Journey to the Off-Center of the Earth

It was the greatest engineering project in history. The Americans started from New York, the Chinese from the Indian Ocean, both racing toward the center. The public cheered when John Henry, the largest tunnel boring machine ever built, switched on and plunged into the earth. Children across America told their parents they wanted to be miners when they grew up.

But as months turned to years and budgets ballooned, excitement faded. Some questioned if the project was worth it—trillions of dollars spent, thousands of workers toiling day and night, a decade at the dig. Yet even when the cooling system failed and tons of molten rock breached the tunnel walls, work persisted. Even when a tsunami burst the dams and the Indian Ocean rushed into the tunnel, smashing mighty boring machines into twisted wrecks, work persisted. Ten years to the day from when John Henry first tasted earth, the two sides met. Deep in the iron core workers shook hands. The tunnel was complete.

What was this all for? Lightning fast global transportation with zero energy cost. The plan was simple: the tunnel would be vacuum sealed, and capsules would be dropped into one side of the hole. They would fall straight down, accelerating until they reached the center of the earth, where they would then fall up, slowing as they climbed. Once the capsules reached the other side of the planet, they would be caught. It would be stunningly fast and basically free. The tunnel would pay for itself in shipping contracts alone.

At least that was the theory.

Drop day arrived. This would prove that it works, that it was all worth it. All around the world people tuned into the broadcast with bated breath.

Release! The capsule detached from the scaffolding and plunged into the eight-thousand-mile hole. It screamed through the vacuum, the camera on its nose streaming video to the world. The control room celebrated, erupting in cheers and hugs. But a lone scientist beckoned the chief engineer to his monitor. They crouched over it, peering at a flashing warning. The chief engineer stood stiffly and called the room to attention. It grew deadly silent. The capsule was off course.

Everyone watched as the capsule drifted toward the wall of the tunnel. Yards became feet. Feet became inches. The video cut.

The moment it made contact, the capsule turned to scrap. Its wreckage bounced, rebounded, and tore itself all the way to the center of the earth. It never came out the other side.

The Standard Model

What went wrong? Why did the capsule crash?

The idea of falling through a tunnel that runs straight through the center of the earth is a classic physics thought experiment discussed in countless articles and videos. And no wonder—it’s an intriguing question and an excellent demonstration of energy and oscillation. It’s beautiful physics.

The standard experiment runs as follows:

The capsule starts at a standstill with the full force of the earth’s gravity pulling it down. Once it is released, the force of gravity accelerates it into the hole. Where does gravity pull? To the center of the earth, toward which the capsule falls. As the capsule falls, less of the earth is beneath it, and the force of gravity weakens. When the capsule reaches the center, the force has zeroed out. At this point, though, the capsule is moving incredibly fast. Even though the sum force is canceled, the capsule doesn’t stop moving—its inertia carries it on.

Now the capsule heads away from the center. Gravity pulls against it. This exactly mirrors the way down, only now gravity pulls backward. The force of gravity gets stronger as the capsule moves away from the center and slows the capsule more and more as it approaches the surface. Assuming perfect conservation of energy (no loss to heat, sound, or air resistance), the force of gravity won’t stop the capsule until it exits the other side of the earth. At this point the capsule pops out of the hole and hangs suspended for a moment above eight-thousand-miles of thin air. Either the capsule is now caught and stopped or it falls back in and the process repeats.

It looks a little something like this:

0x Earth's Rotational Speed
Demo 1: Demonstration of the capsule (red dot) falling through a cross section of the earth without rotation.

We need a number of simplifying assumptions to make this work.

First, no air resistance. In fact, no loss of energy of any kind—perfect conservation. We’re happy to make this assumption since it is a standard in physics thought experiments.

Second, an idealized earth. Our world must be a perfect sphere of uniform density. This is not true, but we are also happy to run with this assumption since the story won’t change too much if density isn’t uniform (the acceleration would be more complicated) or the earth is not a perfect sphere (the capsule just might not make it to the surface).

Third, no planetary rotation. This assumption, though, we cannot accept. The earth’s rotation has a huge impact on this thought experiment. Unlike the previous assumptions, admitting rotation deepens the story and reveals new insights. By considering the earth’s rotation, we gain nuance without excessive complexity, and the picture becomes more beautiful.

The Twist

The earth spins. So what?

First, unless the hole starts at the poles and runs down the earth’s axis of rotation, the capsule is not standing still when it enters the hole. Sure, it is still with respect to the earth’s surface, but the earth is spinning. In fact, at the equator, the capsule would be moving at approximately 1000 miles per hour. This velocity has to be considered when analyzing the dynamics of this system. Because of the initial velocity, the capsule won’t fall straight to the center.

Second, the tunnel also moves, but different points in the tunnel move at different speeds. Points of the tunnel close to the earth’s axis move slower than points farther away. This is because the tunnel moves with a constant angular velocity, not a constant linear velocity.

Let’s walk through what would happen if the capsule fell into a straight-line hole at the equator. At the top of the hole, both capsule and tunnel move at about 1000 miles per hour east. But as it nears its center, the tunnel slows down. The capsule doesn’t. It hits the wall.

The Question

This raises two questions:

Suppose you dig a tunnel so that the capsule does not collide with the wall:

  1. What path would the capsule follow?
  2. What shape would the tunnel be?

These questions may sound the same, but there’s a crucial difference. The path the capsule follows is not in a rotating frame of reference—it is in a static frame of reference. In a static frame of reference, the earth’s rotation is visible. Fixed points on the earth’s surface are moving.

This frame of reference is convenient because it means that we do not have to think about Coriolis forces. We simply consider the earth’s rotation as an initial velocity on the capsule and move from there.

The tunnel is in the earth’s reference frame. Thus, if you dig a tunnel, it rotates with the planet. Once we figure out the path, we can change our frame of reference and determine the tunnel’s shape.

The Force of Gravity

Gravity is the only force at play in this problem. The capsule starts at some initial speed (set by the earth’s rotation) and falls. It does not touch the wall and thus does not experience any normal force (forces due to contact). We’ve assumed no air resistance, thus no friction. By choice of the right tunnel, no collision occurs. Therefore, the capsule is in true free fall.

This is very nice. To understand how the capsule’s velocities change at any given moment, all we need to do is to determine the strength and direction of the force of gravity.

The force of gravity is easy to calculate on the surface because the planet can be treated as a point mass. For a point mass, you calculate the gravitational force with:

F=Gm1m2r2F = G\frac{m_1 m_2}{r^2}

where m1m_1 and m2m_2 are the masses and rr is the distance between them.

Things could get much messier when one is inside the planet, with each bit of the earth pulling with a different strength in a different direction. It’s not necessarily obvious that you are always pulled toward the center, nor is it obvious how the magnitude of the force will depend on your distance from the center. Thankfully, Newton has us covered.

Newton’s Shell Theorem states that if you are anywhere inside a perfect spherical shell, which, like a balloon, is hollow on the inside, the forces of gravity from each piece of the shell will cancel out. No matter where you are in the shell—close to the wall or close to the center—you don’t feel any attraction or repulsion from the wall of the shell.

A corollary of the Shell Theorem states that if you are inside a solid ball—for instance, somewhere deep within the earth—then you can calculate the force of gravity as if all the matter that is farther from the center than you does not exist. Each layer of matter above you cancels itself out.

We can thus conclude that the force of gravity always points to the center and calculate its magnitude based on the capsule’s distance from the center of the earth. Thanks Newton.

Say the capsule is of mass mm and at a distance rr from the center. Let mrm_r be the amount of mass within a distance rr from the center. Using our corollary of the Shell Theorem, we know that the force of gravity, FF, will be:

F=Gmmrr2.F = G \frac{m m_r}{r^2}.

Let MM be the mass of the earth and RR be its radius. Let ρ\rho be the density of the earth. Since we have assumed uniform density, we can solve for ρ\rho by dividing the earth’s mass by its volume.

ρ=M43πR3\rho = \frac{M}{\frac{4}{3} \pi R^3}

This lets us solve for mrm_r because mrm_r will be the volume times the density:

mr=43πr3ρ.m_r = \frac{4}{3} \pi r^3 \rho.

Plug in for ρ\rho. The 43π\frac{4}{3}\pi cancels and

mr=Mr3R3m_r = M \frac{r^3}{R^3}

which, plugged back in for FF, gives

F=GmMR3r.F = \frac{GmM}{R^3} r.

That is, the force of gravity increases linearly as you move away from the center.

Gravitational Force

Figure 1: A graph showing the magnitude of the force of gravity as the distance from the center of the earth grows. Within the earth, the force grows linearly. It peaks at the surface and decays from there.

The Simulation

We have the force of gravity, but that’s only half the battle. We needed the force of gravity because that force controls the capsule’s acceleration (i.e., its velocity’s rate of change). By Newton’s Second Law, F=maF = ma, where aa is acceleration.

The force of gravity depends on where you are in the earth, which depends on your how you are moving (your velocity), which depends on the force of gravity. Therefore, solving for the capsule’s path requires solving a differential equation. Differential equations are often impossible to solve analytically.

So maybe we don’t solve it at all. Well, let’s do something else first, then come back to this. Now that we know how the force of gravity behaves, it’s fairly straightforward to simulate this system.

1x Earth's Rotational Speed
Demo 2: Demonstration of the capsule falling through the equatorial cross-section of the earth with true rotation. The blue trail is the path the capsule follows.

What a beautiful, symmetric shape! Notice that it loops back on itself. With what we know so far, that wasn’t guaranteed to happen. We now have some intuition about the path of the capsule: it follows an orbit within the planet. Better yet, we might have a guess what the shape of this orbit is. It looks like an ellipse. Does it make sense that the capsule would travel along an elipse?

Planetary orbits are elliptical (according to Kepler and your high school physics teacher), but planets orbit outside of one another. This orbit is within a planet. Outside a planet, gravity decays quadratically as distance increases. Inside, gravity decays linearly. There’s no way both of these orbits could be ellipses when they are governed by such drastically different dynamics. Right?

To know for sure, we have to tackle the differential equation.

The Dreaded Differential Decoupled

First, lets set up a coordinate system. We’ll use a two-dimensional Cartesian coordinate system with the origin at the center of the earth. We point the xx-direction in the direction the capsule is moving at its initial location. The yy-direction will point at the capsule’s initial location from the center of the earth.

Demo 3: Our reference plane for the problem. The capsule is the red dot; vary its starting latitude to see how the reference plane behaves. Click and drag to change your perspective.

We will call the distance from the capsule to the center in the xx-direction xx and in the yy-direction yy. Call the component of the gravitational force in the xx-direction FxF_x and in the yy-direction FyF_y. The total force, FF, is a sum of these components. The orthogonal forces, FxF_x and FyF_y, act as the legs of a right triangle, with FF the hypotenuse.

Triangle Forces

Figure 2: A schematic representing the force due to gravity and its components. The capsule is represented with the grey circle. Its $x$ and $y$ coordinates are marked relative to the center of the earth with dashed lines. Its distance to the center of the earth, $r$, is marked above the hypotenuse of the triangle formed by your $x$ and $y$ coordinates. The force due to gravity is shown by the purple arrow labelled $F$. The $x$ and $y$ components of $F$ are shown in purple. It is clear from this picture that the purple triangle associated with the force due to gravity and the black, dashed triangle associated with the capsule’s position are similar triangles.

If you think about similar triangles, you’ll realize that we can cleanly decompose these forces:

Fx=FxrFy=Fyr\begin{aligned} & F_x = F \frac{x}{r} \\ & F_y = F \frac{y}{r} \end{aligned}

The force FF equals GmMR3r-\frac{GmM}{R^3} r, so the xx and yy components of the force are:

Fx=GmMR3xFy=GmMR3y\begin{aligned} & F_x = -\frac{G m M}{R^3} x \\ & F_y = -\frac{G m M}{R^3} y \end{aligned}

Let this sink in for a second. The gravitational force in the xx-direction only depends on your distance in the xx-direction. Likewise, the force in yy only depends on the distance in yy. This is really convenient. We reduced a two-dimensional problem to two one-dimensional problems. One-dimensional problems are a lot easier to solve.

First divide these equations by the mass of the capsule, mm. Since F=maF = ma, dividing by mm gives us the accelerations in the xx and yy components. We could have done this ages ago and could just have been working with the accelerations, but forces are nicer to talk about. Since the acceleration is the second derivative of position, we get the decoupled system of differential equations:

d2dt2x=GMR3xd2dt2y=GMR3y\begin{aligned} & \frac{d^2}{dt^2}x = -\frac{G M}{R^3} x \\ & \frac{d^2}{dt^2}y = -\frac{G M}{R^3} y \end{aligned}

These equations are solvable; they describe oscillation. The frequency, ω\omega, of this oscillation is the square root of the coefficient, GMR3\sqrt{\frac{G M}{R^3}}.

Solving the Differential

Given our initial condition (initial position and velocity), we will be able to write xx and yy as a function of time.

Think back to our coordinate system. The xx-direction points with the spin of the earth and the yy-direction points from the earth’s center. We know that throughout its journey, the capsule will always be within the xxyy plane as its initial velocity is in the plane and the force of gravity always points to a point in the plane (the earth’s center).

In our frame, the initial velocity imparted by the rotation of the earth is entirely in the xx-direction. We’ll call this velocity vxv_x. Conversely, we only have displacement in the yy-direction, which is the distance from the center to the surface, RR.

At time 00, just before the capsule falls, we have:

x(0)=0,ddtx(0)=vxy(0)=R,ddty(0)=0\begin{aligned} & x(0) = 0, \quad \frac{d}{dt}x(0) = v_x \\ & y(0) = R, \quad \frac{d}{dt}y(0) = 0\\ \end{aligned}

From there we solve the differential. Letting ω=GMR3\omega = \sqrt{\frac{G M}{R^3}}, we get:

x(t)=vxωsin(ωt)y(t)=Rcos(ωt)\begin{aligned} & x(t) = \frac{v_x}{\omega} \sin(\omega t) \\ & y(t) = R \cos(\omega t) \end{aligned}

Those formulas might wake repressed geometry memories. They define an ellipse. If x(t)=sin(ωt)x(t) = \sin(\omega t) and y(t)=cos(ωt)y(t) = \cos(\omega t), the path would be a circle. Stretch and squish these by coefficients and you get an ellipse.

Notice that we have now described the path: no matter where on the earth’s surface you start, it is always an ellipse. As the starting point’s latitude increases, the plane that the ellipse is in rotates upward. The initial velocity in the xx-direction decreases as the latitude increases, meaning the ellipse gets thinner and thinner. This makes sense. If we jump from the North Pole, there is no initial velocity, so the ellipse is flat. All the oscillation occurs in the yy-direction.

Demo 4: The plane and path the capsule travels on given its starting latitude. Vary the latitude with the slider and navigate by clicking and dragging.

Hold on a second. Didn’t we agree that the orbit wasn’t an ellipse? Planetary orbits are elliptical but are defined by a different pair of differential equations. How can both of these systems produce the same result if they are governed by different equations?

Just because the equations defining the dynamics are different doesn’t mean that the dynamics themselves must be different. Different equations can have the same solution. Moreover, these dynamics aren’t the same. While both orbits outside and inside planets are ellipses, they are centered differently. When orbiting outside a planet, the center of mass is at on of the foci of the ellipse. In this case, orbiting within a planet, the center of mass is at the center of the ellipse.

Frequent Fallers

The frequency of our orbit is:

ω=GMR3\omega = \sqrt{\frac{G M}{R^3}}

Plugging in the earth’s actual mass and radius, we find ω=1.24103s1\omega = 1.24 * 10^{-3} s^{-1}.

From here, it is trivial to calculate the period of the oscillation (time to return to where we started). The period is 2π2 \pi divided by ω\omega. It takes 85.285.2 minutes to complete an oscillation. Thus it will take 42.642.6 minutes, half the time, to fall through the earth and emerge on the other side.

This number shows up in analyses of these problems that don’t consider rotation, which might surprise you. Shouldn’t our more in-depth analysis get a different answer? Well, the rotational velocity of the earth doesn’t show up in our equation for ω\omega. Therefore the frequency is independent of how fast or slow the earth rotates (which is somewhat unexpected). No matter how fast or slow you spin the planet, the frequency of oscillation will remain the same.

The Tunnel

Time to dig our tunnel! Since the dynamics are essentially unchanged, let’s only consider digging from the equator. This exaggerates the rotational effect and gives the clearest demonstration of the key ideas.

Let’s go back to our old friend, the simulation, and draw the tunnel as we move. The tunnel rotates with the earth. It’s like drawing on a record as it spins. The path the pen takes is the path we have been discussing thus far. The line on the record is the tunnel.

1x Earth's Rotational Speed
Demo 5: Demonstration of the capsule falling through the equatorial cross-section of the earth with true rotation. The blue trail is the path the capsule follows and the black trail is the tunnel it travels in.

Notice that each leg of the tunnel does not come out on the opposite side of the earth from where it started. The capsule goes to the direct opposite point in space, but it takes time to get there, and in that time the earth has spun. So it actually travels less than halfway around the world—42 minutes of longitude less because it took 42 minutes to fall there.

If we call the rotational frequency of the earth ωe\omega_e (remember that the frequency of the falling oscillation is ω\omega), then each leg of the tunnel exits at (1ωe/ω)π(1 - \omega_e/\omega) \pi radians, or 12(1ωe/ω)\frac{1}{2}(1 - \omega_e/\omega) turns, later.

This ratio of ωe/ω\omega_e/\omega controls the star pattern of the tunnel. As the ratio grows, each spoke is farther from directly half-way across the circle.

If and only if ωe/ω\omega_e/\omega is rational, then eventually the tunnel will truly loop back on itself because each leg will start at a rational fraction of a turn. Eventually, these rational fractions will sum to a whole number, meaning we are back where we started.

It looks an awful lot like the earth tunnel loops back on itself. Wouldn’t that be amazing? Unfortunately, there is no reason that the ratio ωe/ω\omega_e/\omega should be rational. The first frequency, ωe\omega_e, is defined by the length of a day and the second, ω\omega, is defined by the earth’s radius and mass and the gravitational constant, GG. These physical parameters have nothing to do with each other, so there is no reason the ratio should be rational.

Nevertheless, the ratio is almost a rational number. For the earth, ωe/ω\omega_e/\omega is very close to 1/171/17, meaning that after 1717 passes, we end near to where we started. In fact, ωe/ω\omega_e/\omega is only 0.000230.00023 off from 1/171/17, giving an error of 0.4%0.4\% if we approximate with 1/171/17.

Playing God

That’s Earth. But it’s not the limit of our imagination. What if we could control how fast the earth spins? What sort of orbits could we achieve? It’s time to play god.

1x Earth's Rotational Speed
Demo 6: Use the slider to spin the earth at different rates. Click FALL to release the capsule. The blue trail is the path the capsule follows and the black trail is the tunnel it travels in.

Hopefully you’ve noticed some odd behaviors as you ramp up the speed of the earth. Initially, the ellipses just get wider. This is consistent with our formula, since

x(t)=vxsin(ωt)x(t) = v_x \sin(\omega t)

and the ellipse gets wider as the initial velocity vxv_x increases.

You may also notice that the distance between the start and end of a tunnel leg shrinks as the ellipse widens. This is also consistent with our equations. ωe\omega_e grows proportionally with the rotational velocity, so (1ωe/ω)π(1 - \omega_e/\omega)\pi continually shrinks.

Once you’ve spun the earth up to 1717 times its normal speed, something strange happens. The capsule doesn’t fall; it just stays on the surface and rotates with the earth.

Where did we see seventeen before? The ratio ωe/ω\omega_e/\omega was very close to 1/171/17. If we spin the earth seventeen times faster, our elliptical internal earth orbit becomes a circle and our tunnel distance at the surface zeroes out. Functionally, this means that if the earth spun seventeen times faster, we would all be at orbit standing at the surface. This would be awfully strange to say the least—a planet spinning fast enough that gravity turns off.

It’s even stranger. You would be at orbit at any depth within the earth, not only on the surface. Both the force of gravity and the translational velocity decay linearly with distance from the center of the earth, so they balance out at any depth.

17x Earth's Rotational Speed
Demo 7: The earth spinning between 17 and 30 times faster than normal.

This raises the question, what if you spin the earth faster? Then we cannot orbit on the surface anymore. Instead, the earth spins so fast it literally throws us off the planet. We are hurled into the vacuum of space, along with the rest of the disintegrated planet. Oops.

Let’s assume the earth remains in one piece; you just throw all of us off it and into orbit, no different from the normal picture of planetary orbit. We are entirely outside the planet, so the planet behaves as a point mass. We travel the same sort of paths satellites travel, only our path skims the surface of the earth.

The “tunnel” becomes additionally noteworthy at this speed. We are all going the same horizontal velocity as the surface of the earth when we leave and when we return. From the perspective of the earth, we don’t fly away horizontally when we are flung off the planet. Instead, we all float up and away, peacefully into the vacuum of space (where we perish somewhat less peacefully). This can be seen in the simulation. Notice that if the planet spins fast enough, the tunnel extends (nonsensically) into outer space, and when the line leaves the earth it starts straight outward.

The orbit also returns us to the planet. This is even stranger to think about. Not only do we take off vertically from the earth’s perspective, we also land vertically. We float down, out of the sky, until our feet make contact with the ground. Then it’s up and away again, back to our orbital dance.

Ramp up the speed of the earth even more. As the dot flies farther and farther away from the planet and the tunnel turns into a tighter and tighter spiral, you might get the suspicion that we are not going to land back on the planet. We won’t. You have spun the planet so quickly that its surface, and us along with it, moves faster than escape velocity. Without enough gravity to hold us down we are all flung away, this time with no hope of return. What goes up doesn’t always come down.

An Energetic Perspective

This entire time we have understood this problem through forces and their impact on velocity—that’s how we simulated the problem and how we built our differential equations. But with a different perspective we can reach novel insights. Let’s talk energy.

From the start, we’ve assumed no loss of energy. The capsule begins with some amount of potential and kinetic energy and conserves their sum. But the amount of kinetic and potential energy the capsule has changes throughout its fall—as it descends, potential is converted to kinetic. When it returns to the surface, the opposite is true.

What does this exchange look like? We can start by adding a graph to our simulation.

Oscillating EnergiesEnergy (mega-joules)102030Time (minutes)246810Potential EnergyKinetic Energy
1x Earth's Rotational Speed
Demo 8: The graph shows the potential and kinetic energy of the capsule.

If you experiment with this demo, you may notice some interesting dynamics. First, the curves are mirror images of one another. This is derived from conservation: any amount of energy added to one curve is subtracted from the other.

At speeds where the capsule remains inside the earth (less than 1717 times the earth’s rotational speed), the peaks and valleys of the two curves seem to be at the same values. Since this phenomenon does not hold when the capsule leaves the earth, it cannot come from simple conservation. What is going on here?

The formulas for kinetic and potential energy are:

K=12mv2K = \frac{1}{2} m v^2


U=0rGmMR3sds=12GMR3mr2U = \int_{0}^{r} \frac{GmM}{R^3} s ds = \frac{1}{2}\frac{G M}{R^3} m r^2

where rr is the capsule’s distance to the center. Using our definition of ω\omega, we can rewrite UU as:

U=12m(ωr)2U = \frac{1}{2} m (\omega r)^2

Let’s recall our equations that describe the path:

x(t)=vxωsin(ωt)y(t)=Rcos(ωt)\begin{aligned} & x(t) = \frac{v_x}{\omega} \sin(\omega t)\\ & y(t) = R \cos(\omega t) \end{aligned}

Take a time derivative to get the velocities:

vx(t)=vxcos(ωt)vy(t)=Rωsin(ωt).\begin{aligned} & v_x(t) = v_x \cos(\omega t) \\ & v_y(t) = -R \omega \sin(\omega t). \end{aligned}

When falling from the equator, vx=ωeRv_x = \omega_e R. Remember that ωe\omega_e is the rotational frequency of the earth and ω\omega is the frequency of the falling orbit. So the capsule’s initial velocity (all in the xx-direction) will be ωeR\omega_e R and its initial displacement from the center (all in the yy-direction) will be RR.

Using the formulas for kinetic energy and potential energy, we get:

K(0)=12m(ωeR)2U(0)=12m(ωR)2\begin{aligned} & K(0) = \frac{1}{2} m (\omega_e R)^2 \\ & U(0) = \frac{1}{2} m (\omega R)^2 \end{aligned}

The capsule is closest to the center at t=π2ωt = \frac{\pi}{2 \omega}. At this point, vx(π2ω)=0v_x(\frac{\pi}{2 \omega}) = 0 and y(π2ω)=0y(\frac{\pi}{2 \omega}) = 0. So the velocity is entirely in the yy-direction and the displacement is entirely in the xx-direction. The velocity is RωR \omega and the displacement is ωeRω\frac{\omega_e R}{\omega}.

This gives us

K(π2ω)=12m(ωR)2U(π2ω)=12m(ωeR)2\begin{aligned} & K\left(\frac{\pi}{2 \omega}\right) = \frac{1}{2} m (\omega R)^2 \\ & U\left(\frac{\pi}{2 \omega}\right) = \frac{1}{2} m (\omega_e R)^2 \end{aligned}

There’s our answer: K(0)=U(π2ω)K(0) = U(\frac{\pi}{2 \omega}) and U(0)=K(π2ω)U(0) = K(\frac{\pi}{2 \omega})!

So what would it mean for the kinetic energy at the surface to equal the potential energy at the surface? Well, the potential energy at the nearest point to the center is the kinetic energy at the surface, so it would be . . . the same as the potential energy at the surface. The potential energy wouldn’t change. The capsule wouldn’t fall. When the initial potential and kinetic energy are equal, the capsule doesn’t fall at all. It is at orbit sitting on the earth’s surface.

From this, we can see that everything we care about in this system is controlled by ωe/ω\omega_e/\omega. Square it and you get the ratio of the initial kinetic energy to the initial potential energy. It is also the ratio of the major and minor axes of the path. And it controls the fraction of a rotation completed on each path through the earth. Thus, the fundamental dimensionless quantity that controls both the shape of the path and the shape of the tunnel is the square root of the ratio of the initial kinetic and potential energies.

A Tunnel Redug

It was the greatest engineering project in history.

The first failure had been crushing, and the whole world remembered. The image of the wall inching closer and closer to the capsule had been seared into public consciousness. All that work wasted.

But the engineers pleaded. Give us a second chance—we’ll do it right this time. To their credit, when the funding finally came through, they did. The world never looked back.

Now there are countless tunnels arcing deep within the earth. And they don’t merely deliver energy-free travel to the opposite side of the planet, but all around it as well.

© 2021, Built by Alex and Jack Strang