# What percentage of a lunar meteor strike is blown back into space?

Reading a discussion on another website, it was proposed that if a large meteor struck the moon, that a few days later, Earth would receive a meteor shower.

I thought about that and it seemed to me that one of the big unknowns would be what percentage of the impact would reflect back into space, away from the lunar gravity well.

The moon, having no atmosphere, then velocity out, could be pretty close to the velocity in, minus whatever's lost in heat and pressure waves and perhaps phase transition, though that could increase the explosive rebound and maybe other energy changes I'm not considering.

My question is simply, is there a fairly simple formula or estimate of what percentage would blow back into space following a lunar impact. I'd suspect that it would depend on both the velocity and mass of the meteor and perhaps, composition. If it was large enough and a direct hit, it could blow the entire moon into space, for example, but I'm thinking more standard asteroid size impacts not giant impacts.

Say, for example, a fairly large impactor like 99942 Apophis, about 27 billion kg, 370 meters in diameter.

• depends rather on where the impact occurs relative to the earth-facing side's center. – Carl Witthoft Feb 28 '19 at 20:24
• @CarlWitthoft Thanks. That's a good point. I know that the near-zero atmosphere makes it much easier, so I'm thinking any large collision would eject a percentage, but I still wonder if any observations or estimates have been made, though there might be too many variables. – userLTK Feb 28 '19 at 20:36

There are some invariances that actually lend themselves well to estimating an answer to this question. From the Deep Impact mission white paper:

A 1 mm projectile striking a target at 10 km/s will yield the same result as a 1 m projectile striking at the same speed, provided all distances are scaled by the same ratio of 1,000 = 1 m/1mm and all times are multiplied by the same factor. Thus, if the 1mm projectile makes a crater 6 cm in diameter in 300 ms, the 1 m projectile will create a 60m diameter crater in 300 s.

The same paper gives us a set of three equations to estimate crater diameter. These equations are the result of experimental research combined with modeling the Navier Stokes differential equations governing the conservation of energy and momentum in materials. The first equation is: $$\pi_2=\frac{1.61gL}{v_i^2}$$ Where $$\pi_2$$ is the inverse of the Froude number, $$g$$ is the local surface gravity, $$L$$ is the projectile diameter, and $$v_i$$ is the impact velocity. We will use an impact velocity of 12.6 kms taken from this paper. The second equation is: $$\pi_D=C_p\pi_2^{-\beta}$$ Where $$\pi_D$$ is a dimensionless measure of crater diameter, and $$C_p$$ and $$\beta$$ are material constants. The third equation is: $$\pi_D=D(\frac{\rho_t}{m})^{-1/3}$$ Where $$m$$ is the mass of the asteroid and $$\rho_t$$ is the density of the lunar surface. The top 10 or so meters of the lunar surface are regolith at about 1.5g/cm³, but the next several km are mixture of bigger, more dense ejecta from billions of years of impacts, combined with finer grained regolith, according to this paper. Given the overall moon density of 3.34 g/cm³, we'll estimate the density of this top layer as 2.5 g/cm³. Solving for $$D$$ and then plugging in all the values of our variables gives us a crater radius $$R$$ of approximately 2.5 km.

In an asteroid impact, the fastest moving ejecta comes from near the center of the crater. The ejecta closer to the edge of the crater has much lower velocity:

The equation governing the velocity of the ejecta $$V_{ej}$$ is therefore partly a function of $$r$$, the distance from the center of the crater. $$V_{ej} = \frac{2\sqrt{Rg}}{1+\epsilon }(\frac{r}{R})^{-\epsilon}$$ Where $$\epsilon$$ is a material coefficient of 1.8.

We can solve the above equation for $$r$$ and solve using $$V_{ej}$$ as 2.38 km/s, the lunar escape velocity. This yields a value of 275m for $$r$$. That means that from the crater with a radius of 2500m, the only ejecta to reach escape velocity comes from the interior most part of the crater with a radius of 275m.

The depth of the crater can be estimated as $$d=L\sqrt{\rho_p/\rho_t}$$ where $$\rho_p$$ is the density of the projectile (asteroid). Since the density of the projectile (2.6g/cc) and the estimated density of the lunar surface (2.5g/cc) are similar, then the depth would be approximately $$L$$, or 370m.

If we estimate the volume of the crater as a cylinder, then the total volume of lunar ejecta will be $$\pi*R^2d$$ is about 7.21e9 cubic meters, and the lunar ejecta to reach escape velocity will be $$\pi*r^2d$$ or about 8.79e7 cubic meters or a little more than 1% of the total ejecta. This may not seem like much, but the total volume of Apophis is only 2.65e7 cubic meters. That means that the ejecta reaching escape velocity will be more than 3 times the original volume of Apophis!

Notes:

1. The ejecta plume is in an inverted cone, so even given ideal geometry, most won't make it to earth.
2. The ejecta that are just under escape velocity will obtain an orbit that leads them to crashing back into the lunar surface.
3. The field of asteroid impact modeling is still in quite a bit of flux. Also, the models tend to perform less well for very large asteroid impacts since we have no ability to perform experiments on that scale. It wouldn't surprise me if the total escape velocity ejecta calculated above differed by an order of magnitude in either direction to a similarly sized lunar asteroid impact.
4. My codes is available on request.