## Recreational physics: crater size

Posted in Science with tags , , on August 16, 2009 by Grad Student

This morning I had fun trying to estimate the diameter of a “crater” created by dropping a metal ball into a box of sand. The “craters” in the sand don’t actually look like a real crater, it looks like a conical hole. (Okay so the pic below is actually that of an antlion trap).

I got the idea for this calculation from a lab I teach where the students drop metal balls of varying masses (20-150 grams) from the same height. They then measure how the diameter of the craters change with the mass of the ball. The hypothesis is that the diameter of the crater, D, depends on the kinetic energy of the ball, KE, like so:

$D=k (KE)^n$

where k and n are the constants that the students will figure out by doing the experiment. As I’ve been the TA for this lab about 6 times, I remember that usually k = 0.1 and n = 0.3. So this morning, I tried to use basic kinematics to see if I could derive what k and n should be theoreticallly. First, I estimated that all the kinetic energy of the ball is used to throw the sand particles to the edge of the crater. This energy is:
$KE=\frac{N m v^2}{2}$
where N is the number of sand particles displaced to the edge of the crater, m is their mass and v is the velocity imparted to them by the ball. We know that N can be expressed as
$N=\frac{\rho V}{m}$
and the volume, V, of the crater is
$V=\frac{1}{3}\pi R^2 h$
assuming the crater is conical and h is the depth of the cone. How do we find h? Note that h isn’t necessarily equal to how far the ball penetrates into the sand. In many cases you can’t even see the ball at the bottom of the crater because it’s covered with sand. Instead, I think the angle of the slope of the sand may remain roughly constant, which mean that h is proportional to R:
$h=w R$
where w is the cotangent of the opening angle of the cone, which I assume it constant. To find how far the sand particles get kicked, we can use the range formula:
$R\sim \frac{v^2}{g}$
where g is the acceleration due to gravity and v is the initial velocity of the sand particles. This formula is exact if the particles are thrown on level ground at an angle of 45 degrees, which is obviously an approximation in our case. Now we can finally put all this together to express the kinetic energy of the ball as:
$KE=\frac{\rho \pi R^2 w R m R g}{2 \cdot 3 m}$
now we can rearrange the formula, cancel out the sand particle mass, and find out how the diameter, D=2R, of the crater depends on the ball’s kinetic energy:
$D= 2 \left(\frac{6}{\pi \rho w g}\right)^{1/4}(KE)^{1/4}$
Comparing this to the first equation above, we see that my predicted n=0.25 isn’t that far away from the n=0.3 found by experiment! Also, k is equal to all the stuff in front of KE in the above equation. If we say that w ~ 1 and the density of sand is about 3 grams per cubic center and gravity is 980 cm/s^2, then:
$k= 2 \left(\frac{6}{\pi \rho w g}\right)^{1/4}=2\left(\frac{6}{3.14 \cdot 3\cdot 1 \cdot 980}\right)^{1/4}\sim 0.32$
Compare this with experimentally found value of k=0.1. My theoretical value of k doesn’t seem to be as accurate as my theoretical n value, but still it’s the same order of magnitude. Since my predictions of n and k are in the right ballpark (especially n), that suggests that I’m getting at least some of the physics right. A huge simplification I’ve made is in assuming that all the ball’s kinetic energy goes into launching sand particles to the edge of the crater. In fact, if I assume that 1% of the ball’s energy goes into ejecting the sand particles, that will reduce my k to 0.1. I should also mention that this analysis is completely wrong when it comes to craters on the moon or other places in the solar system. When you get a rock the size of house or a city hitting the ground with a speed between 10 and 70 kilometers per second, other physics comes into play (e.g. shock waves, vaporization). You can see the difference by comparing the photo below of a real crater to the photo at the beginning of the post.