r^2}


where a is the acceleration, G is the Gravitational Constant, M is the central mass and r the radial distance. That’s an important equation and I’ll invoke it again, but for the moment it’s more useful to know that the potential energy of an object in such a field varies inversely with r. The gravitational potential energy per unit mass is given by:

\large U_{m}=-\frac{GM}{r}

If we drop an object starting from rest at some distance R, we can figure out its kinetic energy per unit mass at some distance r by subtracting one potential energy from the other. And from that, we can figure out the velocity at any given point during the fall:

\large v=\sqrt{2GM\left (\frac{1}{r}-\frac{1}{R}\right)}

Finally, integrating* inverse velocity against distance gives us the time it takes to fall from a stationary starting point at R to some lesser distance r. For the special case of r=0, this comes out to be

\large t=\frac{\pi }{2\sqrt{2}}\sqrt{\frac{R^3}{GM}}

This calculation, incidentally and remarkably, is a major plot element of Arthur C. Clarke’s 1953 short science-fiction story “Jupiter V”.

Plugging in values for the mass and mean radius of the Earth, we find that t turns out to be 895 seconds. Doubling that value gives a total time to fall from one side of the Earth to the other of 29.8 minutes. That’s our lower bound for the journey time, since it assumes the faller is subject to the gravitational effect of Earth’s entire mass throughout the journey. (We draw a veil over what would actually happen at the centre point, where the faller would encounter a point of infinite density and infinite gravity.)

We can also put an upper bound on the journey time by assuming the Earth is of uniform density throughout. Under those circumstance, instead of the gravitational acceleration getting every higher as our faller approaches the centre of the Earth, the acceleration gets steadily lower, and reaches zero at the centre. This is because of something called Newton’s Shell Theorem, which shows that the gravitational force experienced by a particle inside a uniform spherical shell is zero. (Which rather undermines the premise of Edgar Rice Burroughs’s “Hollow Earth” novels.)

So as our faller descends into the Earth, she is accelerated only by the gravity of the spherical region of the Earth that is closer to the centre than she is. For any given radial distance r, the mass m of this interior sphere will be

\large m=\rho \frac {4}{3} \pi r^3

where ρ is the density.

Plugging that into our equation for the acceleration due to gravity (the first in the post) we get:

\large a=\frac {4} {3} \pi \frac {G \rho r^3} {r^2}=\frac {4} {3} \pi G \rho r

So the acceleration is directly proportional to the distance from the centre. This is the defining property of a simple harmonic oscillator, like a pendulum or spring, for which the restoring force increases steadily the farther from the neutral position we displace the oscillating mass.

Which is handy, because there’s a little toolbox of equations that apply to simple harmonic motion (they’re at the other end of my link), and with a bit of fiddling we can derive our journey time . The basic time parameter for oscillators is the period of oscillation, which I’ll call P. But that would be the time taken to fall from one side of the Earth to the other and back again. So the time we’re interested in is just half of that:

\large \frac {P} {2}=\pi \sqrt {\frac {3} {4 \pi G \rho}}

And plugging in the value for the mean density of the Earth, that shakes down to 42.2 minutes.

Notice how this length of time depends only on the density—it actually doesn’t matter how big our uniform sphere is, the time to fall from one side to the other remains the same so long as the density is the same. This is analogous to the fact that the period of oscillation of a pendulum doesn’t depend on how far we displace it—which is why we have pendulum clocks.

And because the acceleration due to gravity, g, at the surface of a spherical mass of radius R is given by

\large g=\frac {GM} {R^2}=\frac {4 \pi G \rho R^3} {3R^2}=\frac {4 \pi G \rho R} {3}

we can also derive our half-period P/2 as

\large \frac {P} {2}=\pi \sqrt{\frac{R}{g}}

A nice compact formula depending on radius and surface gravity, which is often quoted in this context.

And this is the gateway to another interesting result for spheres of uniform density, if you’ll permit me a brief digression. Suppose we dig a straight (evacuated, frictionless) tunnel between any two points on the surface of the sphere, and allow an object to slide along the tunnel under the influence of gravity alone—a concept called a gravity train. How long will such an object take to make its journey from one point on the surface to the other? It turns out that this time is exactly the same as the special case of a fall through the centre. We can see why by constructing the following diagram:

By constructing similar triangles, we see that the ratio of R (the distance to the centre of the earth) to d (the distance to the centre of the tunnel), is always locally the same as the ratio of g (the local gravitational acceleration) to a (the component of g that accelerates the gravity train along the tunnel. So for any straight tunnel at all, d/a is always equal to R/g, which (from the equation above) we know determines the period of oscillation through a central tunnel.

Remarkably then, we can take a sphere of uniform density of any size, and drill a straight hole between any two points on its surface, and the time it takes to fall from one surface point to the other will be exactly the same, and determined entirely by the density of the sphere.

But back to the original problem. I’ve determined that the fall time is somewhere between 29.8 and 42.2 minutes. Here are plots of the velocity and time profiles for the two scenarios I’ve discussed so far:

Can I be more precise? I can indeed, by using the Preliminary Reference Earth Model (PREM), which uses seismological data to estimate how the density of the Earth varies with distance from its centre.

Taking those figures and Newton’s Shell Theorem, I can chart how the acceleration due to gravity will vary as our faller descends into the Earth. Here’s the result, with density in blue plotted against the left axis, and gravity in red against the right axis:

As our faller descends through the low-density crust and mantle of the Earth, and approaches the high-density core, she actually finds herself descending into regions of higher gravity, reaching a maximum about 9% higher than the gravity at the Earth’s surface when she reaches the boundary between the mantle and the core.

If I take the data from the PREM as representing a succession of shells, across which acceleration rises or falls linearly within each shell, I can integrate my way through each shell in turn, deriving velocity and elapsed time. For each shell of thickness s, with initial velocity v0, initial acceleration a0 and final acceleration a1, the final velocity is given by

\large v=\sqrt{v_{0}^{2}+(a_{0}+a_{1})s}

and the time taken to traverse the shell is

\large t=\int_{0}^{s}\frac{dx}{\sqrt{v_{0}^{2}+2a_{0}x+\frac{a_{1}-a_{0}}{s}x^{2}}}

Handing off v as v0 to the next shell and summing all the t‘s once I reach the centre of the Earth will give me the answer I want.

The solution to the time integral is a bit messy, coming out as an arcsin equation when a0 > a1, and a natural log when a1 > a0.

But it’s soluble, and with steely nerves and a large spreadsheet, the graphs of the solution fall neatly between the two extremes I figured out earlier:

And the summed times for the full journey come out to be 38.2 minutes. And that’s my best estimate of the answer to the question posed at the head of this post.

* Integrating this expression turned out to be a little tricky, at least for me.

\small t=\sqrt{\frac{R}{2GM}}\int_{r}^{R}\sqrt{\frac{r}{R-r}} dr

After mauling it around and substituting sin²θ for r/R, then mauling it around some more, I ended up with this eye-watering equality as the general solution:

\small t=\sqrt{\frac{R^3}{2GM}}\cdot \left [( \frac{\pi }{2}-asin\left ( \sqrt{\frac{r}{R}} \right )+ \sqrt{\frac{r}{R}}\cdot \sqrt{1-\frac{r}{R}}\right ]

Exact solutions for this integral look like this:
For a0>a1:

\tiny t=\sqrt{\frac{s}{a_{0}-a_{1}}}\cdot \left [ asin\left ( \frac{a_{0}}{\sqrt{a_{0}^2+v_{0}^2\left ( \frac{a_{0}-a_{1}}{s}\right )}} \right )-asin\left ( \frac{a_{1}}{\sqrt{a_{0}^2+v_{0}^2\left ( \frac{a_{0}-a_{1}}{s}\right )}} \right ) \right ]

For a1>a0:

\small t=\sqrt{\frac{s}{a_{1}-a_{0}}}\cdot ln\left [ \frac{\sqrt{\frac{a_{1}-a_{0}}{s}\left ( v_{0}^2+sa_{0}+sa_{1} \right )}+a_{1}}{\sqrt{\frac{a_{1}-a_{0}}{s} v_{0}^2}+a_{0}} \right ]

(If you can simplify any of these, be sure to let me know.)

## 4 thoughts on “Falling Through The Earth”

1. BigDon says:

Wait, wait, wait, what’s this about Pelucidar, (sp), Dr. Grant?

(I read everything by Burroughs as a youth. Dad was an avid science fiction reader so as I was growing up I already had a huge library to choose from. Later I had the great pleasure of introducing Dad to Allen Dean Foster, one of my all time favorite authors.)

Wouldn’t you experience at least the gravitation of a large asteroid the diameter of the thickness of the shell your standing on? Assuming a shell a couple of hundred kilometers thick.

Oddly enough a Burroughs story also prompted me to want to ask you, “How do you heat a cave without killing yourself?” from a story where an adventurer survived a night in terrain populated by short faced bears and smilodons by building a huge fire at the entrance to his cave every night.

The exact technique I later learned bushmen use to kill dangerous game they have cornered in dens. (In particular, warthogs, but also sometimes problem leopards, which also den in caves.)

I’ve been wanting to ask you that for weeks now! 🙂

1. Oikofuge says:

No, Pellucidar would be gravity-free, sadly.
What Newton figured out was that the strong gravitational attraction of the asteroid-sized bit of the shell right next to you is precisely balanced by the sum of all the weak gravitational pulls from the many, many similar asteroid-sized bits on the far side of the shell.
You can hand-wave your way to seeing how that might be by thinking of the areas involved. There’s a useful diagram here. For any point inside the shell, you can draw two oppositely directed equal-angled cones, extending out until they hit the shell. The area “captured” by each of these cones is proportional to the square of the distance between the point and the shell. Which means the mass of each circular cap is also proportional to the square of the distance. And since gravity works in proportion to the mass and as the inverse square of the distance, the effects cancel out—the gravitational force on point P from (large, distant) area A1 is equal and opposite to the force from (nearby, small) A2. And the same holds true in any direction.
You can make the handwaving precise by integrating the whole volume using infinitesimal angles and an infinitesimally thin shell, and then summing infinitesimally thin shells to make a proper shell of finite thickness (isn’t calculus grand?) and it all works out to an exact balance, everywhere within the shell.

Caves, I dunno. Why would you want to heat a whole cave? In the open, you make your fire and then find a comfortable sleeping distance, which creeps towards the fire until you get so close you have to add more wood and then crawl away again. I guess I’d be tempted to do the same thing with a small fire in the cave entrance. Wild-camped in Africa, we were always told to sleep with our heads towards the fire, on the grounds that we had a fighting chance to fight off a carnivore that sneaked out of the night and grabbed our feet, but would fare less well if one had the opportunity to do the customary “grab the primate by the skull” manoeuvre.

2. BigDon says:

The thought of a large theropod adapted to free fall is somehow less comforting than one that has to walk…

Probably start to resemble marine “Godzilla crocs” from Miocene west coast of South America. Not very big as extinct orders of crocodilians go, but had heads and teeth like misdrawn tyrannosaurs.

As to the original article…

If granted frictionless walls, wouldn’t evacuation of the air be *required*
to prevent an “atmospheric re-entry effect on our inlabornaut? I would assume transonic speeds to cover that distance in that amount of time.

3. Oikofuge says:

Yes, if the tunnel isn’t evacuated, our faller soon reaches terminal velocity, which would then decrease as she fell down the pressure gradient into denser air. (Interesting problem to work on, though.)
And, yes, the faller is moving pretty fast at the centre of the Earth. One of the other neat things about the “uniform density” scenario, which I didn’t go into, is that the velocity at the centre is equal to the orbital velocity at the surface. So we’re talking close to 8km/s for a uniform Earth.

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