# Rogue RKKV impacts the Moon

Well, this is my first post here. Been lurking for a while and have read a lot of the RKKV (Relativistic Kinetic Kill Vehicle) questions in my search for a solid answer to my question. Alas, no such luck. So time to stop lurking...

At some point in the past a relativistic weapon was fired during a some war somewhere by someone that had no idea Earth or its solar system even existed. The shot itself was a miss and the triggerman likely got a proper chewing out. The misfire was noted, reports sent up, but for one reason or another, the top brass never followed up on that one rogue RKKV that got launched off towards a distant arm of the galaxy.

Jump forward in time to near present day Earth. The RKKV has been flying along, minding its own business, and by fluke or bad luck, its path takes it into our solar system. Then it hits the Moon. Good thing too, maybe, because it was heading straight for the Earth. The Moon takes the hit in line with the Earth --- becoming the impactor that stops the RKKV from hitting us dead on.

This is where my question comes in:

Just how bad would this be?

I've played around with Universe Sandbox² a few times trying to see just what would happen [to no real avail] and I've read up on RKKVs on Atomic Rocket. I've even done some back of the envelope maths to try and see just what would happen. However, I'm not entirely sure what would happen.

This particularly RKKV is made of a material with a density of 1000g/cm³ that has been shaped into a kilometer long rod with a diameter of 20 meters [314,159,265,358.98cm³]. It was travelling at a speed of .2c at the time of impact.

My math-fu simply isn't good enough to properly figure out what this would do, but the event itself is a key event in a story I've been working on for a bit now. The event itself is aptly named the "Oops" event and the fact that there is anyone on Earth left to name the event demands that it be a survivable scenario for the critters on Earth.

• There is no material with a density of 1000 g/cm³, and most importantly there cannot be any such material which is stable under normal-ish conditions. The densest material we know of is osmium at 22.6 g/cm³; what you want would be 44 times as dense! (For reference, the gravitational binding energy of the Moon is 1.2E29 J. Is this the value which you were looking for?) Sep 23 at 16:48
• If my calcs are right, that's a 540,000 teraton explosion. Likely the biggest bang seen since the formation of the solar system. And Chicxulub was only like 70-80 teratons...
– BMF
Sep 23 at 17:41
• 1: that's not really a relativistic KKV, it's just a very fast one. At 0.2c, relativistic KE is only 3% higher than Newtonian KE. 2: the point of a relativistic KKV is the speed, not the mass. There's little point in making one of some hyper-dense substance, if you can make it lighter you can make it faster. In fact, you're probably going to want a lower density to ensure it delivers the energy to the target instead of losing most of it to overpenetration. Consider that it'll basically be interacting as a very dense packet of particle radiation, rather than a solid object. Sep 23 at 22:40
• – JBH
Sep 23 at 22:45
• "It was travelling at a speed of .2c at the time of impact" - Are these normally fired at planets or other supermassive objects with no real maneuverability? Barring "point-blank" usage (say, under 1 light-second from your target), .2c is slow enough for a maneuvering target to detect the projectile and get out of its way. If these projectiles are intended as planet-killers, missing should be near impossible and that "chewing out" was very warranted. Sep 25 at 2:17

# You take a piece off of the Moon and kill everyone

I'm rounding up the mass of the projectile to 3 billion kilograms because at the scales involved, accuracy to the gram doesn't matter.

I'm also ignoring relativity just for now.

So:

$$E = \frac{mv^2}{2} = \frac{3kg \times 10^9 \times (6 m/s \times 10^7)^2}{2} = 5.4 \times 10^{24} joules$$

Give or take.

For comparison, the binding energy of the Moon is in the order of $$10^{29}$$ joules. So you are five orders of magnitude short of obliterating her completely.

But with a mass of close to $$7 \times 10^{22}$$ kg... if we were using Batman physics, that would give her a push enough to change her orbital speed up close to 100m/s. But you don't push a planet or similar body like that, and your bullet is three orders of magnitude denser than the Moon. Most likely it penetrates the Moon and comes out the other side. Might still hit the Earth, although with less speed, but is more likely to miss. We're not off the hook though, even if it misses us.

As for the Moon, a considerable portion of her mass will be ejected like it always happens when something opens up a crater. But in this case an incalculable portion of that will escape the Moon, possibly the Earth-Moon system altogether. I say incalculable because things like the angle of impact and direction it's coming matter. Whatever is left of the Moon will collapse into a spheroid again, and a considerable portion of her surface will be molten for quite a while.

As for the ejecta that doesn't leave the Earth-Moon system, they will make for a bombardment that might end human civilization and life. This has potential to be worse than the dino killing asteroid, and that one is believed by some scientists to have killed most dinos by heating up the atmosphere over water boiling temperatures globally through ejecta falling over half the globe.

If you want humanity to survive, you should consider a smaller KKV.

Edit: there are some interesting comments here about whether the projectile survives the impact after passing through the Moon in 57 milliseconds or less. I am just assuming that whatever technomagic is keeping a 1,000 g/cm3 object (about 6x the density of the core of the Sun) from spontaneously exploding like a mini-nova was keeping it intact too. But whether it keeps its shape or becomes a plasma should be the least of humanity's worries.

• I would say that travelling at 0.2 c such an impossibly dense impactor will go through the Moon in 57 milliseconds, like a bullet through an apple... Sep 23 at 21:49
• @AlexP if the only thing special about it is its density, I strongly disagree. At those speeds, the liberation of huge amounts of KE will very rapidly disintegrate the impactor. On the order of nanoseconds. Like a relativistic baseball, nothing withstands those forces.
– BMF
Sep 23 at 22:08
• @BMF: it won't necessarily be as intact as a bullet that has passed through an apple, but it has about as much mass as a 300 km column of rock. With its momentum deposited into a core going through the entire moon, the resulting plume of plasma would still have a velocity relative to the moon of several thousand km/s. I suspect very little of its total yield would be deposited in the moon itself. Sep 23 at 22:22
• @BMF what part of "resulting plume of plasma" sounds like a surviving projectile? Sep 23 at 22:28
• I meant surviving to penetrate deeply. I think it'd be vapor well before. Relativistic missiles don't make great penetrators. (Despite what Seveneves might lead you to believe)
– BMF
Sep 23 at 22:37

Newton's Impact Depth approximation states that impact depth is approximately equal to Projectile Length (L) x Projectile Density (A) / Target Density (B).

So, with L = 1km, A = 1000 g/cm³, B = 3.34 g/cm³ we get an approximate depth of 299.4 km.

The moon's diameter is 3,474.8 km. The impact depth of 299.4 km is about 8.6% of the total diameter of the moon, so the RKKV will cease its forward motion within the moon.

Now, if we assume that the kinetic energy (KE) of the RKKV is on the order of 5.5E24 J (thanks, The Square-Cube-Law), and given that the gravitational binding energy (GBE) of the moon is on the order of 1.25E29 J and the moon's mass is 7.3E22 kg, the amount of the moons mass that could be ejected is:

KE/GBE*Mass = 5.5E24 / 1.25E29 x 7.3E22 = 3.1E18kg.

That's 3 quadrillion tons of matter that we can expect to achieve escape velocity from the moon. That's an awful lot of mass. Of course, this is just a ballpark figure, but it gives an order of magnitude. It obviously can't be much greater than this.

There is also an interesting little effect that was used to good effect in WWII, that only fell into disuse when spaced and composite tank armour became common, and that was the use of HESH warheads. Why am I talking about tank shells? Because it has been shown that the application of a large amount of kinetic energy to an object can cause 'spall' to be ejected from the opposite side of the object, and:

HESH was found to be surprisingly effective against metallic armour as well as concrete structures

and

In general, the higher the armour thickness, the higher the scab weight will be.

So, since the RKKV is a long-rod penetrator, not a broad pancake charge as a HESH round, we can expect that a significant amount of the ejecta will be expelled backwards from alongside the shot path, possibly as much as half of the total mass of ejecta. This is also subject to a significant amount of error, it may be as little as a quarter or as much as three-quarters of the total escaping mass.

That still leaves very roughly 1.5 quadrillion tons of lunar material headed directly away from the moon and in the general direction of the Earth at at least the escape velocity of the moon, 2.38 km/s. Let's say about 5 km/s, which is about the speed of sound in the body of the moon.

The only thing that might save us is that the moon orbits the earth at around 1 kilometre per second, and the moon is around 384,000 km away on average. This will lead to the vector of the ejecta being not directly toward the Earth even though the RKKV was headed directly toward the Earth.

The ejecta might take 21 1/3 hours to reach Earth, but in that time would travel 76,800 km away from a direct line to the centre of the earth. Since Earth's radius is 6371 km, this is around 6 times the diameter of the earth.

So, we can expect that the bulk of the 1.5 quadrillion tons of ejecta would miss earth.

That doesn't mean that Earth won't be hit at all. If even only 0.01% of the remaining ejecta from the moon hits the Earth, that's still 150 billion tons of rock raining down across the entire surface of the earth.

Nowhere would be safe. Lots of living things would die... though since humans are hard to kill, like cockroaches scurrying for cover, there may be some survivors... maybe quite a few.

• Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Worldbuilding Meta, or in Worldbuilding Chat. Comments continuing discussion may be removed. Sep 25 at 4:45
• I think you might be overestimating the ejecta. The impactor has enough KE to accelerate 0.0001% of the moon's mass to escape velocity, meaning that mass will escape if it is the only mass affected by the impact. I expect the energy will be dissipated by deformation and acceleration of matter through far more than a millionth of the moon's mass. You could accelerate 0.0001% of the moon's mass to escape velocity, or 0.001% of the mass to a tenth of escape velocity, and I don't see why it wouldn't be the latter. Sep 25 at 15:38
• One question does come to mind: is the moon sufficiently rigid that the compression shockwave could transmit in a way to cause spalling? Sep 25 at 17:44
• @JoelAelwyn It ought to be around the consistency of concrete, at least, which means that, yes, it ought to spall. Sep 26 at 0:44
• @JoelAelwyn, we've got a few examples of bodies being hit by large objects. The best-known is Mimas, where the area antipodal to the impact point shows features that may be the remnants of a spallation event. That said, I think this answer seriously over-estimates the amount of spallation ejecta and underestimates the amount of same-side ejecta.
– Mark
Sep 26 at 2:49

I'd like to respond to Monty Wild's analysis. He correctly invokes Newton's impact depth approximation to conclude the impactor would make it around 10% of the way through the moon:

Newton's Impact Depth approximation states that impact depth is approximately equal to Projectile Length (L) x Projectile Density (A) / Target Density (B).

So, with L = 1km, A = 1000 g/cm³, B = 3.34 g/cm³ we get an approximate depth of 299.4 km.

The moon's diameter is 3,474.8 km. The impact depth of 299.4 km is about 8.6% of the total diameter of the moon, so the RKKV will cease its forward motion within the moon.

However, the following analysis is dubious:

Now, if we assume that the kinetic energy (KE) of the RKKV is on the order of 5.5E24 J (thanks, The Square-Cube-Law), and given that the gravitational binding energy (GBE) of the moon is on the order of 1.25E29 J and the moon's mass is 7.3E22 kg, the amount of the moons mass that could be ejected is:

KE/GBE*Mass = 5.5E24 / 1.25E29 x 7.3E22 = 3.1E18kg.

I don't think so. Most of the energy will go into heat, not into overcoming the moon's gravitational potential. Because the impactor penetrates hundreds of km into the moon, most of the energy will be deposited too deeply inside to knock material loose. Most of the material that does get knocked loose will fall short of escape velocity. Maybe 1% of the energy might go into knocking rocks loose of the moon. 1% would be 3 * 10^16 kg.

Additionally, a rain of rock from the moon onto the Earth wouldn't actually be that damaging. The rocks would be spread out into a plume, so it would not be a single impactor, it would be a lot of dust and pebbles. Almost all of it would burn up in the atmosphere without reaching the surface.

There would be weather and climate effects. The ejection volume from the Chicxulub crater was in the area of 10^18 kg, so in terms of the cooling effect from dust in the atmosphere, the plume of moon dust might be 30x less than that, assuming it all hits (which it won't). The climate effect of Chicxulub was in the area of a 20% reduction in incoming sunlight for a few years to decades. The climate effect of the 3*10^16 kg moon plume might thus be expected to be around a 1% reduction in incoming sunlight. We would notice, but most things would continue as they were.

• He correctly invokes Newton's impact depth approximation... Newton's approx. fails at supersonic speeds, let alone hypervelocity impacts (orbital speeds), let alone whole fractions of lightspeed.
– BMF
Sep 24 at 21:39
• @BMF The main reason Newton's approximation is said to be less applicable for hypervelocity is that a crater is also formed due to the energy deposited, and the crater could be deeper than the projectile itself travels. The impact depth of the projectile itself should still follow Newton's approximation from momentum considerations. The rock in front of the projectile acts like a gas at such speeds, bouncing off the front of the projectile until the projectile slows to a stop, which is a momentum consideration. Sep 25 at 0:00
• @BMF What can be said definitely is that the projectile itself can't go beyond the Newton's approximation depth. It may deform and fall short, and it may blast out a crater that goes beyond the projectile itself, but it can't go through the moon and come out the other side, because of momentum considerations. Sep 25 at 0:23
• (cont) But it's my understanding that at such high speeds material strengths become irrelevant, and the release of enormous energy (from fission & fusion events) causes practically instant disintegration.
– BMF
Sep 25 at 0:23
• Impact leads to heat. Heat leads to vaporisation. Vaporisation leads to gas. Gas has pressure. Pressure pushes on things, and so do shock waves. Therefore things will move... some of it quite fast. Fast-moving things tend to go up and not come back down. So there will be ejecta. How much and where it is headed is debatable. Sep 25 at 4:57