Grav Train

We all love flying: it’s both an exhilarating experience and a fast way of getting from place to place. But jet planes are gas guzzling beasts and produce immense amounts of CO2. Plus, it can take an entire day to fly around the globe; way too long for impatient businessmen. Here I would like to show you a stupendous way of transportation, one that can take you to anywhere in the world in just 40 minutes. On top of that, it consumes absolutely no energy whatsoever, giving it a zero-emission carbon footprint.
The concept is simple, and is based on a question that little kids really like to ask: if there were a hole that goes straight through the center of Earth, all the way to the other side, what would happen if you jumped in? The nitpickers will say that you will definitely burn to ashes at the core, but let’s ignore such petty things for a moment. You will fall right through, accelerating all the while. Even though gravity is 0 at the center of Earth, you will not stop there, because at that time you already have an immensely large speed from all the falling. Having passed the center, you’d continue through, no longer “falling” but “rising”, until you reach the surface on the other side.
Voila! You have successfully reached the opposite side of the world.

How long would the whole trip take you? How fast will you be going? Can you survive such speeds? What happens if the hole doesn’t connect between two opposite spots on Earth, but between any two arbitrary points? To answer these questions, we will have to look at a bit of physics.

We know that gravity pulls on you towards the center of the earth. How strong you are pulled depends on two things: the mass of what’s pulling you, and how far away you are from it. The equation looks like this:

F = G\dfrac{Mm}{r^2}

Where G is the universal gravity constant, m is your mass, M is Earth’s mass, and r is the distance between you and Earth’s center.
This equation is very useful for calculating the forces between the planets and the Sun, in order to figure out how they move. All you have to do is plug in the known masses and distances, and you can start solving. But what happens if you move inside Earth?
As you go deeper into the ground, there is still a lot of mass that pulls you in closer to the center, just like you were on the surface. However, there is also now mass behind you, which tries to pull you out. How does this affect the strength of the force that pulls on you? It turns out that if you are inside the Earth, you can divide it into two parts: one is a sphere, of which you are right at the edge. The other is an outer shell, which surrounds you. It turns out that the gravitational effect of the outer shell is completely eliminated; it cancels itself out. This means that all that is left to pull you is the central sphere on which you are standing.

As you dig down into Earth, the distance from you to the center gets smaller. Since the force varies as \dfrac{1}{r^2} , this makes it stronger. However, the mass of the sphere gets smaller as well, and since the gravitational force varies according to the mass, this makes it weaker. Which factors wins out?
Suppose that our inner sphere has a uniform density, ρ. Then its mass as a function of radius is:

M = \dfrac{4}{3} \pi \rho r^3

We can write this instead of M in the force equation. This gives us:

F = G\dfrac{Mm}{r^2} = \dfrac{4}{3}G \pi \rho m \dfrac{r^3}{r^2} = \dfrac{4}{3}G \pi \rho m r

So the force actually varies linearly as a function of sphere radius, or, distance from the center. Of course, this is only assuming that the density is constant; in real Earth, it too is a function of distance, as the core is much denser than the surface. Still, this is a good first approximation.

Now that we know how the force changes according to how far away we are from the center, we can imagine the following high-tech transportation system: we dig a tunnel between any two cities in the world, say, London and Los Angeles. This tunnel might be dug at an angle, and not just “straight down” as in the previous example we discussed. We have a special train that drives through this tunnel. “Drive” is a bad word, actually. “Fall” would be much better. So we “drop it” at London, letting it fall through the tunnel, and “catch” it again at Los Angeles, as it slows down due to gravity.
At least, that’s the plan. Let’s see if it works.

At any given time, the train has some position, x(t). We would really love to find out what it is, since if we know its position as a function of time, we can figure out its velocity and acceleration as well. We already know the force acting on it, so we can use Newton’s second law, which tells us the relationship between the acceleration and the forces acting on an object:

F = ma

Lets ignore other forces for now, like friction and air drag, and look at how gravitation makes us move. The force acting on our train is always acting at some angle θ. We can split it into two components:
– Perpendicular to motion: the force acting to pull the train down. We can ignore it for now, knowing that the train does not fall through the floor.
– Tangential to motion: the force acting to pull the train in the direction of its motion. This is what gives the train its speed in the first half of its journey, and pulls it back towards the center after it crosses the center. This is what interests us.

We already know the total force: \dfrac{4}{3}G \pi \rho m r , and now we just have to take the tangential component. This means multiplying the force by \cos \theta . But from the picture, we can see that r \cos \theta is just the horizontal position of the train on the line on which it is travelling, shown as x. We also note that there is a minus sign, because gravity tends to return the train towards the center.

-\dfrac{4}{3}G \pi \rho m r \cos \theta = m a_{tangential}
-\dfrac{4}{3}G \pi \rho m x = m a
-\dfrac{4}{3}G \pi \rho x = \ddot{x}

If we define

\omega = \sqrt{\dfrac{4}{3} G \pi \rho}

Then we can write our equation as:

\ddot{x} + \omega^2 x = 0

This is a linear ordinary differential equation, which tells us of the relationship between the train’s position and its acceleration. Remarkably, it is exactly the same equation which describes how pendulums swing at small angles: the simple harmonic motion equation.
The simplest solution to this equation is well known:

x(t) = A \cos {\omega t}

In our case, A is half the direct distance between the two cities.
There are several interesting things to note here.
The first is the time it takes to complete the trip. The position is determined entirely by the cosine function, so all that determines the trip time is ω. After \frac{2 \pi}{\omega} seconds, the cosine will have completed one full period. In train terms, that means from London to Los Angeles and back. Let’s see how long that is, assuming the average Earth density is 5515 [kg/m3]:

\omega = \sqrt{\dfrac{4}{3} G \pi \rho} = 0.00124 [\frac{1}{s}]
T = \dfrac{2 \pi}{\omega} \approx 5060 [s] \approx 84 [min]

That’s what physicists call “totally crazy”. 42 minutes from London to Los Angeles, another 42 minutes to get back. But the brilliant thing is, that ω doesn’t depend at all on the locations you wish to travel to. All it depends on is the density of Earth. So it doesn’t matter from where to where you are going: your trip will always take 42 minutes. That’s faster than a space shuttle at low orbit.
What about speeds? The maximum distance you can cover is twice Earth’s radius – 12,800 kilometers. You do this in 42 minutes, for a total of about 18,285 [km/hour] on average. But the train starts at velocity 0, and finishes with velocity 0, so the maximum speed is even higher than that. Deriving velocity from position, we get:

v(t) = -A \omega \sin {\omega t}

The speed is maxed when sin = +-1, at the center of the path taken by the train. The value:

v_max = A \omega = 6400000 [m] \cdot 0.00124 [\frac{1}{s}] = 7936[\frac{m}{s}] = 28569 [\frac{km}{h}]

Not half bad. Can our bodies and machines withstand such velocities? Certainly. We don’t care about velocities.
Apart from the incredibly fast speed at which we arrive to any point on Earth, we get another cool bonus: energy efficiency. The train doesn’t care how much mass it carries: there is no difference in going empty, and transporting a full load of people. This stems from the fact that the train doesn’t “use up” any energy at all. No fuel is required to move it. All the energy comes from potential gravitational energy which is turned to kinetic energy. We are just dropping things into holes. So in the theoretical ideal case, this transportation costs us no energy whatsoever. Hurray for global warming.

Problems, anyone?

Despite the attractiveness of such a transportation system, there are indeed a few problems which make it impractical under current technology.
The first problem I’d like to address is friction and air drag. The theoretical 28,500 kilometers an hour is only valid if nothing acts to slow the train down. However, if it’s a real train on wheels, there would be energy loss due to friction between the wheels and the track. If there is air in the tunnel, then drag would slow the train down significantly; so much, in fact, that it would very soon reach a constant speed rather than keep on accelerating. Thus, with friction and air drag, we would “drop” the train down at London, but it wouldn’t have a high enough speed to reach Los Angeles. It would stop somewhere in the middle of the Earth, and happily go up and down around the core. This would not generate satisfied customers. Friction forces are not that large, and can be overcome by installing some kind of engine on the train, either mechanical or rocket. The train can also be levitated by magnets, like the maglev. The drag force is more of an issue, and the tunnel would probably have to be kept in vacuum, which requires energy to generate and keep.
Other external forces which act on the train are the Coriolis effect, which will cause trains to crash against the side walls unless held in place, and magnetic forces, which are stronger near Earth’s core. All sorts of techniques can be used to prevent this lateral movement, including magnets.

A more difficult problem to solve would be the heat inside the tunnel. The temperature in Earth’s core is estimated to be around 6,000 C0. It’s quite obvious that building a tunnel that goes through molten and solid iron at those temperatures would be very hard for us. Evidently, going directly through Earth’s core is out of the question. But what about tunnels which don’t cut across the core?
Well, the temperature on Earth’s surface is about 15 C0. If the temperature gradient were linear, then the tunnel would get slightly less than 1 C0 hotter for every 1 kilometer down. But the gradient is not linear, and Earth’s not-so-small core has roughly the same temperature all round, which means that the temperature increase is larger. It turns out that every kilometer dug down will increase temperature by about 20 C0, at least in the upper parts of Earth’s crust. This might be surprising; when I first saw this fact, I thought to myself, “since when is Earth so hot?”. But then I remembered pictures like these:

and remembered that yes, it is quite hot in there.
So our tunnels and equipment will have to withstand quite a high temperature if we want to gain any considerable distance. Maybe our trains would be ok, if they were indeed kept in vacuum (that means that only radiation heat transfer occurs, which is slow and can be reduced using reflective materials), but the tunnel walls would certainly have to resist the heat.
The Soviets attempted to dig a really deep hole once. They got to about 13 kilometers, before it became too hot for the drill to function properly. Suppose that that is our limit: what is the maximum distance between two points on Earth, which have a tunnel that is no more than 13 kilometers straight down from the surface? In the following picture, it means that K = 13 kilometers.

From the picture, we get:

\cos \alpha = \dfrac{R-K}{R}
\alpha = \arccos{\dfrac{R-K}{R}}
S = R \cdot \alpha

D = 2 \cdot S

Putting in the data of K=13 km and R=6400 km: D = ~815 km. So if we don’t dig any deeper, we’ll only be able to get 815 kilometers in 42 minutes, which is 1164 [km/h]. That’s faster than the fastest train, but not much of an improvement over conventional planes. Our digging technologies aren’t yet up to par to make this worthwhile.

There is also pressure to consider. Your transportation customers certainly won’t like it if the tunnel collapsed on their heads while they were in it. But your tunnel has to support the weight of all the rock and dirt which is found on top of it.
The pressure is strongest in the center of the tunnel. We can integrate the force from the tunnel upwards, in order to find out what the pressure is:

Consider an element of mass dm found at distance h from the center. It has a cross section of a, and height dh. So in total,

\mathrm{d}m = \rho \cdot a \cdot \mathrm{d}h

Earth is pulling it at

\mathrm{d}F = \dfrac{4}{3}G \pi \rho \cdot h \cdot \mathrm{d}m

Integrating from h to H, the total force of all the rock sitting on top of the center of the tunnel is:

F = \int_{h}^{H} \dfrac{4}{3} G \pi \rho^2 a \cdot h \cdot \mathrm{d}h = \dfrac{2}{3} G \pi \rho^2 a (H^2 - h^2)

So the pressure is:

P = \dfrac{F}{a} = \dfrac{2}{3} G \pi \rho^2 (H^2 - h^2)

Plugging in the numbers for a tunnel just 13 kilometers deep, we get about 7060 atmospheres. You would need very thick tunnel walls if you don’t want them to break under the strain.

One last issue I would like to talk about is the sheer amount of material you have to excavate out of the ground. Digging a large tunnel requires moving away a lot of rock. How much of it? Let’s assume that the tunnel has a cross section of 4×4 meters. The surface distance between London and Los Angeles is about 8795 kilometers; a tunnel between them would only measure about 8120 kilometers. Digging this tunnel would require moving away about 130,000,000 cubic meters of material. Most of it is very hard rock. Taking Earth’s average density of 5515 [kg/m3] we have to move 716,500,000 tons of rock. That’s a lot (even though it’s only about 0.00000000000012 of Earth’s total mass).
If you decided to bring all of that rock to the surface, you would be in for a treat. Let’s look at how much energy you need to invest in order to fight gravity and bring it all up.

Let’s look at a small element of mass at the bottom of a well of length H. The well is perpendicular to the line connecting the Earth’s center and the well bottom; in other words, the bottom of the well is in the vertical center of the earth; in other words, the well is half as long as an entire tunnel. The element is found at height h in the well, its height is dh, its cross section is a, and its density is ρ. Thus:

\mathrm{d}m = \rho \cdot a \cdot \mathrm{d}h

As we lift it up, gravity pulls it down, with the same tangential force we discovered earlier.

F = \dfrac{4}{3} G \pi \rho \mathrm{d}m x

By definition, the work required to lift that one element, dm, is equal to:

\mathrm{d}w = \int_h^H F \mathrm{d}x
\mathrm{d}w = \dfrac{4}{3} G \pi \rho \cdot \mathrm{d}m \int_h^H x \mathrm{d}x = \dfrac{2}{3} G \pi \rho (H^2 - h^2)\mathrm{d}m

That’s just the energy needed to take out one small dm. To take out everything, we need to take dm over all of the well, which means over all of h. Putting in dm, we get:

\mathrm{d}w = \dfrac{2}{3} G \pi \rho^2 (H^2 - h^2) \cdot a \cdot \mathrm{d}h

Going over all of h, from 0 to H, we get:

w = \int_0^H \mathrm{d}w = \int_0^H \dfrac{2}{3} G \pi \rho^2 (H^2 - h^2) \cdot a \cdot \mathrm{d}h

w = \dfrac{4}{9} G \pi \rho^2 a H^3

Plugging in x = 4060 kilometers (half the tunnel from London to Los Angeles) and average Earth density, we get about 842,519 GWh. According to Wikipedia, that’s about the yearly electric consumption of India, and we didn’t even consider heat and kinetic energy. So good luck.

A conclusion from all of this? First of all, physics is cool. Second, this is an interesting way to travel. It’s not currently feasible or economical on Earth, but perhaps one day our technologies will be powerful enough; further, some of the problems discussed, such as heat and air resistance, pose less of a problem on places like the Moon.

2 thoughts on “Grav Train

  1. Interesting! I don’t know why I thought about this about this and worked out the same equation as yours (of course this interesting transport system was discussed long ago as well I discovered). What I was thinking is also passenger comfort, in fact for passengers, assuming no friction and drag, the gravity acceleration component in the direction of travel would be compensated by the apparent acceleration of the train, in other words passengers would be in free fall in the direction of travel and would feel only the g component in their local apparent vertical (head to toe), which turns out to be – after some calculation – constant during the trip and dependent on the “slope”, e.g. travelling to a city some 9000 km it would be around 70% of g at the surface. So a comfortable trip with no “felt” accelerations!

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s