A la this image I made.

The "how" (nuclear pulse drive "adjusting" Eros's orbit at its aphelion) and "why" (uber-sized Rods from God) isn't particularly relevant — all that's relevant is that 433 Eros's perihelion has been changed from 1.1334 AU to 1 AU, and its orbit just happens to intersect with the Moon at that perihelion. Oh, and none of this is to scale, but the angles in it (especially the first image) are intended to be accurate.

Anyway, in order to calculate the presumably-unfortunate effects of this on the Earth, I've done some math (see the bottom of this post). However, as I can't find hard data which would let me make fairly accurate guesses, I'm currently making the following assumptions:

  • 10% of the energy released by the impact is converted into ejecta velocity; the remaining 90% is absorbed by the Moon.
  • 10% of the ejecta energy is in ejecta that escapes the Moon at 2,380 m/s or more; the remaining 90% falls back into the Moon. The Moon's escape velocity of 2,380 m/s provides a benchmark for determining the mass escaping the Moon; 10% of the energy dumped into flinging rocks out of the Moon will fling them fast enough to escape the Moon's gravity. All the rocks flung by that 1% of the impact energy will be going ≥ 2,380 m/s, letting me determine the maximum mass escaping the Moon.
  • 25% of the ejecta escaping the Moon falls into the Earth over time, and 25% of it will fall into the Moon over time, with the remainder coming back under the gravitational influence of the Moon. This relies on the next assumption...
  • ...that Eros hitting the Moon tail-on at a right angle relative to its surface means none of its fairly impressive velocity will carry over into propelling ejecta, meaning none of the ejecta the impact produces will be going quickly enough to escape the Moon/Earth system entirely. Sure, there's Newton's Third Law, but much of the "equal and opposite" reaction to Eros's impact/disintegration will go into crushing/compressing lunar rock, not throwing it away from the Moon. Also, crushed rock can be crushed further or pushed out of the way to crush more rock, whereas rock that's been thrown away can't have any more energy dumped into it.
  • Of the 50% of the ejecta that escapes the Moon, ends up under its gravitational influence again, but does not directly fall into it, the Moon will act as a shepard moon towards it, slingshot half of it (the ejecta orbiting slightly inside the Moon's orbital path) out of the Earth/Moon system, and slingshot the other half of it (the ejecta orbiting slightly outside the Moon's orbital path) into the Earth.
  • 10% of the ejecta pulled into the Earth actually makes it through the atmosphere to impact; the rest is too small to make it to the surface and disintegrates in the atmosphere.

Basically, if we look at the image below (which I got from the 9th page of this), assume the asteroid represents the Moon, and assume the arrow represents the trajectory of a piece of ejecta...

  • 90% of the impact's energy is in that little explosion and doesn't go anywhere.

  • 9% (90% of 10%) of the impact's energy propels Class I ejecta.

  • 0.25% (25% of 10% of 10%) of the impact's energy propels Class II ejecta.

  • 0.25% (25% of 10% of 10%) of the impact's energy propels Class IV ejecta that gets slingshotted out of the Earth/Moon system.

  • 0.25% (25% of 10% of 10%) of the impact's energy propels Class IV ejecta that gets slingshotted into the Earth.

  • 0.25% (25% of 10% of 10%) of the impact's energy propels Class V ejecta that reaches the Earth directly.

  • In regards to the Moon, Class III doesn't exist; any ejecta that ends up in Class III will, over time, get pulled out of orbit and crashed into the Moon by lunar mascons, essentially putting it in Class II.

Or, for another visual, this image:

enter image description here

  • This is a "top-down"/"birds-eye" view of the Earth/Moon system; the Moon's orbit takes it towards the bottom of the page, and the entire thing is moving to the left as it orbits the Sun.

  • If ejecta is in the blue section, it's orbiting faster and at a lower altitude than the Moon, so it'll either fall into the Earth or "outrun" the Moon over time, eventually catch up to it again, and then slingshotted around the "bottom" of the Moon and out of the Earth/Moon system.

  • If ejecta is in the red section, it's orbiting slower and at a greater altitude than the Moon, so the Moon will "outrun it" over time, meaning the ejecta will either fall into the Earth or have the Moon eventually catch up to it again, slingshotting it around the "top" of the Moon and into the Earth.

  • If ejecta is in the green section, it's not going fast enough to escape the Moon's gravity, so it'll fall back into the Moon.

Yet another image, from the same perspective:

enter image description here

The Moon acts as a shepard moon; any particle orbiting to the left of or behind the Moon in the same direction gets thrown into the purple outer ring, whereas any particle orbiting to the right of or in front of the Moon in the same direction gets thrown into the orange inner ring, and likely within the Earth's Roche limit for particles of that size, meaning tidal forces will eventually tug the particle into the Earth.

Why am I asking this question?

Simple: I need to make better assumptions. Remember all those percentages? I need to know whether, for instance, assuming the Moon will absorb 90% of the impact energy is accurate or not, or whether or not the Earth will actually be hit with half of what escapes from the Moon.

To reiterate — I'm currently assuming that:

  • 10% of the energy released by the impact is converted into ejecta velocity

  • 10% of the ejecta energy is in ejecta that escapes the Moon

  • 50% of the ejecta escaping the Moon hits the Earth

  • 90% of the ejecta hitting the Earth burns up and becomes airborne particulates; the remaining 10% physically hits the Earth's surface

Are these accurate assumptions to make in regards to modeling an impact between the Moon and 433 Eros?

Anyway, here's that math I mentioned:

6.687 quadrillion kg 433 Eros impact w/Moon @ 1 AU from Sun, prograde Moon velocity
  • 34,148 m/s velocity relative to Sun - 1,022 m/s Moon orbital velocity = 33,126 m/s collision.
  • (((33,126 m/s)^2) x 6.687 x (10^15) kg)/2 = 3.66892913 x (10^24) J impact energy, 3.66892913 x (10^23) J net ejecta energy, and 3.66892913 x (10^22) J of energy in the ejecta escaping the Moon.
  • ((((3.66892913 x (10^22) J) x 2) / ((2380 m/s)^2)) = 1.29543434 x (10^16) kg total ejecta escaping the Moon, of which...
  • ...3.2385859 x (10^15) kg falls back into the Moon, 3.2385859 x (10^15) kg is flung out of the Earth/Moon system by the Moon, 3.2385859 x (10^15) kg is flung into the Earth by the Moon, and 3.2385859 x (10^15) kg falls directly to the Earth by itself.
  • Of the 6.4771718 x (10^15) kg of ejecta hitting the Earth, 5.8294546 x (10^15) kg burns up into atmospheric dust and 6.4771718 x (10^14) kg physically impacts the Earth's surface.

It would be a lot worse if Eros hit the Moon head-on as the Earth was coming around the Sun at a right angle to its trajectory. Conversely, it'd be a lot less dangerous if it came in in the same direction Earth was orbiting in. As a compromise, I had Eros broadside the Earth/Moon system halfway between those two extremes.

Also, note that these calculations don't include the velocity incurred by Eros accelerating as the Moon's gravity pulls on it, nor the velocity incurred by the nuclear pulse drive attached to it firing off its last H-bombs to add some extra ΔV right before impact for maximum energy release. They're also fairly inaccurate in that they try to divide the types of ejecta trajectories into a few neat little categories rather than the mess that n≥3 orbital physics usually are, as well as because they portray 2,380 m/s as the velocity of all the ejecta escaping the Moon (some would go that speed, much would go faster, meaning that 1.29543434 x (10^16) kg is probably too much). However, these are the bare minimum; I'm basically going for a Fermi estimate here.

  • 3
    $\begingroup$ There might be some heuristics out there you can use, but this is the sort of problem that you see answered in a master's thesis and as such it seems unlikely you'll get a useful, succinct answer. If you're doing the sort of postgrad research that requires you to accurately answer this sort of question, you're probably asking on the wrong site. If you're not doing actual academic work here, then you don't need a super accurate answer and handwaving is fine... if nothing else, practically no-one will be able to tell you that you're wrong. $\endgroup$ Commented Jan 6, 2023 at 8:10
  • $\begingroup$ @StarfishPrime I agree; in retrospect, I think my question's problem is it's too broad, not that it can't be answered well. After posting, I found how to find the net energy (~2/15x initial kinetic energy; ~4.8919055E23 J in this case) and mass (~4.75x impactor mass; ~3.176E16 kg in this case) of ejecta resulting from a ~32.5 km/s achondrite/achondrite impact with lunar gravity. And if Eros hits the Moon at a certain angle relative to the Earth, all escaped ejecta falls to the Earth, axing that variable. My problem now is finding the quantity of ejecta escaping the Moon in the first place. $\endgroup$
    Commented Jan 6, 2023 at 23:21
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    $\begingroup$ An important note. You want an estimate. At the same time, you put the collision velocity as 33,126 m/s. 5 digits of precision is too much, 2 would be enough. But this is your choice, not an error. The error is made here: “(33,126 m/s)² × 6.687×10¹⁵ kg / 2 = 3.66892913×10²⁴ J” The lowest precision of all numbers that went into the formula had 4 digits; of the 9 digits in the result, 5 are simply random! 4 in, 4 out! The correctly rounded answer is 3.669×10²⁴J. Rule: add 1 to the last digit if the first chopped one was 5 or greater (it was a 9, thus 3.668+0.001). $\endgroup$ Commented Jan 7, 2023 at 0:07
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    $\begingroup$ “I'm currently making the following assumptions:” — a source they are based upon would be helpful. 10% in ejecta and 90% absorbed does not sound plausible. The “absorbed” part is what melts and vaporises rock, and the Moon doesn't have a lot of gravity to contain ejected material. Don't make a mistake: 10²⁴J is a huge amount of energy, not a little pop. Very active volcanism at the directly opposite point is evident in large impacts, ejecting even more stuff over a short (~10⁴…10⁶ years) geological time. I'd look for papers on similar-sized past Lunar impacts—she has a few quite painful scars. $\endgroup$ Commented Jan 7, 2023 at 0:26
  • 1
    $\begingroup$ Sure, the process is complex, and multiple factors are at play. A lot of ejecta is formed when the wave in the molten pool reflects and rebounds—just like a small jet shoots up where a drop of water falls into a water-filled Petri dish near the center. This part is little affected by the impact angle, counterintuitively (but most craters are neatly round!). A lot depends on the composition of the impacted body and the impactor. I'm sure there are formation models for particular large Lunar craters in the literature—large impact study is an enough-funded area, relevant to the Earth safety. $\endgroup$ Commented Jan 7, 2023 at 0:57

1 Answer 1


Well, let me attempt to give a few clues to what the right answer might be. However, before let me stress;

Just do what is good for the story. In this case you have a situation where any outcome is reasonable. From a World Ending event to Virtually no debris. The entire range is within reason.

It depends

On a lot of factors. The two biggest of which are the angle of incidence and impact velocity. These two factors will contribute the most to your question.

First, the Velocity. Using Universe Sandbox 2 (So a Newtonian Simulator that will have inaccuracies), the 1AU Velocity of Eros should be 33800 m/s. Now, Earth's velocity is 29600 m/s and Lunar orbits between 1080 and 970 m/s. Now the big question is what the relative velocity will be. For the biggest effect we can assume that Eros slams into Lunar, so Lunar moves towards Eros at 1080m/s. This means Lunar and Earth's velocity add up (since by definition Luna is at its lowest point in the orbit. So, the Impact velocity of Eros is 33800m/s - (29600m/s - 1080m/s) = 5280m/s. Or 4200 if the orbit is the other way around, but again we assume the most energetic impact possible.

Now, the Angle of incidence will have to be equal to the tilt of Eros´s orbit. Which is 10.8 Degrees. Which for us translates to 90-10.8 = 79.2 Degrees. When you run the numbers of such an impact at the equator it looks kinda like this;

enter image description here

All of the red lines show you where the debris are going. As you can see a grand total of 0 come back to the Moon, which is because this image assumes an equal velocity for all of them, which is of course not true.

But that is not to important, for now anyways, the important part is that all of these Debris are moving at about 4000-5000 m/s. Which means all of them are faster than escape velocity at Lunar orbit. Which intern means the supermajority of ejecta is bound to leave the Earth Lunar System. Now, these Debris are move Retrograde relative to the sun. So they are moving at about 25km/s. So the debris are all in an orbit around sun and will come back about 8.55 months later.

So, my conclusion would be that Initially not a lot of ejecta would hit Earth. The impact is extremely violent which in this case kinda works in favor for us. At least initially. The issue is that all of this ejecta is still there. And every 8.55 months it will reach Earths orbit. So we can expect a Shower of this ejecta every few years. How many Years ? My sim´s say the first intersect happens about 10 years later. However in reality the Ejecta will form a ring on this orbit so with each year passing impacts will become more and more frequent.

In saying that, other gravitational bodies still exist so it is likely a lot of the large pieces would get thrown out.


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