# How Would You Build a Pile Driver on the Moon?

Who cares about pile drivers — especially on different planets? Well, apparently Zach Weinersmith does.

This seems like a great physics question — so I’m going to find a solution. But first, let’s take a moment to recognize that I actually had a picture of a pile driver. Yes, I try to take pictures of things I might need in the future. It’s a win in this case.

OK, what about a pile driver? From what I understand, it’s basically a large mass that is lifted above a post. The mass falls down to hit the post and knock it into the ground. I assume this is useful in cases where you can’t just dig a hole and put in the post — like with the posts in a dock where the bottom is underwater. But seriously, what do I really know about pile drivers? That’s about it.

Let’s assume that we have a pile driver on Earth. Here are the important parameters.

Really, there are only three things that we can change for this pile driver. The value of the dropping mass (m), the height that it’s dropped (h) and the local gravitational field.

What about the piling? Is that even what they are called? You know, the thing that’s being pile driven into the ground. But this piling is sort of a problem. I’m not exactly sure how these things get pushed down. Is there some constant frictional force pushing up on it as its pushed down by the mass? Does the upward pushing force depend on depth? I assume so.

But still, I think we can model this piling-ground as just a spring. So, when the pile driver drops down it will hit this “spring” and compress it. If you model the piling this way, it seems like we could repeat the process on another planet (assuming the ground would be similar — which it totally would not be).

With that, we have the following physics problem (on Earth).

In this situation, the mass (m) starts at position 1. It then falls a distance h and hits the spring (with spring constant k). After the impact, the mass moves down an extra distance (s) until it’s in position 2.

We need a relationship between all these variables — with that, I’m going to use the work-energy principle (since we only care about changes in position). OK, here is my super quick crash course on the work-energy principle. Basically it says that the work done on some “system” is equal to the change in energy of that system.

What is work? What is energy? What the heck is a system? Yes, there’s a lot here. Let’s start with the system. Imagine that I draw a dotted line around the stuff that I care about — that’s my system. Any external forces that push on the system (from the outside) would be work. But don’t worry about work, we aren’t going to need it in this example.

What about energy? Well, that also depends on the system. So, let’s say we have the system of the pile driving mass, the piling (spring) and the planet. In that case there’s no external forces and there can be three types of energy.

- Kinetic energy: the energy due to the motion of an object.
- Gravitational potential energy: energy stored in the gravitational interaction between the objects and the planet.
- Spring potential energy: energy stored in a compressed (or stretched) spring.

We can calculate these energies as:

Great. Now we are ready to set this thing up. Let’s look at the changes in energy as the mass drops from position 1 (see the diagram above) to position 2. Since there’s no work done, I can write the work-energy equation as:

Remember, the work-energy principle deals with the CHANGE in energies. OK, with that — there are some of these energy terms that are zero (Joules). Since the mass starts and finishes with a zero velocity, the two kinetic energies are zero Joules.

Also, the initial spring potential energy (U_s1) is also zero. Why? Well, when the mass is released the spring is neither compressed nor stretched. So, no energy there.

Finally, for the gravitational potential energy — we need to pick a location where y = 0 meters. It really doesn’t matter where this is, we just need to stick with that location. I’m going to pick the lowest point of the mass as y = 0. That means that the final gravitational potential energy (U_g2) is also zero. Oh, but that also means that the mass starts at a y position of h+s and not just s.

Now we have a much simpler work-energy equation.

Now I can solve for h:

This is actually shows one of the possible answers. Suppose that I have a pile driver on Earth with a gravitational field of g (where g = 9.8 N/kg). Now I want to use the same thing on the moon with g_m = g/6 (approximately). If I want the value of s to be the same (the pile driven the same distance) then I need to drop from the following height (h_m — for h on the moon).

Now I’m going to substitute g_m = g/6.

That’s the answer. Notice that you don’t drop from 6 times the height as on Earth. Why not? It’s because of that value of s. If the pile is driven over a significant distance then the change in gravitational potential energy won’t be too much and you will have to drop from an even larger height. Of course, if the pile driven distance is small, then we can ignore this “s” term. In that case, yes — you need to drop the mass from a height of 6 times greater on the moon.

Wait. Is there another answer? What if I want to use the same pile driver with the same height but a different mass? If I solve the work-energy equation for the mass of the driver, I get:

Notice that in this case, the h and s are grouped together. Now if I move this pile driver to a planet (or moon or no, that’s no moon — that’s a space station) with g_m = g/6, the mass would need to be 6 times as large as on Earth to have the same impact.

That’s it. Have fun pile driving on the moon. I can’t wait to see your dock on the edge of you moon lake.