Secondary Craters: Kinetic Energy & Ejection Velocity

[Kreizhn]

Homework Statement

Consider the impact between an iron meteoroid (density = 7000 kg/m3) with a diameter of 300 m and the Moon.
1)Calculate the kinetic energy involved if the meteoroid hits the Moon at a velocity of 12 km/s.
2) If the rocks are excavated from the crater with typical ejection velocities of 500m/s, calculate how far from the main crater one may find secondary craters.

Homework Equations

The Attempt at a Solution

So part 1) is pretty simple, just using the basic E= \frac{1}{2} m v^2. I did the work and found the energy is roughly 5.7x10^19 Joules

Now for part 2) I'm wondering if there isn't an obvious way of doing this that I'm somehow missing.

It wouldn't be too hard to say that the maximum distance will occur when ejecta leave the crater at an angle of \frac{\pi}{2} radians, and then use the moon's gravitational pull to find the maximal distance, but I'm wondering if it isn't somehow more obvious than that...

 

[Nick89]

I think you mean \pi / 4 radians, pi/2 would be straight up ;)

I can't think of any easier way to do this. It's not that hard to do is it? You can see it as a simple projectile motion. With a known velocity and angle you can work out the distance.

 

[Kreizhn]

Yes, you are right about \frac{\pi}{4}. Thanks.

 

[dynamicsolo]

The maximum range expression is probably all they want you to use for the question and that is the simplest way to get an estimate. It will be a bit of an underestimate since the escape velocity of the Moon is 2380 m/sec, so ejecta moving at 500 m/sec are traveling at over 20% of escape velocity, meaning that they are actually on suborbital elliptical trajectories, rather than simple parabolas. (Another way of saying this is that the fragment flies "high enough" that the local acceleration of gravity is no longer essentially constant.) So, properly speaking, we would look for the intersections of elliptical paths with the lunar "sphere".

So there are certainly ways to make finding the answer harder. What you've chosen is what most people would do to get an adequate estimate. (We know that some fragments leave much faster than 500 m/sec. Some few have left the Moon entirely and landed on us! [see, for instance, http://en.wikipedia.org/wiki/Lunar_meteorite ] )