# Help with calculating an impact on the moon

1. Apr 9, 2012

### John Clayborn

Hello everyone, I'm new here. I'm relatively familiar with some astrophysics calculations, but this particular equation is kicking my proverbial butt. I'm currently writing a sci-fi book series and there's an impact that happens on a moon, but I want to be able to realistically calculate what the damage would be.

I have all of the data on the impactor; size, mass, velocity, density (7.85 kg/cm^3: steel), etc. I even have the data on the moon itself. (I already have an excel spreadsheet set up to aid in those calculations). What I cannot determine is the correct equation to use to calculate the diameter of the transient crater or the depth of the crater. I had found one paper, but I can't seem to get to the calculation to work.

The original paper that I found was an academic paper written by the authors of the "Impact:Earth!" program. The calculation are quite detailed, but I'm coming up with an error. Obviously, as this moon is without an atmosphere I can completely skip the section about determining atmospheric effects on the impactor.

Here are my questions;

The formula they use is as this: Dtc (Diam. Transient Crater) = 1.161(Pi/Pt)^1/3 L^0.78 Vi^0.44 gE^-0.22 sin^1/3∅

Question 1: Where are they getting the constant of 1.161 from? They state that it should be changed if the impact occurs in water. This isn't the Earth's denisty (5.52 g/cm^3), so what is it? and should it change if the impact is on the lunar surface as opposed to the rocky surface of the Earth?

The rest I get more or less..Pi - Desity of the impactor divdided by Pt - density of the target raised to the 1/3 power. L is the diamter of the impactor raised to 0.78 power. Vi is the velocity of the impactor raised to the 0.44 power. gE is originally Earth's gravity, however, since the impact is on a moon then I would just substitute this value for the lunar gravity and raise that to the negative 0.22 power. But the next section is where I get lost. They want me to take sin to the 1/3 power and multiply it by the angle of attack. However, you can't take 1/3 of sine since sine itself is dependent upon ∅. Right? So shouldn't that last part be ∅Sin^1/3? (or something).

Any help that you guys could offer would be most appreciated.

2. Apr 9, 2012

### cepheid

Staff Emeritus
Welcome to PF,

I can at least address the very specific math question at the end of your post. Sine is a function, which you can think of as being an operation that acts on a number. It is not itself a number, and therefore it makes no sense to raise it to the one-third power (or any power).
The expression $\sin^{1/3}(\phi)$ means $[\sin(\phi)]^{1/3}$. It's just a shorthand notation. You can think of the expression on the right-hand side as saying, "first take the sine of phi, and then take the result of that and raise it to power of one-third."

As for the impact physics, I'd have to think about that some more. I imagine a general approach would be to compute the kinetic energy of the object just before impact, and then equate that to the gravitational binding energy of the surface material that it has to displace. The latter will depend on the volume of material you consider, which will tell you how big a crater you can make.

That's just a guess on my part.

3. Apr 9, 2012

### John Clayborn

Thanks for clarifying that, Cepheid. I had a feeling that's what they meant, but it's always nice to have my logic validated by someone else. ;)