# The moon will fall to earth in 100 years, how fast is the asteroid that knocked it down?

We have at most 100 years to abandon earth. The 10-km asteroid which killed the dinosaurs has hit the moon head-on instead. As a result, the moon has slowed down and it’s orbit will decay by 400,000 km after 100 years (it will fall to earth).

# How fast was the asteroid moving relative to the moon’s forward motion?

Assume a metallic asteroid of 80% iron, 10% nickel, 10% magnesium.

More specifically: the obit becomes elliptical brushing within 250km of earth, causing drag which will slow it to a collision in 100 years. Assume a circular initial orbit.

The image below represents what the orbit is doing:

The calculation extends this chart out such that the surface of the moon and earth collide in 100 years $$\pm$$ 5 years.

Bonus if an orbital plot like the Tiangong-1 is in your answer

• Orbital decay does not work like that. It'll either crash in under a single orbital period, or it will repeatedly brush the Earth's atmosphere (starting within a single orbital period) causing immediate apocalyptic destruction. And will probably crash in under 100 years anyway. You can't hard-science this one. – Starfish Prime Oct 27 '19 at 13:09
• Agreed. Also,the Moon will just take a dinosaur killer impact, basically get a new Tycho-like crater, and continue in its orbit. It's been there for the past four billion years, it's done that. Life on Earth cared about the dinosaur-killer. Earth didn't. – notovny Oct 27 '19 at 13:33
• An 10 km asteroid moving fast enough to cancel that much of the Moon's momentum instantly would probably liquify the Moon, if not totally ionize it. – BMF Oct 27 '19 at 14:05
• @PatriciaShanahan my answer covers that idea. 1) you can't effect such an orbital change via an asteroid impact (you'll smash the moon) and 2) you'll exceed the Roche limit of the moon and smash it up as it approaches earth. You can't do what the OP wants, the way the OP wants it. – Starfish Prime Oct 27 '19 at 15:19
• If the key requirement is simply that the Moon gets hit with something that causes a catastrophe in 100 years, how about a glancing impact that knocks a big chunk of lunar material into an Earth-crossing elliptical orbit around the Sun, and scientists work the orbit forwards in time and discover it will impact the Earth in 100 years? – Dan Hanson Oct 28 '19 at 21:20

Surprisingly, I find myself going against the obvious answer.

You talked about decaying orbits in your question, everybody looked at that and quite correctly said no. However, if we remove the "decaying" part it becomes possible. Note that the catastrophic splash from the impact that Starfish Prime mentioned remains--I doubt there would be anyone left on Earth by the time it comes down. Likewise, my answer needs a relativistic impactor.

Instead of a decaying orbit, the impact comes from behind. The moon's apoapsis is raised beyond the edge of Earth's Hill Sphere, it begins to wander--it's basically in Earth's orbit about the sun, but almost perfectly sharing Earth's orbital period. That's a recipe for a future encounter and the moon's gravity will tend to pull it in. Note that exactly what happens will be extremely sensitive to the initial conditions.

• This is exactly what I was looking for in the meta question. An event with no forewarning, ever-present omen of destruction (moon stays visible - no scientists lecturing about calculations), no escape, 100 years to annihilation. Did this run on an orbit simulation? – Vogon Poet Oct 28 '19 at 3:20
• @VogonPoet I didn't run a simulation and that would be impractical--there's no way to calculate an orbit that comes back in 100 years, it would take simulating a lot of trajectories to find one that did it. I would be surprised if none exist, though. – Loren Pechtel Oct 28 '19 at 5:17
• I dunno about "obvious answer"... more requested answer ;-) The OP specifically stated that the moon had been struck by an asteroid, and its orbit had slowed and decayed. Still, +1 for the neat idea. Probably nigh-on impossible to work out the initial conditions though... most gravity simulations just involve things being flung away. Its really depressingly hard to get a good earth-shattering kaboom :-( – Starfish Prime Oct 28 '19 at 11:07
• The Hill Sphere radius for Earth is approximately 1.5 million kilometres. The moon's orbit is currently 384,000 kilometres. Assuming ideal circumstances so you hit the moon at exactly the right angle so you'd get apogee out to the Hill Sphere, you're talking energy at the same sorts of magnitudes as to to send the moon into the Earth, ie totally unrealistic. – Keith Morrison Oct 28 '19 at 17:22
• @KeithMorrison I fully agree on the energy level--note that I said it still needs a relativistic impact. I was just looking for some way for it to come down in 100 years. – Loren Pechtel Oct 28 '19 at 22:51

TL;DR: What you want is impossible for multiple reasons.

Lets be honest: what you want is clearly impossible. In order to have a slowly decaying orbit, you need to have either a repeatedly applied or continuously applied force, such as a rocket engine, atmospheric drag, solar wind drag, multiple asteroid impacts... whatever. You cannot cause the moon's orbit to gradually decay via a single asteroid-sized impact alone.

You suggested atmospheric drag, but this requires drastically lowering the perigee of the moon's orbit. Doing so with a single asteroid impact is almost certainly not going to work, because moon-sized masses of rock and relativistic kill vehicles are not snooker balls. You will have to handwave in some other force.

You might be able to do this with gravity via a close encounter with a massive third body (eg. some kind of extrasolar rogue planet passing through the inner solar system), but most such interactions will result in the moon being ejected into solar orbit. You are welcome to play with a gravity simulator to find a suitable solution to this particular issue yourself, but you are unlikely to get a solution to the three body problem on worldbuilding.SE.

Lets take your original request at face value, though... lower the perigee of the moon's orbit such that its surface comes within 250km of the earth's surface.

Lets break out the vis-viva equation, which is useful when thinking about the energy needed to change orbits:

$$v^2 = GM\left( \frac{2}{r} - \frac{1}{a} \right)$$

Where $$v$$ is the relative orbital speed, $$r$$ is the distance between the orbiting bodies and $$a$$ is the semi-major axis of the orbit.

We want to change the moon's orbit from an approximately circular one of radius 3.84x108m to an elliptical one with a perigee of 8.35x106m (which leaves a ~250km gap between their surfaces, more or less). Such an orbit will have a semi major axis of ~1.96x108. You can throw these figures into the vis-viva equation, to get you the change in velocty the moon needs: ~817m/s.

You can throw this into the old $$E=\frac{1}{2}mv^2$$ with the mass of the moon to get a required energy of 2.457x1027J, or about half an exatonne, TNT equivalent. That's a substantial bang, and it isn't immediately obvious what that would do to the moon. The gravitational binding energy of the moon is about 1.2x1029, so the energy released isn't enough to reduce it to a vast cloud of rubble smeared out into a ring around earth, but it also isn't obvious that it will nicely knock the moon onto a new trajectory.

There's an approximation for crater size, variously used for asteroids hitting planets or hypervelocity objects hitting whipple shields and the like. Roughly speaking, it says the volume of the crater caused by an impact is $$V_c \approx E_p/S_c$$, where $$E_p$$ is the kinetic energy of the projectile and $$S_c$$ is the "cratering strength" of the target, handwaved as three times its tensile strength. Moon rock has a tensile strength of maybe 180MPa, so the cratering strength ends up as 5.4x108J/m3, giving an impact crater volume of something like 4.55x1018 cubic metres. If this were a hemispherical crater on a vast flat surface, it'd have a depth of 1.3x106 metres. The moon has a radius of about 1.7x106 metres and so a volume of about 2.1819m3. This means about a fifth of the moon will excavated by the impact, and much of the energy of the impact will go into the kinetic energy and heat of all that excavated material. Quite a lot of this material will fly towards earth, at considerable speed, probably sterilising the hemisphere facing the moon. The other hemisphere will suffer the after effects in fairly short order. Quite a lot of this material will escape into solar orbit (or beyond) but plenty more will end up as a new ring system. Quite a lot of it will probably fall onto earth, be ejected from the earth-moon system or re-accrete with the larger chunks of the moon over the next few million years.

The remaining large chunk(s) of the moon will almost certainly experience an orbital change, but the odds of it being the change that you wanted are pretty slim.

Lets imagine, though, that this impact did indeed inject most of the moon, which remains more or less intact into the desired orbit. The new orbital period of the moon is a little under 10 days, and within half that time the moon will reach its perigee and brush the atmosphere of the earth.

But wait, you've forgotten the Roche limit. Once the moon gets close enough to the earth, gravitational tidal forces will simply tear it apart. The moon is tough, but the earth's gravitational field is quite substantial. The badness will start at about 18000km separation, and get progressively worse. How many orbital passes are needed I'm not sure, but the moon will break up under tidal stresses, and given that its orbit is brushing the atmosphere, generous quantities of the moon will rain down upon the earth, starting with its very first pass.

Your 100 year limit is almost entirely irrelevant; the surface of the earth will be subject to a series of impacts not seen since the Late Heavy Bombardment. I shan't bother looking at what happens to the atmosphere and oceans of earth when the moon comes that close, because it doesn't really matter any more. Rocks fall, everyone dies.

Anyway, the meat of your question: how fast does your asteroid need to be travelling?

Lets break out the relativistic kinetic energy equation: $$E = mc^2\left[ \frac{1}{\sqrt{v^2/c^2} }- 1 \right]$$

I'm not going to bother to work out the mass of your example asteroid, but I'll just use the low-end estimate for the Chixulub impactor, a nice round 1015kg. With a bit of deft re-arranging, you can see that it will need a velocity of about 0.99997c.

• Well this provides a lot of the calculations needed to come to an answer, but I think it inserts an assumption that the moon achieved 250 km orbit in the first period. In doing so, this completely avoids calculating a trajectory which would leave a collision 100 years in the future. I appreciate the work, however the goal is not to simply turn a circular orbit into our lips instantaneously. Achieving 250 km separation at some point, through the various forces at work, is a corollary assumption Which may or may not be wrong, but shouldn’t invalidate the question entirely.. – Vogon Poet Oct 27 '19 at 14:52
• @VogonPoet You cannot have such a trajectory via a single asteroid impact. Orbits do not work that way. What other forces are you imagining will continue to lower the moon's orbit? – Starfish Prime Oct 27 '19 at 14:59
• Ok but I’d rather not completely invalidate the work here by changing the question. And again avoiding a guess (since I don’t have the references at hand to solve the problem myself), the goal is simply making the moon do what Tiangong 1 did, but falling to earth after 100 years rather than 2. If a series of impacts achieves this then that is done simply by an interplanetary asteroid collision which left a stream of éjectorate spraying the moon. This is too complicated, it was hoped that one known asteroid in some way = 100 years left until moon destroys earth. That’s the meta question. – Vogon Poet Oct 27 '19 at 15:49
• @VogonPoet and therein lies a lesson about asking how a goal might be accomplished, rather than deciding on a way to accomplish it and asking others to retrospectively justify it ;-) It wouldn't be unreasonable to ask a new question, to be honest. – Starfish Prime Oct 27 '19 at 15:52
• @BMF yeah, but working out that is yet another monstrously difficult issue, and given how many other issues there are with the scenario it isn't really one worth investigating much more closely. – Starfish Prime Oct 27 '19 at 16:20

Your mechanIsm for making the Moon collide with Earth in 100 years would not work.

LONG ANSWER PART ONE: ANOTHER SOLAR SYSTEM OBJECT COLLIDES WITH EARTH

As other answers have said, you can not hit the Moon with a single impact that will change the orbit of the Moon so that the Moon will crash into Earth in 100 years.

But it is perfectly possible for astronomers in the present time or some future time to discover that an astronomical body is going to impact the Earth at some specified time in their future.

Some present day astronomers are working on projects to discover and catalog as many asteroids and comets with Earth crossing orbits as possible. And they are calculating the orbits of those asteroids and comets as precisely as they can and re observing them to get more data to recalculate their orbits better. The hope is that if some asteroid or comet is calculated to be on a course that will eventually impact on Earth with devastating results, the calculated impact date will be far enough in the future for the asteroid or comet to be diverted in time.

For example, Biela's Comet had an orbit that crossed the orbit of Earth:

The comet appeared as predicted during its 1832 apparition, when it was first recovered by John Herschel on 24 September.1 The orbital elements and ephemeris calculated by Olbers for this return created something of a popular sensation, as they showed that the comet's coma would likely pass through the Earth's orbit during a close approach on October 29. Subsequent predictions, in the media of the time, of the Earth's likely destruction overlooked the fact that the Earth itself would not reach this point until November 30, a month later, as pointed out by François Arago in an article designed to allay public fears.2 Despite this, the fact that Biela's Comet was the only comet known to intersect the Earth's orbit was to make it of particular interest, both to astronomers and the public, during the 19th century.

In the 1820s it was calculated that Biela's Comet would collide with Earth in the year AD 4339.

The Year 4338: Petersburg Letters (Russian: 4338-й год: Петербургские письма) is an 1835 novel by Vladimir Odoevsky. It is a futuristic novel, set in the year 4338, a year before Biela's Comet was to collide with the Earth as computed in the 1820s although the comet burned up later in the nineteenth century. This work was originally conceived as the third part of a trilogy, which was also to have featured depictions of Russia in the time of Peter the Great and in the author's contemporary period, the 1830s. The first part was never written and the second and futuristic parts remained unfinished. Fragments were published in 1835 and 1840, with the fullest version appearing in 1926.

Biela's Comet split in 1845 and the fragments were last seen in 1852.

And on the other hand, in recent decades, a few of the Earth-crossing asteroids that have been discovered have been discovered only after they passed very close (by astronomical standards) to Earth and were moving farther away. Thus it is certainly possible that an asteroid or comet headed for Earth might not be discovered before the impact.

So in a fictional story where an asteroid or comet is discovered to have an orbit that will result in an impact on Earth, the characters might possibly have hours, days, weeks, months, years, decades, centuries, or millennia to prepare for the impact.

A writer can choose an arbitrary length of time for humans to try to divert an incoming asteroid or comet or to evacuate the Earth if the impact can't be averted. Of course it would be better if the writer got someone with expert knowledge to calculate an orbit for the object that makes it reasonable for it to be discovered his chosen length of time before the impact.

And of course the writer should make an effort to chose a size and mass for the incoming object that will make it impossible for any reasonable technology to divert the object in time to prevent an impact.

I believe there was recently an question about how big an incoming object would have to be to make the impact unavoidable by anything humans could do. Here it is:

I note that there are objects classified as "centaurs" orbiting beyond Jupiter, that seem to have characteristics of both asteroids and comets. There orbits are usually rather unstable due to perturbations by the giant planets, so is it is easy to imagine that if some extrasolar dwarf planet passed through the other solar system it might perturb a number of centaurs into orbits that pass into the inner solar system, including Earth-crossing orbits.

And some of the centaurs are very large, tens and even sometimes hundreds of miles in diameter, so a large centaur calculated to impact with Earth would be very difficult to divert.

I note that billions of comets are believed to orbit the Sun in the Oort Cloud far beyond the planets. Occasionally comets are diverted into the inner solar system by the gravity of nearby stars, etc, and any comet that enters the inner solar system for the first time might possibly have an orbit that will make it collie with Earth.

Comet nuclei can have dimensions of less than one kilometer or mile up to several miles. In extreme cases a comet can have a radius of 30 kilometers (19 miles).

So it is certainly possible for a comet from the Oort Cloud on a collision course with Earth to have thousands of times the mass of a normal sized comet and thus be thousands of times more difficult to deflect.

Furthermore, some of the centaurs might have originated in the Oort Cloud, since they have similarities to comets. And the largest known centaur, 10199 Chariko, probably has a diameter of about 250 kilometers, which might be about 50 to 100 times the diameter of a typical size comet, making it have about 125,000 to 1,000,000 times the mass of a typical comet.

If there are Chariko-sized objects in the Oort cloud, and one of them has been perturbed into a collision course with Earth, it would be many thousand times more difficult to divert than a typical sized comet.

Anyway, it is the job of good writer to make the readers understand what it would take to divert an object from a collision course with Earth, and when the object in the story is too massive to be diverted, make the readers understand how hopeless it would be to try diverting it.

LONG ANSWER PART TWO: A ROGUE PLANET COLLIDES WITH EARTH

Remember that it is always possible for a writer to create a larger incoming astronomical body than can be diverted by any human effort. Our Sun and solar system orbits around the center of the galaxy like many other stars do. And just like it is possible for two solar system objects to collide as the both orbit the Sun, it is possible for two different solar systems to collide as they both orbit the center of the galaxy.

Of course astronomers have calculated the courses of all the nearby stars and know how fast the distances between them and the Sun are increasing and decreasing, and they know that none of them is going to collide with the Sun in the next ten million years.

And surveys of the sky in infrared light have detected some very dim brown dwarfs close to the solar system, and it now seems very unlikely that any brown dwarfs close enough to become a danger in the next few million years would still be undiscovered.

For decades, science fiction writers have imaged there might be "rogue planets" that don't orbit any stars but travel alone though interstellar space. In When Worlds Collide (1932,1933) by Philip Wylie and Edwin Balmer, a rogue planet from interstellar space collides with and totally destroys Earth, so that plot has been used in fiction for at least 87 years.

Astronomers have actually detected rogue planets that don't orbit any star orbiting the center of the galaxy by themselves.

Astrophysicist Takahiro Sumi of Osaka University in Japan and colleagues, who form the Microlensing Observations in Astrophysics and the Optical Gravitational Lensing Experiment collaborations, published their study of microlensing in 2011. They observed 50 million stars in the Milky Way using the 1.8-meter MOA-II telescope at New Zealand's Mount John Observatory and the 1.3-meter University of Warsaw telescope at Chile's Las Campanas Observatory. They found 474 incidents of microlensing, ten of which were brief enough to be planets of around Jupiter's size with no associated star in the immediate vicinity. The researchers estimated from their observations that there are nearly two Jupiter-mass rogue planets for every star in the Milky Way.[9][10][11] Other estimates suggest a much larger number, up to 100,000 times more rogue planets than stars in the Milky Way.[12] A 2017 study by Przemek Mróz of Warsaw University Observatory and colleagues, with six times larger statistics than the 2011 study, indicates an upper limit on Jupiter-mass free-floating or wide-orbit planets of 0.25 planets per main-sequence star in the Milky Way.[13]

Nearby rogue planet candidates include WISE 0855−0714 at a distance of 7.27±0.13 light-years.[14]

Rogue planets could form in interstellar space similarly to how stars form, or else be ejected from solar systems by gravitational interactions. And of course meteoroids, asteroids, comets, moons, dwarf planets, etc., can also be ejected from solar systems into interstellar space just as well, and much more often, than planets are.

Suppose that an interstellar object is detected approaching the solar system and Earth at a relative speed of 1 kilometer per second, and is detected 100 Julian calendar years before calculated impact with Earth.

The object will travel 60 kilometers in each minute, 3,600 kilometers in each hour, 86,400 kilometers in each day, and 31,557,600 kilometers in each Julian calendar year. So it will travel 3,155,760,000 kilometers, or about 21.0949 Astronomical Units (AU), in exactly 100 Julian calendar years.

That distance would be between the orbits of Uranus and Neptune, though an extra solar object would probably not be approaching from the plane of the ecliptic.

If the rouge planet orbits the galaxy in the same direction as the Sun, it would probably have a relative speed compared to the Sun of less than about 100 kilometers per second, I guess.

Thus the rogue planet could travel about 315,576,000,000 kilometers, or 2,109.4952 AU, in exactly 100 Julian calendar years between being discovered and collision with Earth.

Of course if the rogue planet was orbiting the galaxy in the opposite direction to the Sun, it might have a relative velocity of several times 100 kilometers per second and might possibly be detected at several times a distance of 2,109.4952 AU.

A writer would want his approaching rouge planet to be small enough not to have been detected yet, but large enough to have been detected with the lead time he wants for the story - in your case 100 years.

Astronomers have searched for additional planets in our solar system and the lack of success so far has establish limits on on large and how close such hypothetical planets could be.

As of 2016 the following observations severely constrain the mass and distance of any possible additional Solar System planet:

An analysis of mid-infrared observations with the WISE telescope have ruled out the possibility of a Saturn-sized object (95 Earth masses) out to 10,000 AU, and a Jupiter-sized or larger object out to 26,000 AU.7 WISE has continued to take more data since then, and NASA has invited the public to help search this data for evidence of planets beyond these limits, via the Backyard Worlds: Planet 9 citizen science project.[92]

Using modern data on the anomalous precession of the perihelia of Saturn, Earth, and Mars, Lorenzo Iorio concluded that any unknown planet with a mass of 0.7 times that of Earth must be farther than 350–400 AU; one with a mass of 2 times that of Earth, farther than 496–570 AU; and finally one with a mass of 15 times that of Earth, farther than 970–1,111 AU.[93] Moreover, Iorio stated that the modern ephemerides of the Solar System outer planets has provided even tighter constraints: no celestial body with a mass of 15 times that of Earth can exist closer than 1,100–1,300 AU.[94] However, work by another group of astronomers using a more comprehensive model of the Solar System found that Iorio's conclusion was only partially correct. Their analysis of Cassini data on Saturn's orbital residuals found that observations were inconsistent with a planetary body with the orbit and mass similar to those of Batygin and Brown's Planet Nine having a true anomaly of −130° to −110° or −65° to 85°. Furthermore, the analysis found that Saturn's orbit is slightly better explained if such a body is located at a true anomaly of 117.8°+11° −10°. At this location, it would be approximately 630 AU from the Sun.[95]

Many of these limitations also apply to a possible rogue planet approaching from interstellar space.

But astronomers have mostly searched for planets orbiting in the ecliptic plane or relatively close to it. A rogue planet from interstellar space could approach from any angle, including angles far from the ecliptic plane.

Thus it may be possible for your rogue planet to be closer when detected than a planet in our solar system might be while remaining undetected.

• Then the audience just keeps saying “Hurry up and send Bruce Willis up with a nuke!!” A 100 year-off comet strike is not inevitable in our mind’s eye today. – Vogon Poet Oct 27 '19 at 17:04
• @Vogon Poet I have reorganized and revised my answer. It is the duty of a science fiction writer to make the audience understand why some astronomical bodies are too massive to be diverted from a collision course with Earth, even if there is 100 years of warning. It is a trope that: tvtropes.org/pmwiki/pmwiki.php/Main/… But you don't have to follow it. So I have revised my original discussion of solar system objects colliding with Earth, and have added a discussion of objects from outside the solar system colliding with Earth. – M. A. Golding Oct 27 '19 at 18:48

It is probably not possible to achieve such a scenario, but exact calculation is extremely difficult due to many assumptions that would need to be made. Firstly because the Moon is so much more massive than the asteroid, the asteroid must be moving at very close to the speed of light to have a comparable momentum. In fact the asteroid would need to have a very much greater momentum than the Moon because most of the Moons rotational velocity around the Earth would need to be cancelled by the impact. The collision would be so energetic and so much energy would be imparted to the Moon in such a short period that it is hard to be certain of the outcome.

I suspect the asteroid would be converted into subatomic particles and radiation on impact and a lot of the radiation might be “wasted” in carrying away energy of the impact in all directions. But a substantial part of that high energy beam would be absorbed by the Moon and some might even pass right through the Moon. It would be nothing like any kind of natural collision between astronomical bodies. However much energy was wasted by re-radiation it would always be possible to increase the energy of impact to allow for that, but in all probability there is a limit as to how much energy can be imparted to the Moon by such means as the material of the Moon is not strong enough to be pushed in that way. So it would be a bit like trying to move a giant marshmallow (house sized) by firing a bullet at it. No matter how high velocity the gun the marshmallow is not going anywhere.

If the velocity of the Moon could be reduced sufficiently (and I don’t think it can) there would be nothing to stop the Moon falling into a closer orbit. An orbit that was too close would lead to atmospheric drag which would lead to catastrophic impact. An orbit which was much higher could be stable or semi stable. Between the two there must be an orbit such that the decay time is around one hundred years. But see below*

Tidal drag could easily be responsible for lowering the orbit of the Moon over a century if the Moon was in a low enough orbit. But the effects on both the Earth and the Moon would be extreme with huge tides and enormous gravitationally induced tidal volcanism that would probably destroy the surfaces of both the Earth and the Moon long before impact.

*One final issue however would be the Roche limit at around 11,000 miles distance the tidal forces acting on the Moon would exceed the Moons gravitational forces and the Moon would disintegrate forming a ring of debris most of which would probably fall to Earth over a short period leading to a series of massive catastrophic collisions.

• Not very marshmallowy... the impact is inevitably going to leave a Very Big Hole with lots of ejecta. I don't think there's an orbit above the moon's Roche limit where tidal forces could cause it to decay in under 100 years, but doing the maths on such a thing is Quite Hard... – Starfish Prime Oct 27 '19 at 16:08
• (I also don't think that the radiation released by the impact will shine though the moon... there's not enough to break the moon up entirely, and a few thousand km of rock and metal is quite effective at absorbing most particle and EM radiation) – Starfish Prime Oct 27 '19 at 16:10
• I don't think you appreciate the energy involved in a object that has a momentum much greater than the Moon traveling effectively at light speed concentrated into an asteroid. It would be a totaly alien event quite beyond the realm of crates and ejecta – Slarty Oct 27 '19 at 19:11
• I'm not sure what crates have to do with it, but you can look at the actual kinetic energy of each individual proton or neutron, and work out their mean free path through 2000km of rock and iron. Spoiler alert: they don't make it :-( Also, the impactor has about a fiftieth of the moon's momentum. – Starfish Prime Oct 27 '19 at 19:25
• Ahh, you meant craters, apologies. But no, you're quite wrong. You'll get a lot of exotic particles coming out of the collision, but they're largely unstable and for the most part will decay into more boring and conventional particles and gamma rays, which will in turn simply go into heating up the moon and the cloud of ejecta (producing a normal, boring fireball) or shooting off into space. The exotic region will be small and short lived compared to the larger fireworks show. – Starfish Prime Oct 27 '19 at 19:40