It's possible that the Earth might soon gain another temporary "mini moon" as a sun-orbiting object comes close enough to perhaps become captured and orbit the Earth along with our giant moon. This mini moon (technically it has the name 2020 SO) might be an asteroid—or it might even be a rocket booster back from the 1960s. But whatever it is, it looks like it will be in Earth's orbit around October 15, 2020.

There's a lot going on with the motion of this mini moon. Of course there is a gravitational interaction with both the Earth and the moon, but it's also interacting with the sun. Not only that, but the Earth and moon are accelerating as they move in a mostly circular orbit around the sun. But let's start off with something simpler. Suppose it's just the Earth, the moon, and the mini moon. Can we model the motion of these three objects? The answer: Yes we can. Also, is it very easy for an object to get trapped in the Earth-moon system? Let's find out.

Imagine the Earth, moon, and the mini moon are in the following positions.

This diagram looks terrible. It looks so bad because it's fairly realistic. Yes, the moon is that far away from the Earth and it's very difficult to see. Also, I made the mini moon WAY too big—but that's the only way you can see it. This is why many textbooks show the Earth-moon system without the correct scale. It gets even worse if you try to include the sun, since it's even further away and would make the size of the moon and Earth like tiny little invisible ants. So, now that I've shown this system with the correct scale (except for the mini moon), I'm going to make a more useful diagram.

Yes, there's a bunch more stuff on this diagram, so let me describe what's going on. What about these arrows? These are representations of the gravitational interactions between the three bodies (Earth, moon, mini moon). Whenever you have two objects that have the property we call "mass" (which is pretty much everything), there is an attractive gravitational force pulling these two objects together. The magnitude of this force is proportional to the product of the masses of the two objects and inversely proportional to the square of the distance between them. Since it's painful to write that relationship in words, we can also describe it with the following equation.

In this expression, *G* is the universal gravitational constant (with a value 6.67 x 10^{-7} Nm^{2}2/kg^{2}), *M _{E}* and

*M*are the masses of the Earth and moon and

_{m}*r*is the distance from the center of the Earth to the center of the moon. But there are two very important things to notice with these objects and the forces. First, the forces come in pairs. If the moon pulls on the Earth (labeled F

_{m-E}) then the Earth pulls back on the moon with the same magnitude force (labeled F

_{E-m}). Second, for each object there are two gravitational forces acting on it. The total force is just the vector sum of these two forces (called the net force).

But what do these net forces do to an object? The net force on an object changes that object's momentum, where the momentum is the product of the object's mass and its velocity. Yes, we use the symbol p for momentum—that's just the symbol we always use (we can't use m since that's already the mass).

Putting this together with the net force, you get the momentum principle. It's kind of a big deal in physics.

So, it seems like you could solve this problem for the motion of the mini moon. Just calculate the force, use that to find the change in momentum and then use that momentum (and the velocity) to find the new location of the mini moon. Yes, this would work—but it's actually impossible to obtain single equation for the position of the mini moon. The difficult part is that the mini moon pulls on BOTH the Earth AND the moon. This means that their momentums also change. All three objects interact with each other and it's just not solvable unless you make some approximations (like deciding that the force on the Earth is too small to worry about).

This problem is actually so famous that it has a name. It's called the Three Body Problem—and we can solve it. I know what you are thinking. I just said you can't solve, right? No. I said you couldn't get an equation of motion for the three objects. However, I CAN find the position of the stuff at certain times. The way to find out how these things move is with a numerical calculation. In numerical calculation, the problem is broken into a whole bunch of short time intervals. During each time interval, we can assume the gravitational forces are constant (even though they aren't). With constant forces, it's fairly easy to find out where the objects are at the end of the time interval. Then, moving into the next time interval, I can just find the new force (since all the objects moved) and assume it's constant again.

This might seem like you are getting a solution at no extra work, but this method comes with a cost. If you break the motion into 1 second time intervals and you want to find out where the stuff is after 100 seconds then you would need to do all these calculations 100 times. So, instead of one impossible problem of finding the equation of motion, you get many simple problems. But at least it's possible.

Personally, I don't want to do endless calculations for the motion of these three objects. However, I don't mind making my computer do it for me. In fact, no one does these kind of calculations by hand anymore. Many people might even call it a computational physics solution. I think it's important to keep the "numerical calculation" name so that no one thinks you HAVE to use a computer—you don't.

OK, I'm not going to go over all the details because I would rather focus on the results. If you want to get into the process of building a numerical calculation using python, I've got a quick tutorial that should get you started.

But don't worry. I'm not just going to show you the result. I'm going to show you what happens WITH the code. Here is the motion of the mini moon in a reference frame in which the center of mass is at rest (so, ignoring the motion around the sun). If you want to run the calculation again, just click the "play"—to see the code, click the "pencil" icon.

I just picked the initial position and velocity values of the mini moon based on this excellent animation on the 2020 SO Wikipedia page. However, you can see that my version of the mini moon doesn't really get trapped in the Earth-moon system. It's not even a temporary moon. In this system with a stationary Earth, it's just not going to get trapped. It's all about energy. Imagine that you have a ball rolling on flat ground—but there is a hole that it's moving toward (maybe more like a depression in the ground). When the ball enters the depression, it rolls downhill and speeds up. But then when it reaches the other side, it goes uphill and slows down.

If this is a perfect ball with a perfect ground, then there would be no energy loss due to friction. That means that the ball will leave the hole with the same speed it entered. It wouldn't get "trapped." This is just like the mini moon moving near a stationary Earth—but it's not an actual depression, it's just a change in gravitational potential energy due to the interaction between the Earth and moon.

So, how could you get a moving ball to both get to the depression, and then stay there? One answer—have the depression accelerate. If this depression is accelerating away from the ball, the relative velocity between the ball and depression will be such that it won't have enough velocity to climb back up out of the hole. Oh, this is exactly what happens with the Earth and the mini moon as it becomes at least temporarily trapped near the Earth. The Earth is in fact NOT stationary. It's orbiting the sun, which means that it's accelerating as the direction of motion changes. Yes, it's true that the acceleration of the Earth seems minor compared to the gravitational interaction with other objects—but this is why it's so difficult for objects to get trapped near the Earth. So, the mini moon has to come it in with a low relative velocity and at just the right angle to be trapped. But our solar system is old enough that most of the objects that fit this criteria have already been trapped. All the moons are used up—mostly.