# Science Sunday: Falling Through the Earth

Now that Mindy’s teaching everyone about gravity, I thought it would be fun to show the classic gravitational problem, “What happens if you fall though the center of the Earth?” It’s a very common problem, and the math isn’t too difficult. So let’s run through it, shall we?

(If you can’t follow the math, it’s ok. I explain everything at the end.)

First things first, let’s draw a picture. The big circle is the Earth, which, as you can see, has a radius of R. The smiley face is you, falling down through the Earth. (I used a smiley face instead of a stick figure because I couldn’t find that clipart. Sue me.) Anyway, we can write your position at any time as r, where r is your distance from the center of the Earth. Obviously, r is a function of time, so our goal is to develop an equation for r in terms of things we know. We can also say that the total mass of the Earth is M, and your mass is m. Now, if you remember way back two weeks ago, you know we can write the gravitational force like this: You’ll also remember that G is a constant, and it equals 6.67 x 10-11. Of course, if you remembered that, you also remember, from one week ago, that the gravitational acceleration can be written like this: Remember that g is the gravitational acceleration that you feel on the surface of the Earth. Its value is roughly 9.8 m/s2.

Still with me? Good. Because this is where stuff gets a little complicated.

Turns out, when you’re falling through the Earth, the force of gravity doesn’t depend on the entire mass of the Earth. In fact, it only depends on the mass between you and the center.

That’s a bit confusing, so I’ll draw another picture: When we’re looking at gravitational forces, only the stuff inside the red circle matters. Why? Because, in the end, all the outside mass ends up cancelling out. I don’t remember exactly how, though. I proved it once, and then promptly forgot.

So how do we calculate the mass inside the red circle? Well, first we assume that the Earth has uniform density, which isn’t true, but we’re doing physics not geology so shut up.

Next, we find the total density of the Earth. Density is defined as mass per volume, so the density of the Earth is its total mass divided by its total volume. Like this: In physics, density is represented by a lowercase rho. The volume of the Earth is, of course, the volume of a sphere with radius R. Technically, the Earth is not exactly a sphere, but we’re already assuming the Earth has a uniform density, so why stop there?

So now that we have the density, we can multiply by the volume of the red circle to get the mass contained within it.When we do that, we get this: Substituting this value for M in the first equation, we get: Ok, almost done. Still with me?

Next, we use Newton’s Second Law, F=ma, to get the acceleration you feel as you fall down the hole. That looks like this: And cancelling out the mass on both sides of the equation, we get this: One last step. If you go all the way back to the beginning of this post, you’ll see that g=GM/R2. Conveniently, there’s also a GM/R2 in the previous equation as well. So lets replace all those terms with a g, shall we? And we’re done. This gives us a fundamental relationship between where a falling object is and how much it’s accelerating. If you followed all the math to this point, congratulations, you get a gold star. If you didn’t, well, you can still look at this equation and read the rest.

So what does this actually tell us? Well, right off the bat, it tells us that the motion is periodic, that is, it repeats at a regular frequency. In fact, that frequency is the square root of g/R. Secondly, if we know the frequency, we know the period, which is 2π divided by the frequency. In this case that’s the time it takes to make one round trip through the Earth. If you do the math, that comes out to about an hour and a half. At the center of the Earth (where your speed is greatest) you’d be travelling at about 18,000 mph.

But wait. Here’s where things get really interesting.

What happens if you fall through a hole that doesn’t pass through the center of the Earth?

In other words, what if the hole looked like this? Here, r and R are still the same, but now we introduce three new variables: y, the vertical distance between the hole and the center of the Earth; x, the horizontal distance between you and the center of the Earth; and theta, the angle between r and y. Surprisingly, it turns out that the equation of motion is exactly the same in this case as in the previous case. This means that the motion through the hole is still periodic, it still has the same frequency, and it still has the same period. Or, to put it another way: The amount of time it takes you to fall through a hole in the Earth is the same no matter where you start or end. Think about that for a bit.

And we can even go one step further. What if, instead of falling through the Earth, you were to orbit it at the surface? How long would it take you then? If you guessed the same amount of time, you’d be correct! This is a very interesting feature of this kind of gravitational motion, in that it’s largely path-independent. You can get from one place to another in the same amount of time regardless of the path you take, and if you make a round trip, all paths are roughly equal. I’m fairly certain this holds even for curved paths through the Earth, although I haven’t proven it, so I’m not positive that’s true. In physics, if we don’t know the answer, we say “This is an exercise for the reader.” So if anyone wants to try their hand at proving this in the comments, be my guest.

Anyway, that’s it for what was probably going to be tomorrow’s Physics Philes. Sorry Mindy.

### 2 Comments

1. Gravity is awesome 🙂

2. I’d seen this before, but that other article didn’t tie this to the orbital speed at the surface… or, with a little margin or error, to the orbital speed of the ISS a few hundred miles up. Awesome.