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If the Earth was hit by an asteroid having diameter of about 5 km and moving with a speed close to the speed of light? What would happen?

Would it instantly evaporate? Could it make the Earth evaporate? Could it simply fly through it due to its large momentum?

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Some Student
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    Where would an "asteroid" have been accelerated to this speed? – ProfRob Feb 20 '22 at 07:02
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    @ProfRob - By aliens that don't like people asking nosy questions, – Valorum Feb 20 '22 at 08:25
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    Well, the question is actually unanswerable unless you specify what you mean by "close to the speed of light". The difference in energy between 0.5, 0.9, 0.99, and 0.99999c is tremendous. – ProfRob Feb 20 '22 at 10:08
  • There are some informative youtube videos on the subject. Such as this one: https://www.youtube.com/watch?v=DwgMjr-Qu1Y – polfosol ఠ_ఠ Feb 20 '22 at 11:25
  • According to a Startrek Voyager book, where they discussed how to destroy a planet at any cost, they considered accelerating the USS Voyager as close as possible to the speed of light and ramming the planet. The predicted result: A Voyager sized hole straight through the planet. Pro: They have 200-300 years more knowledge about physics then we do and would know better. Con: It's a book, it's not real :-) – gnasher729 Feb 20 '22 at 11:41
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    Can you please pick a number for "close to light speed? Say 99%? – Bohemian Feb 20 '22 at 23:27
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    5km is small??? – DKNguyen Feb 20 '22 at 23:38
  • @DKNguyen: By asteroid standards, yes. – Vikki Feb 21 '22 at 00:47
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    @DKNguyen I guess 5 km isn't that small. According to the table on Wikipedia there are only ~90,000 asteroids >5 km and ~10,000 >10 km (and 25 million > 0.1 km). Here's a graph from Walsh, Michel & Richardson (2008) – PM 2Ring Feb 21 '22 at 02:38
  • 5km is not a small asteroid. For comparison the Chicxulub asteroid was 10km. I.e. your small asteroid is one eight the size of the one that killed the dinosaurs. – Taemyr Feb 21 '22 at 10:47
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    There is no such thing as "close to the speed of light". It's like saying "close to infinity" - you can pick any number you want with effects ranging anywhere between mostly harmless and "will destroy the entire universe". – J... Feb 21 '22 at 13:52
  • @Bohemian: In the Startrek universe it seems space ships go straight "past" speed of light normally for long distance travel, and don't ever try to get close. But I think the big question is not how much energy gets stored, but how much of that energy can be transferred to the planet. At these speeds, spaceship or asteroid would go through the planet within 25-50 milliseconds. – gnasher729 Feb 21 '22 at 14:24
  • @gnasher729 how would you push a column of $>10^{18}$ kg of Earth material out of the way without significant deceleration? (For reference, the asteroid has a mass of $\sim 10^{14}$ kg). – ProfRob Feb 21 '22 at 17:11
  • @Struggling_Student I very much like your questions including this one! They are simple and thought provoking and lead to active, high quality answering. Unfortunately sometimes people forget to +1 such questions' author and so far there are no ways to add a reward bounty to questions. – uhoh Feb 28 '22 at 22:48

3 Answers3

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If the Earth was hit by an asteroid having diameter of about 5 km and moving with the speed close to the speed of light? What would happen?

According to https://what-if.xkcd.com/20/

The momentum would be enough to knock the Earth into a different orbit—but the Earth is no more. The energy deposited is ten thousand times greater than the planet’s gravitational binding energy, and the planet is blown into an expanding cloud of plasma, with a particularly energetic streamer extending away from the far side of the impact site, out into space.

The Sun hiccups and flares as it absorbs waves of dust. The surfaces of Mars and Venus are scoured clean by the waves of incredibly high-energy plasma.

But that is for a trivial 30 meter diameter asteroid. Presumably the results for a 5000 meter diameter asteroid would be fairly inconvenient.

RedGrittyBrick
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    The conclusion doesn't depend on the asteroid mass at all. The momentum would not be enough to significantly perturb the earth's orbit without a ludicrously high speed. Maybe you could give some (any) detail about what speed has been assumed and what calculations have been done – ProfRob Feb 21 '22 at 00:13
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    @ProfRob: That particular quote is for a 30-m 'roid at 0.9999999999999999999999951c (the same speed as the Oh-My-God particle). Pretty sure that counts as a "ludicrously high speed". – Vikki Feb 21 '22 at 00:46
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    @Vikki Well, it's the last section of a What-If, so the ludicrosity has been turned up to 11. ;) It's hard to explain how the OMG particle managed to travel with that speed. A macroscopic object has no chance, even if you can come up with some mechanism for giving it that much KE. – PM 2Ring Feb 21 '22 at 02:57
  • Ethan Siegel has some info on the GZK cosmic energy limit for particles. – PM 2Ring Feb 21 '22 at 03:06
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    The linked article has conclusions for ranges of speeds. That's for the most absurd one. You should specify how "close to the speed of light" you mean, because many people would call even 0.99c "close". Of course it's possible to get arbitrarily close in theory. +1 for linking the XKCD article which answers the question, but -1 for a poor job of specifying which part you're quoting and adapting that quote into an answer to this question. – Peter Cordes Feb 21 '22 at 07:46
  • XKCD is not a valid reference, it a comic. – Starship - On Strike Feb 23 '23 at 21:10
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It might be hard to find sources where such collisions have been rigorously modeled, but certainly the amount of kinetic energy available is enough to do some damage.

The amount of kinetic energy is unlimited since $\gamma = 1/\sqrt{1 - v^2/c^2}$ can go to infinity, but let's just use $mc^2$ to define what "close to the speed of light" might mean.

With a density of 2 g/cm^3 a 2500 meter radius sphere will have a mass of 1.3E+14 kg and moving at our "close to the speed of light" a kinetic energy of 1.2E+31 Joules in the Earth's frame.

That a reduced energy of 2E+06 Joules per kilogram of Earth. That is less than the energy necessary to completely disassemble the Earth's mass to infinity, but way more than is necessary to completely destroy it as a solid planet and convert it to a gas or plasma.

But like I said, it might be hard to find citeable sources where someone has done a rigorous simulation including all necessary hydrodynamic transport physics to model such an explosion to see exactly how this would unfold.

My hunch is that it would not simply just "punch a clean, cold hole" through Earth

and keep going, leaving the earth to fill in a 5 km diameter cylindrical perforation somehow. Due to the high density and relativistic speeds I think there would be so much radiation pressure that the planet would be quickly and completely heated to some mixture of gas and thermal plasma.

But that's just my hunch.

There's a slightly related question in Space SE with some slightly related "target will be vaporized" answers:

uhoh
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    You probably should mention that the KE equals the rest mass energy at 0.866c – PM 2Ring Feb 20 '22 at 01:20
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    @PM2Ring when I was doing my Ph.D. I got into an argument with a professor during an oral exam. He asked me to calculate some special relativistic problem and I said I didn't know how. He said any Ph.D. candidate should be able to to it. I said that I didn't understand special relativity, I was not comfortable with its foundations and so I won't simply apply equations with which I am not yet comfortable. That's not (doing) Physics. He said "Yes it is" and we got into a Python-like argument. – uhoh Feb 20 '22 at 01:29
  • @PM2Ring since that time (from the previous century) I've continued to shy away from anything more than hand-waving, vague references to special relativity. Please feel free to edit my post; my guess is that that 0.866c is actually a nice, pretty rational number or its square root. – uhoh Feb 20 '22 at 01:31
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    Yes, it's $\sqrt{\frac34}$, which gives $\gamma=2$ – PM 2Ring Feb 20 '22 at 01:33
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    FWIW, in https://physics.stackexchange.com/a/595175/123208 I have a graph that shows that the Newtonian equation for KE is still fairly accurate at 0.1c – PM 2Ring Feb 20 '22 at 01:37
  • @PM2Ring Oh that looks interesting and helpful! As soon as my morning coffee fully kicks in I'll revisit. Thanks! – uhoh Feb 20 '22 at 01:45
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    Nobody has done the calculation because asteroids are not accelerated to close to the speed of light by any known physical process. – ProfRob Feb 20 '22 at 10:10
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    @ProfRob: I wouldn't be surprised if someone with an interest in science fiction and the relevant expertise has taken a stab at it, perhaps as expert advice to an SF author on a plot point involving chucking asteroids at planets via some kind of reactionless drive. Most beta readers wouldn't go into that rigorous detail, but I wouldn't be surprised if someone somewhere has done it. – Peter Cordes Feb 21 '22 at 07:54
  • Nothing is going to punch a nice, clean hole. This is actually a result from classical physics. The basic problem is that there is way more mass in the cylinder you're trying to create than there is in the asteroid. This can easily be a factor of a million. – MSalters Feb 21 '22 at 11:00
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The kinetic energy of a relativistic mass is given by $(\gamma -1)mc^2$, where $m$ is the mass of the object and $\gamma$ is the Lorentz factor $(1 - v^2/c^2)^{-1/2}$, where $v$ is the speed.

Asteroids (obviously it wouldn't be an asteroid from our Solar System) have a range of densities from about 1 to 6 g/cm$^3$. Let's leave that as a variable - $\rho$.

The mass of the asteroid is then just its volume multiplied by its density and then the kinetic energy is $$K = (\gamma -1)\frac{\pi d^3}{6}\rho c^2 $$ The momentum of the asteroid is $$ \gamma mv = \gamma \frac{\pi d^3}{6} \rho v$$

How much damage this will do will depend on the size of $\gamma$ and $\rho$ and the question isn't answerable without at least specifying what the former is.

The gravitational binding energy of the Earth is approximately $3GM_E^2/5R_E = 2\times 10^{32}$ J. In an inelastic collision, roughly all of the kinetic energy would be transferred. If we equate the binding energy with $K$, then we find that the value of $\gamma$ that gives enough energy to "unbind" (i.e. explode) the Earth is $$ \gamma_{\rm explode} > 2\times 10^{32}\frac{6}{\pi d^3 \rho c^2} +1 = 12.3 \left(\frac{d}{5{\rm km}}\right)^{-3} \left(\frac{\rho}{3 {\rm g/cm}^3}\right)^{-1}$$

$\gamma =12$ corresponds to a speed of $0.9965c$. This is about 10,000 times faster than the asteroid hypothesised to have killed off the dinosaurs and caused mass extinction.

Note that the asteroid cannot simply "punch through the Earth" because the column of material it would have to displace to do so has a mass that is roughly 2500 times that of the asteroid and is encased in a mass which is many orders of magnitude bigger than that (ignoring glancing blows). It might emerge on the other side, but only after having lost most of its kinetic energy.

But this raises another possibility. Suppose that $\gamma$ was a bit less than this - rather than being destroyed, would the Earth be knocked out of orbit?

Conservation of momentum suggests the change in velocity of the Earth would be $$ \Delta V_E = \frac{\gamma m v}{M_E + m} \simeq \gamma c\left(\frac{m}{M_E}\right) $$ If we say that a "significant orbital perturbation" is a $\Delta V_E > 1$ km/s (the Earth's orbital speed is about 30 km/s), then the $\gamma$ required to achieve this is $$\gamma_{\rm perturb} > 10^5 \left( \frac{\Delta V_E}{1 {\rm km/s}}\right) \left( \frac{d}{5{\rm km}}\right)^{-3}\left( \frac{\rho}{3 {\rm g/cm}^3}\right)^{-1} $$

Thus it seems that the Earth will not get "knocked out of orbit" before it gets totally destroyed by the deposition of kinetic energy. Note, that since $\gamma_{\rm explode}/\gamma_{\rm perturb}$ is independent of asteroid size and density, this conclusion is independent of those factors.

ProfRob
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