<h3>Possibly not much</h3>
<p>From <a href="https://ntrs.nasa.gov/citations/19870027368" rel="nofollow noreferrer">Vickery (1986)</a>:</p>
<blockquote>
<p>The characteristics of three primary extraterrestrial craters and the associated craters were examined to generate a size-velocity distribution for large ejecta fragments. The lunar craters Copernicus and Aristillus and the Martian crater Dv on Olympus Mons were used. Attention was focused on the radial distances between the primary and secondary crater centers and the diameters of the secondaries. The primary craters selected are all relatively young, which avoided contamination of the data from secondaries from other primaries. Attempts were made to account for the speed of the hypervelocity impacts and the elemental compositions of the impactors. An apparent velocity cutoff of about 1 km/sec was observed for the secondaries, <strong>which implies that no meteoroid impacts can accelerate ejecta to escape velocities from the moon or Mars.</strong></p>
</blockquote>
<p>This was based on direct observation of Mars and the Moon. However, of course, the asteroids observed were smaller than Apophis.</p>
<p>Although, this applied only to detectable secondary craters formed by large ejecta fragments. There might have been faster ejecta that was not detected because it wasn't large enough to form secondary craters.</p>
<hr />
<p>But see also <a href="https://www.sciencedirect.com/science/article/abs/pii/0019103585900211" rel="nofollow noreferrer">Hartmann (1985)</a> (you can look for it on sci-hub) which summarizes a variety of experimental impacts at speeds up to a couple km/s. The result seems to be that in the first few microseconds after a hypervelocity impact, there is a thin jet of ejecta that moves at a substantial fraction (&lt;1) of the hypervelocity impactor speed. The vast majority of ejecta occurs later in the impact and moves much slower, down to a few m/s. So it is possible that the initial thin jet could reach escape velocity, but there isn't much of it by mass.</p>
<hr />
<p>It's questionable whether the formulas used in Connor's answer are applicable to very high-speed ejecta. The formula for <span class="math-container">$V_{ej}$</span>:</p>
<p><span class="math-container">$$V_{ej} = \frac{2\sqrt{Rg}}{1+\epsilon }(\frac{r}{R})^{-\epsilon}$$</span></p>
<p>has a discontinuity for small <span class="math-container">$r$</span>. <span class="math-container">$V_{ej}$</span> tends to infinity as <span class="math-container">$r$</span> tends to 0. So for small <span class="math-container">$r$</span> (and high <span class="math-container">$V_{ej}$</span>) this is clearly not valid. The author cited for this formula, Kevin Housen, published <a href="https://sciencedirect.com/science/article/abs/pii/S0019103510003623" rel="nofollow noreferrer">Housen and Holsapple (2011)</a> (check sci-hub) explaining the limitations of the formula. It does not give valid results for <span class="math-container">$r &lt; n_1 \frac{L}{2}$</span> (<span class="math-container">$n_1 \approx 1.2$</span>). At these small radii, the regolith will be driven down underneath the impactor and will not emerge at all - the ejecta velocity is 0 (with the exception of some small amount of mass that may be jetted at a high speed, which is dependent on the geometry of the impactor, and which Housen does not treat).</p>
<p>Plugging in the same values, I get the same answer Connor got, 275m for the Apophis-Moon impact. This is greater than 1.2 * 370 * 0.5 = 222 m, which indicates ejecta between a radius of 222m and 275m would reach escape velocity for the Moon.</p>
<p>Housen gives a formula for the mass of ejecta under a given radius: <span class="math-container">$M(x) = k \rho x^3$</span> where x is the radius, <span class="math-container">$\rho$</span> is the material density, and <span class="math-container">$k$</span> is a constant, between 0.4 and 0.75 depending on soil type, 0.75 for dense soil. So the volume ejected by this formula at a speed sufficient to escape the Moon's gravity would be 0.75 * ((275 m)^3 - (222 m)^3) = 7,391,870 m^3.</p>
<p>This is sensitively dependent on the value of <span class="math-container">$\epsilon$</span>. The above calculation is for <span class="math-container">$\epsilon = 1.8$</span>. If instead <span class="math-container">$\epsilon = 1.6$</span> or lower, then no ejecta will reach escape velocity according to Housen's formula. Though apparently from (Housen 2011) 1.8 is about as low as <span class="math-container">$\epsilon$</span> tends to get (<span class="math-container">$\epsilon$</span> in Housen 2011 is replaced by <span class="math-container">$1/\mu$</span>, and the highest <span class="math-container">$\mu$</span> seems to be around 0.55 from looking at the various graphs).</p>
<p>Here is some Python code to calculate this.</p>
<pre><code>#!/usr/bin/python3
# ----08-23-2024 21:14:34----
import math

# units = m, kg, s
L = 370 # impactor length, m
rho_i = 3000 # impactor density, kg/m^3
v_i = 12600 # impactor speed, m/s
m = 27e9 # impactor mass
# m = 4/3 * math.pi * math.pow(0.5 * L, 3) * rho_i

Vej = 2380 # moon escape velocity, m/s
g = 1.625 # m/s^2
rho_t = 2500 # target density, kg/m^3

epsilon = 1.8
C_D = 1.6
beta = 0.22

pi_2 = 1.61 * g * L * math.pow(v_i, -2)
pi_D = C_D * math.pow(pi_2, -beta)
D = pi_D * math.pow(rho_t / m, -0.33333)
R = D * 0.5
r = R * math.pow(Vej * (epsilon + 1) / (2 * math.sqrt(R * g)), -1/epsilon)

validRadius = 1.2 * L * 0.5 # minimum radius at which the formula is valid

if r &lt; validRadius:
    print(&quot;No ejecta will reach escape velocity.&quot;)
else:
    V = 0.75 * (r**3 - validRadius**3)
    print(&quot;Material between&quot;, validRadius, &quot;m and&quot;, r, &quot;m radius will reach escape velocity.&quot;)
    print(&quot;This is&quot;, V, &quot;m^3&quot;)
</code></pre>

CC BY-SA 4.0
2024-08-23T20:21:38.797Z
2024-08-26T08:00:28.487Z
2024-08-26T08:00:28.487Z
causative
<p>There are some invariances that actually lend themselves well to estimating an answer to this question.  From the <a href="https://www.researchgate.net/publication/226043470_Impact_Cratering_Theory_and_Modeling_for_the_Deep_Impact_Mission_From_Mission_Planning_to_Data_Analysis" rel="nofollow noreferrer">Deep Impact mission white paper</a>:</p>
<blockquote>
<p>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.</p>
</blockquote>
<p><strong>The same paper gives us a set of three equations to estimate crater diameter.</strong> These equations are the result of experimental research combined with modeling the <a href="https://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations" rel="nofollow noreferrer">Navier Stokes</a> differential equations governing the conservation of energy and momentum in materials. The first equation is:
<span class="math-container">$$\pi_2=\frac{1.61gL}{v_i^2}$$</span>
Where <span class="math-container">$\pi_2$</span> is the inverse of the <a href="https://en.wikipedia.org/wiki/Froude_number" rel="nofollow noreferrer">Froude</a> number, <span class="math-container">$g$</span> is the local surface gravity, <span class="math-container">$L$</span> is the projectile diameter, and <span class="math-container">$v_i$</span> is the impact velocity. We will use an impact velocity of 12.6 kms taken from this <a href="https://iopscience.iop.org/article/10.1088/1757-899X/468/1/012023/pdf" rel="nofollow noreferrer">paper</a>. The second equation is:
<span class="math-container">$$\pi_D=C_p\pi_2^{-\beta}$$</span>
Where <span class="math-container">$\pi_D$</span> is a dimensionless measure of crater diameter, and <span class="math-container">$C_p$</span> and <span class="math-container">$\beta$</span> are material constants. The third equation is:
<span class="math-container">$$\pi_D=D\left(\frac{\rho_t}{m} \right)^{1/3}$$</span>
Where <span class="math-container">$m$</span> is the mass of the asteroid and <span class="math-container">$\rho_t$</span> 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 <a href="https://spudislunarresources.com/Bibliography/p/33.pdf" rel="nofollow noreferrer">paper</a>.  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 <span class="math-container">$D$</span> and then plugging in all the values of our variables gives us a crater radius <span class="math-container">$R$</span> of approximately 2.5 km.</p>
<p>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:</p>
<p><a href="https://i.sstatic.net/Cj8ZO.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Cj8ZO.png" alt="enter image description here" /></a></p>
<p>The equation governing the velocity of the ejecta <span class="math-container">$V_{ej}$</span> is therefore partly a function of <span class="math-container">$r$</span>, the distance from the center of the crater.
<span class="math-container">$$V_{ej} = \frac{2\sqrt{Rg}}{1+\epsilon }(\frac{r}{R})^{-\epsilon}$$</span>
Where <span class="math-container">$\epsilon$</span> is a material coefficient of 1.8.</p>
<p>We can solve the above equation for <span class="math-container">$r$</span> and solve using <span class="math-container">$V_{ej}$</span> as 2.38 km/s, the lunar escape velocity.  This yields a value of 275m for <span class="math-container">$r$</span>.  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.</p>
<p>The depth of the crater can be estimated as <span class="math-container">$d=L\sqrt{\rho_p/\rho_t}$</span> where <span class="math-container">$\rho_p$</span> 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 <span class="math-container">$L$</span>, or 370m.</p>
<p>If we estimate the volume of the crater as a cylinder, then the total volume of lunar ejecta will be <span class="math-container">$\pi*R^2d$</span> is about 7.21e9 cubic meters, and the lunar ejecta to reach escape velocity will be <span class="math-container">$\pi*r^2d$</span> 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!</p>
<p><strong>Notes:</strong></p>
<ol>
<li>The ejecta plume is in an inverted cone, so even given ideal geometry, most won't make it to earth.</li>
<li>The ejecta that are just under escape velocity will obtain an orbit that leads them to crashing back into the lunar surface.</li>
<li>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.</li>
<li>My codes is available on request.</li>
</ol>

CC BY-SA 4.0
2024-08-23T17:50:09.007Z
As r approaches 0, the formula given for Vej goes to infinity. $$V_{ej} = \frac{2\sqrt{Rg}}{1+\epsilon }(\frac{r}{R})^{-\epsilon}$$ This seems like a serious flaw as it also means that infinite energy will be ejected from the epicenter.
causative
CC BY-SA 4.0
2024-08-23T18:25:34.547Z
@causative I don't think r can be smaller than the radius of the impactor.
Connor Garcia
CC BY-SA 4.0
2024-08-23T18:26:12.303Z
@causative Yes, you found an error and I approved the edit. I will go back and redo my calculations to see if I get the same results.
Connor Garcia
CC BY-SA 4.0
2024-08-23T18:38:38.367Z
I think you're right about r being larger than the radius of the impactor. From https://www.sciencedirect.com/science/article/abs/pii/S0019103510003623 it's explained "The point-source assumption is appropriate when the phenomena of interest occur at ranges greater than the impactor radius". An issue is that this formula for $V_{ej}$ is for *gravity-dominated* ejecta, tending to be towards the outer edge of the crater: "Housen et al. (1983) noted that gravity should be important only for the furthest launch positions, which are near the crater edge".
causative
CC BY-SA 4.0
2024-08-23T18:42:28.887Z
I think that if your particles are approaching escape velocity, then they are not gravity-dominated and the formula doesn't apply. (Wouldn't gravity-dominated mean that most of the energy is spent just raising them out of the crater against gravity?)
causative
CC BY-SA 4.0
2024-08-23T20:18:21.857Z
Housen indicates in the paper I linked that his formula is invalid for r < n_1 a, with n_1 about 1.2 and a being the radius of the impactor. At those small radii, *no* ejecta is produced; the regolith is driven down underneath the impactor and does not come up again.
causative
CC BY-SA 4.0
2024-08-23T20:57:15.597Z
I think you are right, for Apophis a = 185m so for r > 222m the formula should be valid. So between 222m and 275m the ejecta will reach escape velocity.
causative
CC BY-SA 4.0
2024-08-23T21:28:29.720Z
That's going to affect your volume estimate though, more like 3e7 m^3. But actually, material is not ejected in concentric cylinders. r is not the origin position of the ejecta; r is the radius of the *crater rim* at the time that the particle is ejected. The crater is carved out more like concentric hemispheres than concentric cylinders, initially making a small crater that is excavated to larger and larger ones as the impact continues. Houser gives a scaling law M(r) = 0.75 rho r^3 where rho is density of material and M(r) is cumulative mass of material ejected by radius r.
causative
CC BY-SA 4.0
2024-08-23T21:33:51.280Z
So that suggests only around 7e6 m^3 would reach escape velocity.
causative
CC BY-SA 4.0
2020-11-30T05:28:31.873Z
2024-08-23T18:17:47.617Z
2024-08-23T18:17:47.617Z
Connor Garcia
<p>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.</p>

<p>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.</p>

<p>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.  </p>

<p>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.</p>

<p>Say, for example, <a href="https://en.wikipedia.org/wiki/99942_Apophis" rel="noreferrer">a fairly large impactor like 99942 Apophis</a>, about 27 billion kg, 370 meters in diameter.  </p>

CC BY-SA 4.0
2019-02-28T20:24:29.187Z
depends rather on where the impact occurs relative to the earth-facing side's center.
Carl Witthoft
CC BY-SA 4.0
2019-02-28T20:36:13.630Z
@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
<meteor><impact>
What percentage of a lunar meteor strike is blown back into space?
userLTK
# What percentage of a lunar meteor strike is blown back into space?

(Asked by: username_0 on 2019-02-28)

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](https://en.wikipedia.org/wiki/99942_Apophis), about 27 billion kg, 370 meters in diameter.

**Tags:** meteor, impact

> **Comments:**
> **username_1**: depends rather on where the impact occurs relative to the earth-facing side's center.  
> **username_0**: @username_1  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.  

---

## 2 answers

### Answer by username_3 on 2020-11-30. Score: 9 | Accepted answer

There are some invariances that actually lend themselves well to estimating an answer to this question. From the [Deep Impact mission white paper](https://www.researchgate.net/publication/226043470_Impact_Cratering_Theory_and_Modeling_for_the_Deep_Impact_Mission_From_Mission_Planning_to_Data_Analysis):

> 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](https://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations) 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](https://en.wikipedia.org/wiki/Froude_number) 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](https://iopscience.iop.org/article/10.1088/1757-899X/468/1/012023/pdf). 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:
$$\pi\_D=D\left(\frac{\rho\_t}{m} \right)^{1/3}$$
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](https://spudislunarresources.com/Bibliography/p/33.pdf). 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:

[![enter image description here](https://i.sstatic.net/Cj8ZO.png)](https://i.sstatic.net/Cj8ZO.png)

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.

> **Comments:**
> **username_2**: As r approaches 0, the formula given for Vej goes to infinity. $$V_{ej} = \frac{2\sqrt{Rg}}{1+\epsilon }(\frac{r}{R})^{-\epsilon}$$ This seems like a serious flaw as it also means that infinite energy will be ejected from the epicenter.  
> **username_3**: @username_2 I don't think r can be smaller than the radius of the impactor.  
> **username_3**: @username_2 Yes, you found an error and I approved the edit. I will go back and redo my calculations to see if I get the same results.  
> **username_2**: I think you're right about r being larger than the radius of the impactor. From https://www.sciencedirect.com/science/article/abs/pii/S0019103510003623 it's explained "The point-source assumption is appropriate when the phenomena of interest occur at ranges greater than the impactor radius". An issue is that this formula for $V_{ej}$ is for *gravity-dominated* ejecta, tending to be towards the outer edge of the crater: "Housen et al. (1983) noted that gravity should be important only for the furthest launch positions, which are near the crater edge".  
> **username_2**: I think that if your particles are approaching escape velocity, then they are not gravity-dominated and the formula doesn't apply. (Wouldn't gravity-dominated mean that most of the energy is spent just raising them out of the crater against gravity?)  
> **username_2**: Housen indicates in the paper I linked that his formula is invalid for r < n_1 a, with n_1 about 1.2 and a being the radius of the impactor. At those small radii, *no* ejecta is produced; the regolith is driven down underneath the impactor and does not come up again.  
> **username_2**: I think you are right, for Apophis a = 185m so for r > 222m the formula should be valid. So between 222m and 275m the ejecta will reach escape velocity.  
> **username_2**: That's going to affect your volume estimate though, more like 3e7 m^3. But actually, material is not ejected in concentric cylinders. r is not the origin position of the ejecta; r is the radius of the *crater rim* at the time that the particle is ejected. The crater is carved out more like concentric hemispheres than concentric cylinders, initially making a small crater that is excavated to larger and larger ones as the impact continues. Houser gives a scaling law M(r) = 0.75 rho r^3 where rho is density of material and M(r) is cumulative mass of material ejected by radius r.  
> **username_2**: So that suggests only around 7e6 m^3 would reach escape velocity.  

---

### Answer by username_2 on 2024-08-23. Score: 2

### Possibly not much

From [Vickery (1986)](https://ntrs.nasa.gov/citations/19870027368):

> The characteristics of three primary extraterrestrial craters and the associated craters were examined to generate a size-velocity distribution for large ejecta fragments. The lunar craters Copernicus and Aristillus and the Martian crater Dv on Olympus Mons were used. Attention was focused on the radial distances between the primary and secondary crater centers and the diameters of the secondaries. The primary craters selected are all relatively young, which avoided contamination of the data from secondaries from other primaries. Attempts were made to account for the speed of the hypervelocity impacts and the elemental compositions of the impactors. An apparent velocity cutoff of about 1 km/sec was observed for the secondaries, **which implies that no meteoroid impacts can accelerate ejecta to escape velocities from the moon or Mars.**

This was based on direct observation of Mars and the Moon. However, of course, the asteroids observed were smaller than Apophis.

Although, this applied only to detectable secondary craters formed by large ejecta fragments. There might have been faster ejecta that was not detected because it wasn't large enough to form secondary craters.

---

But see also [Hartmann (1985)](https://www.sciencedirect.com/science/article/abs/pii/0019103585900211) (you can look for it on sci-hub) which summarizes a variety of experimental impacts at speeds up to a couple km/s. The result seems to be that in the first few microseconds after a hypervelocity impact, there is a thin jet of ejecta that moves at a substantial fraction (<1) of the hypervelocity impactor speed. The vast majority of ejecta occurs later in the impact and moves much slower, down to a few m/s. So it is possible that the initial thin jet could reach escape velocity, but there isn't much of it by mass.

---

It's questionable whether the formulas used in username_3's answer are applicable to very high-speed ejecta. The formula for $V\_{ej}$:

$$V\_{ej} = \frac{2\sqrt{Rg}}{1+\epsilon }(\frac{r}{R})^{-\epsilon}$$

has a discontinuity for small $r$. $V\_{ej}$ tends to infinity as $r$ tends to 0. So for small $r$ (and high $V\_{ej}$) this is clearly not valid. The author cited for this formula, Kevin Housen, published [Housen and Holsapple (2011)](https://sciencedirect.com/science/article/abs/pii/S0019103510003623) (check sci-hub) explaining the limitations of the formula. It does not give valid results for $r < n\_1 \frac{L}{2}$ ($n\_1 \approx 1.2$). At these small radii, the regolith will be driven down underneath the impactor and will not emerge at all - the ejecta velocity is 0 (with the exception of some small amount of mass that may be jetted at a high speed, which is dependent on the geometry of the impactor, and which Housen does not treat).

Plugging in the same values, I get the same answer username_3 got, 275m for the Apophis-Moon impact. This is greater than 1.2 \* 370 \* 0.5 = 222 m, which indicates ejecta between a radius of 222m and 275m would reach escape velocity for the Moon.

Housen gives a formula for the mass of ejecta under a given radius: $M(x) = k \rho x^3$ where x is the radius, $\rho$ is the material density, and $k$ is a constant, between 0.4 and 0.75 depending on soil type, 0.75 for dense soil. So the volume ejected by this formula at a speed sufficient to escape the Moon's gravity would be 0.75 \* ((275 m)^3 - (222 m)^3) = 7,391,870 m^3.

This is sensitively dependent on the value of $\epsilon$. The above calculation is for $\epsilon = 1.8$. If instead $\epsilon = 1.6$ or lower, then no ejecta will reach escape velocity according to Housen's formula. Though apparently from (Housen 2011) 1.8 is about as low as $\epsilon$ tends to get ($\epsilon$ in Housen 2011 is replaced by $1/\mu$, and the highest $\mu$ seems to be around 0.55 from looking at the various graphs).

Here is some Python code to calculate this.

```
#!/usr/bin/python3
# ----08-23-2024 21:14:34----
import math

# units = m, kg, s
L = 370 # impactor length, m
rho_i = 3000 # impactor density, kg/m^3
v_i = 12600 # impactor speed, m/s
m = 27e9 # impactor mass
# m = 4/3 * math.pi * math.pow(0.5 * L, 3) * rho_i

Vej = 2380 # moon escape velocity, m/s
g = 1.625 # m/s^2
rho_t = 2500 # target density, kg/m^3

epsilon = 1.8
C_D = 1.6
beta = 0.22

pi_2 = 1.61 * g * L * math.pow(v_i, -2)
pi_D = C_D * math.pow(pi_2, -beta)
D = pi_D * math.pow(rho_t / m, -0.33333)
R = D * 0.5
r = R * math.pow(Vej * (epsilon + 1) / (2 * math.sqrt(R * g)), -1/epsilon)

validRadius = 1.2 * L * 0.5 # minimum radius at which the formula is valid

if r < validRadius:
    print("No ejecta will reach escape velocity.")
else:
    V = 0.75 * (r**3 - validRadius**3)
    print("Material between", validRadius, "m and", r, "m radius will reach escape velocity.")
    print("This is", V, "m^3")

```


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