<p>It might be hard to find sources where such collisions have been rigorously modeled, but certainly the amount of kinetic energy <em>available</em> is enough to do some damage.</p>
<p>The amount of kinetic energy is unlimited since <span class="math-container">$\gamma = 1/\sqrt{1 - v^2/c^2}$</span> can go to infinity, but let's just use <span class="math-container">$mc^2$</span> to define what &quot;close to the speed of light&quot; might mean.</p>
<p>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 &quot;close to the speed of light&quot; a kinetic energy of 1.2E+31 Joules in the Earth's frame.</p>
<p>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.</p>
<ul>
<li><a href="https://astronomy.stackexchange.com/q/27916/7982">What is the nature of &quot;rock vapor&quot; in this description of the formation of the Moon?</a></li>
</ul>
<p>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 <em>exactly</em> how this would unfold.</p>
<h3>My hunch is that it would not simply just &quot;punch a clean, cold hole&quot; through Earth</h3>
<p>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.</p>
<p>But that's just my hunch.</p>
<p>There's a slightly related question in Space SE with some slightly related &quot;target will be vaporized&quot; answers:</p>
<ul>
<li><a href="https://space.stackexchange.com/q/34446/12102">A starship is traveling at 0.9c and collides with a small rock. Will it leave a clean hole through, or will more happen?</a></li>
</ul>

CC BY-SA 4.0
2022-02-20T01:20:02.317Z
You probably should mention that the KE equals the rest mass energy at 0.866c
PM 2Ring
CC BY-SA 4.0
2022-02-20T01:29:48.557Z
@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](https://www.youtube.com/watch?v=ohDB5gbtaEQ).
uhoh
CC BY-SA 4.0
2022-02-20T01:31:56.617Z
@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
CC BY-SA 4.0
2022-02-20T01:33:45.467Z
Yes, it's $\sqrt{\frac34}$, which gives $\gamma=2$
PM 2Ring
CC BY-SA 4.0
2022-02-20T01:37:41.260Z
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
CC BY-SA 4.0
2022-02-20T01:45:08.427Z
@PM2Ring Oh that looks interesting and helpful! As soon as my morning coffee fully kicks in I'll revisit. *Thanks!*
uhoh
CC BY-SA 4.0
2022-02-20T10:10:12.213Z
Nobody has done the calculation because asteroids are not accelerated to close to the speed of light by any known physical process.
ProfRob
CC BY-SA 4.0
2022-02-21T07:54:09.150Z
@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
CC BY-SA 4.0
2022-02-21T11:00:33.687Z
_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
CC BY-SA 4.0
2022-02-20T00:45:47.537Z
2022-02-20T01:05:15.693Z
2022-02-20T01:05:15.693Z
uhoh
<p>The kinetic energy of a relativistic mass is given by <span class="math-container">$(\gamma -1)mc^2$</span>, where <span class="math-container">$m$</span> is the mass of the object and <span class="math-container">$\gamma$</span> is the Lorentz factor <span class="math-container">$(1 - v^2/c^2)^{-1/2}$</span>, where <span class="math-container">$v$</span> is the speed.</p>
<p>Asteroids (obviously it wouldn't be an asteroid from our Solar System) have a <a href="https://en.wikipedia.org/wiki/Standard_asteroid_physical_characteristics#Density" rel="noreferrer">range of densities</a> from about 1 to 6 g/cm<span class="math-container">$^3$</span>. Let's leave that as a variable - <span class="math-container">$\rho$</span>.</p>
<p>The mass of the asteroid is then just its volume multiplied by its density and then the kinetic energy is
<span class="math-container">$$K = (\gamma -1)\frac{\pi d^3}{6}\rho c^2 $$</span>
The momentum of the asteroid is
<span class="math-container">$$ \gamma mv = \gamma \frac{\pi d^3}{6} \rho v$$</span></p>
<p>How much damage this will do will depend on the size of <span class="math-container">$\gamma$</span> and <span class="math-container">$\rho$</span> and the question isn't answerable without at least specifying what the former is.</p>
<p>The gravitational binding energy of the Earth is approximately <span class="math-container">$3GM_E^2/5R_E = 2\times 10^{32}$</span> J. In an inelastic collision, roughly all of the kinetic energy would be transferred. If we equate the binding energy with <span class="math-container">$K$</span>, then we find that the value of <span class="math-container">$\gamma$</span> that gives enough energy to &quot;unbind&quot; (i.e. explode) the Earth is
<span class="math-container">$$ \gamma_{\rm explode} &gt; 2\times 10^{32}\frac{6}{\pi d^3 \rho c^2} +1 = 12.3 \left(\frac{d}{5{\rm km}}\right)^{-3} \left(\frac{\rho}{3 {\rm g/cm}^3}\right)^{-1}$$</span></p>
<p><span class="math-container">$\gamma =12$</span> corresponds to a speed of <span class="math-container">$0.9965c$</span>. This is about 10,000 times faster than the asteroid hypothesised to have killed off the dinosaurs and caused mass extinction.</p>
<p>Note that the asteroid cannot simply &quot;punch through the Earth&quot; 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.</p>
<p>But this raises another possibility. Suppose that <span class="math-container">$\gamma$</span> was a bit less than this - rather than being destroyed, would the Earth be knocked out of orbit?</p>
<p>Conservation of momentum suggests the change in velocity of the Earth would be
<span class="math-container">$$ \Delta V_E = \frac{\gamma m v}{M_E + m} \simeq \gamma c\left(\frac{m}{M_E}\right) $$</span>
If we say that a &quot;significant orbital perturbation&quot; is a <span class="math-container">$\Delta V_E &gt; 1$</span> km/s (the Earth's orbital speed is about 30 km/s), then the <span class="math-container">$\gamma$</span> required to achieve this is
<span class="math-container">$$\gamma_{\rm perturb} &gt; 10^5 \left( \frac{\Delta V_E}{1 {\rm km/s}}\right) \left( \frac{d}{5{\rm km}}\right)^{-3}\left( \frac{\rho}{3 {\rm g/cm}^3}\right)^{-1} $$</span></p>
<p>Thus it seems that the Earth will <em>not</em> get &quot;knocked out of orbit&quot; before it gets totally destroyed by the deposition of kinetic energy. Note, that since <span class="math-container">$\gamma_{\rm explode}/\gamma_{\rm perturb}$</span> is independent of asteroid size and density, this conclusion is independent of those factors.</p>

CC BY-SA 4.0
2022-02-20T10:58:45.933Z
2022-02-20T17:17:39.507Z
2022-02-20T17:17:39.507Z
ProfRob
<blockquote>
<p><strong>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?</strong></p>
</blockquote>
<p>According to <a href="https://what-if.xkcd.com/20/" rel="noreferrer">https://what-if.xkcd.com/20/</a></p>
<blockquote>
<p>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.</p>
<p>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.</p>
</blockquote>
<p>But that is for a trivial 30 meter diameter asteroid. Presumably the results for a 5000 meter diameter asteroid would be fairly inconvenient.</p>

CC BY-SA 4.0
2022-02-21T00:13:06.257Z
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
CC BY-SA 4.0
2022-02-21T00:46:20.903Z
@ProfRob: That particular quote is for a 30-m 'roid at 0.9999999999999999999999951c (the same speed as the [Oh-My-God particle](https://en.wikipedia.org/wiki/Oh-My-God_particle)).  Pretty sure that counts as a "ludicrously high speed".
Vikki
CC BY-SA 4.0
2022-02-21T02:57:49.003Z
@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
CC BY-SA 4.0
2022-02-21T03:06:03.803Z
[Ethan Siegel](https://medium.com/starts-with-a-bang/the-universe-has-a-speed-limit-and-it-isnt-the-speed-of-light-543b7523b54f) has some info on the GZK cosmic energy limit for particles.
PM 2Ring
CC BY-SA 4.0
2022-02-21T07:46:42.197Z
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
CC BY-SA 4.0
2023-02-23T21:10:24.643Z
XKCD is not a valid reference, it a comic.
Starship
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2022-02-20T10:54:20.550Z
2022-02-20T10:54:20.550Z
RedGrittyBrick
<p>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?</p>
<p>Would it instantly evaporate? Could it make the Earth evaporate? Could it simply fly through it due to its large momentum?</p>

CC BY-SA 4.0
2022-02-20T07:02:44.207Z
Where would an "asteroid" have been accelerated to this speed?
ProfRob
CC BY-SA 4.0
2022-02-20T08:25:44.907Z
@ProfRob - By aliens that don't like people asking nosy questions,
Valorum
CC BY-SA 4.0
2022-02-20T10:08:04.210Z
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
CC BY-SA 4.0
2022-02-20T11:25:36.530Z
There are some informative youtube videos on the subject. Such as this one: https://www.youtube.com/watch?v=DwgMjr-Qu1Y
polfosol ఠ_ఠ
CC BY-SA 4.0
2022-02-20T11:41:51.810Z
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
CC BY-SA 4.0
2022-02-20T23:27:30.760Z
Can you please pick a number for "close to light speed? Say 99%?
Bohemian
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2022-02-20T23:38:01.477Z
5km is small???
DKNguyen
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2022-02-21T00:47:48.097Z
@DKNguyen: By asteroid standards, yes.
Vikki
CC BY-SA 4.0
2022-02-21T02:38:56.287Z
@DKNguyen I guess 5 km isn't *that* small. According to the table on [Wikipedia](https://en.wikipedia.org/wiki/Asteroid#Size_distribution) there are only ~90,000 asteroids >5 km and ~10,000 >10 km (and 25 million > 0.1 km). Here's a [graph](https://www.researchgate.net/profile/Patrick-Michel-3/publication/45858498/figure/fig1/AS:307363794964481@1450292647586/The-size-distribution-of-Main-Belt-asteroids-with-the-cumulative-number-larger-than-a.png) from [Walsh, Michel & Richardson (2008)](https://arxiv.org/abs/0906.4366)
PM 2Ring
CC BY-SA 4.0
2022-02-21T10:47:11.363Z
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
CC BY-SA 4.0
2022-02-21T13:52:43.373Z
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...
CC BY-SA 4.0
2022-02-21T14:24:06.097Z
@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
CC BY-SA 4.0
2022-02-21T17:11:41.697Z
@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
CC BY-SA 4.0
2022-02-28T22:48:39.153Z
@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
<asteroids><impact>
What if the Earth was hit by a small asteroid having close to the speed of light?
Some Student
# What if the Earth was hit by a small asteroid having close to the speed of light?

(Asked by: username_0 on 2022-02-20)

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?

**Tags:** asteroids, impact
> **Comments:**
> **username_1**: Where would an "asteroid" have been accelerated to this speed?  
> **username_2**: @ProfRob - By aliens that don't like people asking nosy questions,  
> **username_1**: 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.  
> **username_3**: There are some informative youtube videos on the subject. Such as this one: https://www.youtube.com/watch?v=DwgMjr-Qu1Y  
> **username_4**: 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 :-)  
> **username_5**: Can you please pick a number for "close to light speed? Say 99%?  
> **username_6**: 5km is small???  
> **username_7**: @DKNguyen: By asteroid standards, yes.  
> **username_8**: @DKNguyen I guess 5 km isn't *that* small. According to the table on [Wikipedia](https://en.wikipedia.org/wiki/Asteroid#Size_distribution) there are only ~90,000 asteroids >5 km and ~10,000 >10 km (and 25 million > 0.1 km). Here's a [graph](https://www.researchgate.net/profile/Patrick-Michel-3/publication/45858498/figure/fig1/AS:307363794964481@1450292647586/The-size-distribution-of-Main-Belt-asteroids-with-the-cumulative-number-larger-than-a.png) from [Walsh, Michel & Richardson (2008)](https://arxiv.org/abs/0906.4366)  
> **username_9**: 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.  
> **username_10**: 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"*.  
> **username_4**: @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.  
> **username_1**: @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).  
> **username_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.  

---

## 3 answers

### Answer by username_14 on 2022-02-20. Score: 17

> **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.

> **Comments:**
> **username_1**: 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  
> **username_7**: @ProfRob: That particular quote is for a 30-m 'roid at 0.9999999999999999999999951c (the same speed as the [Oh-My-God particle](https://en.wikipedia.org/wiki/Oh-My-God_particle)).  Pretty sure that counts as a "ludicrously high speed".  
> **username_8**: @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.  
> **username_8**: [Ethan Siegel](https://medium.com/starts-with-a-bang/the-universe-has-a-speed-limit-and-it-isnt-the-speed-of-light-543b7523b54f) has some info on the GZK cosmic energy limit for particles.  
> **username_12**: 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.  
> **username_15**: XKCD is not a valid reference, it a comic.  

---

### Answer by username_11 on 2022-02-20. Score: 13

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.

* [What is the nature of "rock vapor" in this description of the formation of the Moon?](https://astronomy.stackexchange.com/q/27916/7982)

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:

* [A starship is traveling at 0.9c and collides with a small rock. Will it leave a clean hole through, or will more happen?](https://space.stackexchange.com/q/34446/12102)

> **Comments:**
> **username_8**: You probably should mention that the KE equals the rest mass energy at 0.866c  
> **username_11**: @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](https://www.youtube.com/watch?v=ohDB5gbtaEQ).  
> **username_11**: @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.  
> **username_8**: Yes, it's $\sqrt{\frac34}$, which gives $\gamma=2$  
> **username_8**: 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  
> **username_11**: @PM2Ring Oh that looks interesting and helpful! As soon as my morning coffee fully kicks in I'll revisit. *Thanks!*  
> **username_1**: Nobody has done the calculation because asteroids are not accelerated to close to the speed of light by any known physical process.  
> **username_12**: @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.  
> **username_13**: _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.  

---

### Answer by username_1 on 2022-02-20. Score: 12

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](https://en.wikipedia.org/wiki/Standard_asteroid_physical_characteristics#Density) 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.


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