Rogue RKKV impacts the Moon

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.