The new book from Randall Munroe is finally out—*What if? 2*. Surely you know Munroe from the excellent xkcd comics as well as the original *What If?*. (You can read an excerpt from *What If? 2* here, as well as a Q&A with Munroe about answering impossible questions.) This time, Munroe uses basic scientific ideas and estimations to answer questions like: "How long would it take for a human to drive a car all the way to the edge of the universe?" or "What would happen if you shrunk Jupiter down to the size of a house and placed it in a neighborhood?" or "If the Earth were a massive eye, how far would it see?"

Those are great questions, but I want to focus on Munroe’s solution to the following query: “I want to lose 20 pounds. How much of the Earth’s mass would I have to ‘relocate’ to space in order to achieve my goal?”

This seems plausible, Munroe writes: “Your weight comes from the Earth’s gravity pulling you down. The Earth’s gravity comes from its mass. Less mass should mean less gravity. Remove mass from the Earth, and you’ll lose weight.” In Munroe’s classic style, he then goes on to show that it would require harvesting all the power from the sun with a Dyson sphere in order to peel off enough of the planet to get the job done.

But he points out that there is another way to reduce your weight by 20 pounds. You don’t need to remove all that surface material from the Earth. You just need to go underground—essentially putting yourself *beneath* some of the planet’s mass and avoiding its gravitational effect. Your weight would decrease as you go further and further below the surface. I know that sounds weird, but maybe this diagram will help:

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But why does that outer part of the Earth not matter when it comes to weight? And exactly how long does that tunnel need to be to subtract 20 pounds from your weight? This gives us a lot of interesting physics to unpack, but let's start with the very basics: weight and the gravitational field.

Real Gravity

The gravitational force is an interaction between any two objects that have mass. For now, let’s consider the following two objects: a human (mh) and the Earth (mE).

The magnitude of this gravitational force depends on the values of the masses and the distance between their centers, which is r, assuming they are roughly spherical. (Yes, I'm assuming the human is a sphere, as this will allow us to calculate the force as if it is all concentrated in a single point.) With that, we have the following expression:

If I replace the human with some other object—say, an elephant—then the attractive gravitational force will increase, since the mass of the elephant is larger than that of the human. This other object will also experience a force (due to its interaction with the Earth) that is proportional to its mass.

We can actually look at the gravitational effects due to the Earth that *only* depend on the Earth. To do this, we can take an object—once again, an elephant—and calculate the gravitational force on it. Then we divide this force by the mass of the elephant. Doing that gives us the force per unit mass due to the Earth. We call this quantity the gravitational field (g) in newtons per kilogram. (Yes, this is the same "g" you see in other physics problems. It's not the "acceleration due to gravity" as it's commonly called—it's the gravitational field.) The nice thing about this gravitational field is that now we can just focus on the Earth and not worry about other objects.

Let's take a look at this gravitational field around the Earth. It's a vector that always points toward the center of the Earth and increases in magnitude when you are closer to the Earth’s surface. Here is a model created in Python:

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Each arrow shows the direction of the gravitational field at that location. You can see that the gravitational field decreases with distance from the Earth’s surface.

But in Munroe's scenario, we are tunneling *inside* the Earth. With that in mind, let's break the Earth into smaller chunks. This is important, since the gravitational field obeys the superposition principle, which says that if you take two of these masses (small pieces of the Earth) and look at the total gravitational field at some location, then the gravitational field would just be the vector sum of the two individual fields.

That seems like a tough concept, but it's not so bad. Let me show you an example. Suppose I have two masses out in space. They are m1 and m2. They each create a gravitational field (g1 and g2), but I want the total field at some location. (I'll call it location "O.") The net (total) field would look like this:

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Since the gravitational fields from both masses are vectors (meaning their direction matters), then it's possible to have these two field vectors pointing in opposite directions, which means that they would completely cancel each other out.

Here is another example with two *identical* masses, again labeled m1 and m2. Now consider three locations (shown by the yellow points) in between them. The colored arrows show the individual (yellow) and net (blue) gravitational fields at the point on the left.

Notice that the point on the left is closer to mass m1 such that the gravitational field pointing to the left has a greater magnitude than the gravitational field pointing to the right due to m2. This makes the net gravitational field still point to the left, and *not* be zero.

But if you look at the point in the middle, then the distance from mass 1 and mass 2 are equal, and they produce equal magnitude (but opposite direction) gravitational fields, for a net field of zero. *Only* at that point in the middle would the gravitational field be zero.

It follows that if you have a spherical distribution of masses, then the gravitational field at the exact center should be zero, because every mass on the sphere has a matching mass on the other side to cancel its gravitational field.

But what about points *other* than the center of the sphere? Let's consider a simpler situation. Here is an arrangement of 10 equal masses in a circle. (I've included an arrow to represent the gravitational field due to each one at each location being a little bit off to the side of the center.)

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If you calculate the net gravitational field, it's essentially zero newtons per kilogram, although a slight rounding error might occur. Notice that the gravitational field location is closer to the left side of the circle. You might think that it would be just like the case shown above of the two masses, in which the net field points to the left. But in this case, although this location is farther from the right side of the circle, there are *more* points to the right of the observation location than there are to the left.

So this creates two competing factors: The location is closer to the left side, but there are more masses on the right side. In the end, these two effects counteract and produce a zero field. If you increase the number of points in the circle (to make it more continuous) then the net gravitational field at *any* point inside is equal to zero.

But there’s another thing to consider: The Earth isn't a circle. It's a sphere. (Well, it’s mostly spherical.) Let's expand this calculation for points in a sphere. I'm going to repeat this with a *bunch* more point masses in a spherical shell—let’s go with about 1,000. Also, since it's a little complicated to get masses evenly distributed over a spherical shell, I'm just going to spread them out in a random pattern. (This is called a Monte Carlo calculation.)

Here are 1,000 random points on a sphere. There's a yellow dot in there, too—that's where I'm going to calculate the net gravitational field.

With all of these points, the net gravitational field is very close to zero. And as you increase the number of points, the net gravitational field gets closer to zero N/kg as the sphere becomes more continuous. Boom. Now you can see the physics concept that Munroe used to answer the 20-pound weight loss question: The planet’s outer mass doesn't add to the gravitational field for an object below the surface. If you go below the surface, all the mass above you doesn't matter. (Sometimes this is called “Gauss' Law for Gravity”—which is also known as the divergence theorem.)

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Now, what if I move *outside* this sphere? It turns out that the gravitational field due to a spherical distribution produces the same gravitational field as if all the mass was concentrated into a single point at the center of the sphere. This is kind of nice, as it allows us to easily calculate the gravitational field from the Earth by just using the distance from the center of the object, instead of worrying about its actual size and its total mass.

Now, we have one more thing to consider: How does the gravitational field (and therefore your weight) change as you get closer to the center of the Earth? We’ll need this information to find out how far a person would have to tunnel to reduce their weight by 20 pounds.

Let's start with the Earth as a sphere of radius (R) and mass (m). In this first approximation, I'm going to assume the Earth's density is constant so that the mass per unit volume of stuff on the surface (like rocks) is the same mass per volume as the stuff at the center (like magma). This actually isn't true—but it's fine for this example.

Imagine we dig a hole, and a person climbs down it to a distance (r) from the center of the Earth. The only mass that matters for the gravitational field (and weight) is this sphere of radius (r). But remember, the gravitational field depends on both the mass of the object and the distance from the sphere’s center. We can find the mass of this inner part of the Earth by saying that the ratio of its mass to the mass of the whole Earth is the same as the ratio of their volumes, because we assumed uniform density. With that, and a little bit of math, we get the following expression:

This says that the gravitational field inside the Earth is proportional to the person’s distance from the center. If you want to decrease their weight by 20 pounds (let's say 20 out of 180 pounds), you would need to decrease the gravitational field by a factor of 20/180, or 11.1 percent. That means they would need to move to a distance from the center of the Earth of 0.889 × R, which is a hole that's just 0.111 times the radius of the Earth. Simple, right?

Well, the Earth has a radius of 6.38 million meters—about 4,000 miles—which means the hole would have to be 440 miles deep. Actually, it's even deeper than that, because the density of the Earth isn't constant. It ranges from about 3 grams per cubic centimeter at the surface up to around 13 g/cm3 in the core. This means you'd need to get even *closer* to the center to get a 20 pound reduction in weight. Good luck with that. If you really want to lose weight, you'd be better off just joining a gym.