What impact is required for a visible (from Earth) ejecta plume on Earth's Moon, and would the Moon survive?

Question

Take the Earth-Moon system as we know it. Now, something causes a large rock to be lobbed in the direction of our moon. Exactly how that happens is deliberately left unspecified; it could be everything from an Earth-Mars war, to a rogue planet passing through the solar system, to something else entirely.

The impact should be sufficiently energetic to cause the resulting ejecta plume to be clearly visible from Earth to any creature with human-like vision who happens to be looking Moonward at the time. Let's put the lower limit for this at a plume height of about 10% of the angle subtended by the Moon itself as viewed from Earth at the maximum height where there is still a reasonable particle density. Bonus points to answers taking into account albedo and particle density, but that's not required.

You may choose the time, point and angle of impact arbitrarily, as long as the condition of visibility from Earth is met. (So hitting the back side of the Moon probably won't do it, but you are free to hit the Apennines from the side if you want to.) I'm thinking a direct hit near the terminator near full moon, but if something else works then feel free. I'm hoping for the spectacular effect, which is why I'm not fixating on any particular location.

You can assume that humans do nothing to counter the threat of an approaching large space rock; as to why, there is every reason from technical or physical inability to international politics or that humans are long gone from the planet.

With the preliminaries out of the way, two very much related questions:

What impactor parameters could give this short-term effect? Impactor mass, velocity (relative to the Moon), strike angle, location if relevant, any other relevant parameters I'm not thinking of?

Will the Moon likely survive the impact in the medium term? Said in one other way, is the impact energy comfortably below the gravitational binding energy of the Moon (which is approximately $1.24 \times 10^{29}$ J)? You don't need to worry about the effect of the impact on the Moon's orbit.

Answer 1

How small of a thing can we see on the moon?

Wikipedia has a list of objects on the moon that can be visible with unaided vision. Doing some poking around Astronomy blogs for real-life experience as opposed to calculations, the mode answer seems to be the crater Tycho, which is 86 km across.

What needs to hit the moon to make a 86 km ejecta plume?

First, lets use kinematics to determine how fast the ejecta must be going to get to that height. We can use $$v_f^2 = v_i^2 + 2ad$$ where $d$ is 86000 meters, $a$ is the surface gravity of the moon, -1.6 m/s$^2$, and $v_f$ = 0. I solve this for $v_i$ = 524 m/s.

Next we can go to a paper to calculate expected ejecta velocity, such as Richardson, et al., 2007. If you look at Fig 7. of this paper, it shows that ejecta velocity is a logarithmic function of distance from the impactor's edge. Now I could not extract any useful information for what mass density of ejecta would be needed to be visible, and this is getting a bit mathematically complicated anyways, so lets make a simplifying assumption that we need the impact to cause ejecta within 1km of the impact edge to be going 500 m/s.

From equation (12) we can calculate the volume of the transient impact crater as $$V_g = K_1\left(\frac{m_i}{\rho_t}\right)\left(\frac{ga}{v_i^2}\right)^\frac{-3\mu}{2+\mu}.$$ The density ratio term from eqn (12) is dropped; we'll assume that the density of the impactor is the same as the moon. For material properties we will use 'soft rock' from Table 1; so $\mu=0.55$, $\rho_t=2250$, and $K_1=0.2$. $m_i$ is the mass of the impactor, and $a$ is its radius. We can express mass in terms of radius as $m_i = 4/3\pi\rho_t a^3$. the gravitational force of the moon is $g$ and equals 1.6 m/s$^2$. I resolve this expression as $$V_g = 1.7a^{2.353}v_i^{1.29}.$$

Now for follow on equations, we want $R_g$, radius of the transient impact crater, which is related in equation (11). I solve that backwards for $$R_g = \left(\frac{3}{\pi}V_g\right)^{1/3} = 1.18a^{.784}v_i^{.431}.$$ As a test run of our model so far, if you plug in a 100m object and a 10 km/s impact, you get a 2.3 km transient impact crater. Great!

Now we move on to eqn (28) which gives us ejecta velocity as a function of distance from impactor rim $$v_e(r) = \frac{\sqrt{2}}{C_{Tg}}\left(\frac{\mu}{\mu+1}\right)\sqrt{gR_g}\left(\frac{r}{R_g}\right)^{-1/\mu}.$$

$C_{Tg}$ is a proportionality constant equal to 1.6, see discussion around Eqns (15) and (20). I simplify these terms to $$v_e(r) = .397r^{-1.82}R_g^{2.32}.$$ We want $v_e$ to be 500 m/s at a distance of r = 1000 m, so plugging these and $R_g$ in we can solve for radius and velocity $$2.47\times10^{8}= a^{1.82}v_i.$$ Obviously, there are infinitely many solutions, but for some reasonable projectiles, if we set radius to be 250 m, then impact speed must be 10.7 km/s. At maximum comet speed of about 70 km/s, we get a radius of about 90m.

Conclusion

To make an ejecta plume visible on the moon's surface to the naked eye, you need to hit it with an object at least 100m if it is moving at the speed of a long term comet, or at least 250m for a glancing blow.

The second question is an obvious yes, those impacts are pretty small on a scale of 'things that have hit the moon', and won't leave a (naked eye) visible impact crater when all is said and done.

Answer 2

A tiny spacecraft, Lunar Prospector (which is also the subject of this paranoid answer) was crashed with the expextation of making a visible plume.

People do see “transient phenomina” probably caused by small impacts, and they don't leave marks that can be seen with the most powerful telescopes. So again, small insignificant bodies.

For survivability, consider the “rays” on the lunar surface coming from some craters. Clearly was a huge plume to make fallout for hundreds or thousands of miles; but just a crater. No big deal.

You calculated the binding energy… do you know what that means? Consider that the impactor will be falling from (approximatly) at-rest infinity, the same as used in the binding energy. So, just from the definition of terms it appears that a falling object would need to be as large as the moon in order to impact with the same energy as the moon’s binding energy.

Answer 3

We have an observation of an impact plume that dates from the middle ages. The moon is still there. We do not know the height of the plume, though.