73
$\begingroup$

Say there is a football sized rock in the path of the ship. Will it create a football sized hole through the ship in the blink of an eye, or will more happen? The ship would be filled with metal, water, some fuel and air for simplicity sake. Will there be a huge shockwave throughout the ship from the impact? Will it collide with enough energy to initiate fusion with the atoms of the hull?

$\endgroup$
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – called2voyage Mar 1 at 19:52
118
$\begingroup$

@Hobbes answered this in a comment.

Your final guess

Will it collide with enough energy to initiate fusion with the atoms of the hull?

is correct.

See the first XKCD What-If comic, "Relativistic Baseball" for details.

The short answer is your entire spaceship is going to suffer a sudden and gratuitous existence failure due to the fusion reaction.

ps - as noted in the comments, the fusion reaction won't begin until shortly after the rock hits the ship. If the ship is bigger than the distance between the pitcher's mound and home plate then it won't really matter much - you'll have an unscheduled/uncontained fusion reaction somewhere in your ship, possibly not in the engine compartment.

If the ship is small enough, then I'd assume Everyday Astronaut's answer would apply.

$\endgroup$
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – called2voyage Mar 1 at 19:52
35
$\begingroup$

I answer this question from a purely structural-mechanical point of view, i.e. not considering fusion as discussed in Dan Pichelman's answer.

Will it create a football sized hole through the ship in the blink of an eye [...]?

That's probably right. The rock hits the structure so fast that no inertial effects have a chance of taking place before the incident is already over. "Communication" of the subjected material with the surroundings would simply be too slow. So, any elasticity or plasticity of the structural material would be irrelevant to the impact. The rock would leave a hole of the approximate shape of its silhouette. As it goes through the layers of the ship, that silhouette will of course change.

Will there be a huge shockwave throughout the ship from the impact?

I guess not as much as you'd expect, from a structural point of view. Except if the hull of the ship contains residual stresses beforehand. Those might relax after the structure has been weakened, causing waves propagating through the ship in the classical way. But that would not be of any significance since the ship would have major leakage of its inner atmosphere (if it was pressurized).

Think of a classical metal shear. There, the surrounding material is mounted somewhere, and only the local part of the material is affected by the "impact". In the case of our starship, the mounting of the surrounding material is its inertia.

$\endgroup$
  • 10
    $\begingroup$ The energy is carried sideways not by sound or mechanical forces but by gamma rays, X rays and other energetic particles. $\endgroup$ – Steve Linton Feb 25 at 16:55
  • 3
    $\begingroup$ As the mass starts interacting with the ship, conservation of momentum would dictate that the center of mass of (football plus affected parts of ship) will in the short term have a constant velocity, which will be somewhere between the velocities of the football and the ship. If the football is heavy relative to the stuff it passes through, all of that mass is going to be squished flat and won't re-expand until the ship is a long way away. If the football passes through enough denser material, however, it might be slowed enough that other effects take over. $\endgroup$ – supercat Feb 25 at 19:10
  • 11
    $\begingroup$ @GiuPiete I think that there is no textbook on relativistic aerodynamics :-) $\endgroup$ – Everyday Astronaut Feb 25 at 20:09
  • 9
    $\begingroup$ I don't think that's true Loren. Imagine shooting a bullet through an empty milk carton. It would be more like that. $\endgroup$ – Nick van der Kroon Feb 26 at 0:03
  • 2
    $\begingroup$ @LorenPechtel Space rocks tend to be extremely dense (at least those reaching Earth's surface are a lot denser than most Earth rocks) while spacecraft are designed to be light weight. $\endgroup$ – Everyday Astronaut Feb 26 at 9:05
20
$\begingroup$

Let us consider a 10 kg rock with a $0.1\,\mathrm{m}^2$ cross-sectional area. Further let’s assume that we are lucky and the rock passes only through one wall of the spaceship (say $3\,\mathrm{mm}$ of aluminium) $10\,\mathrm{m}$ of air and another wall. The total mass of material in this cylinder is them about $$ 2700 \times 3 \times 10^{-3} \times 0.1 \times 2 = 2\,\mathrm{kg}$$ of aluminium and about $1\,\mathrm{kg}$ of air.

All of that material will basically end up moving at the same velocity as the rock (it is trapped in front of it). The momentum of the rock is given by

$$ p=\frac{m_0v}{\sqrt{1-v^2/c^2}}=\gamma m_0v\,, $$

which works out to be

$$\frac{9 c}{\sqrt{0.19}}\approx 20c\,. $$

This must also be the momentum of the rock (or the plasma cloud it may well have turned into) plus the $3\,\mathrm{kg}$ of spaceship and air it will sweep up when it leaves. Using the same equation, we can recover the new $v$. Taking $ c= 1$ we have

$$20 = \frac{13 v}{\sqrt{1-v^2}}$$

so $400 - 400 v^2 = 169 v^2$, $v^2 = 400/569$ and $v = 0.84$

Now let's see (from the ships frame of reference), what this does to the kinetic energy. We find an equation on the same site. $$E=\sqrt{p^2c^2 + (m_0c^2)^2}\,.$$

We can apply this three times. To the incoming rock $E = \sqrt{500}$ (still in units where $c =1$). The $3\,\mathrm{kg}$ of spaceship and air has energy 3 in the same units. The $13\,\mathrm{kg}$ of debris at the end has energy $\sqrt{569}$. So the energy which must be released as heat, radiation, sideways KE, etc. is

$$3 +\sqrt{500} - \sqrt{569}$$ which is about $1.5$.

Converting back to standard units, we find about $130\,\mathrm{PJ}$ ($1.5\,\mathrm{kg}$ of mass). Some of this will be carried away in the internal energy of the ball of plasma which was formerly rock and vital bits of our spaceship, but a good proportion is going to end up as heat in the rest of the spaceship.

It is probably not good news for the crew!

Addition: We can do more. Consider the first impact ("the entry wound"). A similar calculation shows that that liberates $1 + \sqrt{500} - \sqrt{521} kg$ KE. Let's arbitrarily assume that half of that goes into random motion of the 11kg of matter involved (we'll come back to the other half, which is probably mostly gamma rays). So in its own frame of reference we need the average velocity to satisfy $$0.25 c^2 = 11/2 v^2$$ where $v$ is the typical velocity of the random motion (ignoring relativity, since for this kind of rough approximation $0.25 << 11$). This gives us $v = 0.21c$. In this frame of reference the spaceship is about 5m thick, so we'd expect an "exit wound" of radius about 1m. Of course as it spreads out to that size, the fireball with interact with more matter, releasing still more of it's KE and possibly getting bigger and slower and hotter.

So our football size rock comes in, and a 2m diameter nuclear fireball goes out through the other side of the ship 30ns later. Still not looking like good news.

Meanwhile, the other half of the entry energy is coming out as gamma rays (if we're lucky some will be neutrinos, which will cause us no further trouble, but I doubt it's much). A phenomenon called "relativistic beaming" means that the radiation is (from the ship's perspective) concentrated in a cone in front of the fireball. I'm hitting the limits of my astrophysics, but I think it will be about 64 times more intense in front of the fireball than to the side. Let's assume that. We are radiating away about 0.25kg of energy in about 30ns. A square meter of surface 10m away to the side of the impact will receive $$\frac{1}{4\pi 10^2} \frac{0.25c^2}{64} \approx 3\times 10^{12} J$$ How much will absorbed depends on what's there, but this is enough to raise the temperature of a human body by millions of kelvins.

So yes, it looks very much as if the whole ship will be vaporised.

$\endgroup$
  • 1
    $\begingroup$ "All of that material will basically end up moving at the same velocity as the rock" -- actually not clear. The "rock is solid matter" is a lower energy scale fact. At 0.9 c, you have to think about particle cross sections. $\endgroup$ – Yakk Feb 28 at 14:58
  • 1
    $\begingroup$ @Yakk you are essentially asking how far 50GeV aluminium nuclei penetrate into rock. I don't have the figure to hand :-) but i'd be a bit surprised if it's more than 30cm. $\endgroup$ – Steve Linton Feb 28 at 16:03
  • 1
    $\begingroup$ So, RHIC does gold nucleus at 100 GeV. How far does the gold nucleus penetrate the "beam stoppers"? That'd give us an order-of-order-of-magnitude. (Gold/Aluminum? Rock/whatever the Beam Stopper material is? Pretty close.) $\endgroup$ – Yakk Feb 28 at 17:10
  • $\begingroup$ The particles are stopped on a cooled disc of tantalum. $\endgroup$ – Cees Timmerman Mar 2 at 4:17
6
$\begingroup$

On 0.9c, the kinetical energy of an 1g rock is $mc^2(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1)$ (ref).

Substituting 0.9c and 0.001kg, we get 116 TJ.

As a comparison, the Little Boy nuclear bomb released 93 TJ energy.

Also the result would look similar. No known spaceship material could resist this.

$\endgroup$
  • 1
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – called2voyage Mar 1 at 19:53
5
$\begingroup$

Probably the spaceship is destroyed.

The big problem is that the energy of the rock-spaceship collision is large. So large that even if a trivial amount is deposited into the spaceship, the ship is gone.

A football (I'll assume association football, aka soccer) is about 350 cubic inches or 6000 cm^3. At 2 grams/cm^3 that is 12 kg. 12 kg at 0.9 c is $(\gamma - 1)mc^2$, where $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$, or $\gamma = \frac{1}{\sqrt{0.19}}$, so we are talking 1.3 times the rest-mass of the rock, or $1.4 * 10^{18} J$.

In more concrete units, 400 megatonnes of TNT.

The largest bomb ever exploded on Earth, the Tsar Bomba, was a 60 megatonne explosion. So the energy of the collision is 10x the energy scale of the largest nuclear weapon anyone ever exploded, and that was considered so ridiculously huge it was useless for war purposes.

If even a fraction of a fraction of the energy of the collision is deposited into the space craft, it won't be there anymore.

XKCD's what-if covers this for a baseball, but I'll note that Randall underestimated the energy of the collisions. 1.3 times the rest-mass means that Fusion/Fission are operations that occur at the wrong energy scale; talking about Fusion/Fission at these energy scales is like talking about the chemistry of plutonium during a nuclear blast.

(One of the problems of nuclear chain reactions is getting some of the byproduct of decay to go slow enough to reliably induce further fission. Those byproducts are usually travelling near to 0.1 c, not 0.9 c).

Most of the rest-mass of matter is in the quark-gluon binding of the protons and neutrons in the nucleus. At 1.3 times that value, this means that the impact is going to tear neutrons and protons apart, and you'll get a (short-lived) quark-gluon plasma, which will then precipitate into nearly random particles (photons, antimatter, normal matter, leptons -- just random junk). It acts more like a cyclotron impact than a bomb, but cyclotrons don't throw things the size of footballs around.

About the only saving grace is that at these energy levels, "solid matter" is a bit of a misnomer. Matter is solid and "takes up space" partly because the low energy states are occupied, and the Pauli exclusion principle doesn't leave room for more electrons to fit in that volume of space. But at 0.9c, electrons are ghosts, EM itself is a extremely weak force, and the lack of "low energy states" is not stopping anything.

You'll have to examine the cross-sectional areas of nucleus-nucleus interaction at those speeds. I was unable to find sufficient data on cross sections at those intermediate speeds -- most cyclotrons are higher energy, and "high energy" particles like nuclear radiation are pretty uniformly much lower energy.

Regardless, some collisions are going to happen. The nucleus-nucleus collision will result in both nucleus exploding into fountains of quarks moving at various directions away from the center of momentum of the impact (so, roughly 0.45 c). These "low energy" particles are going to have a much higher change of interacting with either the rock football of the space ship, and their secondary collisions even moreso.

Another chunk of the energy will be emitted as xrays and gamma rays.

So the football will smash through the ship. In its wake, at relativistic speeds, a wave of fission, fusion and hard radiation will blossom out. Chemisty, and hence biology, will suffer an outside context event. Both the ship and football will explode in what looks like a nuclear explosion, the exact scale of which I am not qualified to determine, but somewhere between Trinity and 10x the size of Tsar Bomba.

$\endgroup$
2
$\begingroup$

I think it's fair to say this is not really some thing we can describe with any confidence. One thing is clear: the energies involved are stupendous, as @peterh points out, 1g of material at 0.9c is nuke-scale energy.

Not all of this will be released, and a fair amount of the blast will have a high velocity relative to the ship. Depending on the relative thickness of the ship and the rock, and how rapidly the material disperses: the rock/bits-of-the-ship that interact will leave at a fair speed, and the energy release will happen centred around this velocity. This velocity is important. The higher it is the more energy will be 'carried away' from the stricken ship. Without understanding how the material spread, and what bits of the ship will get caught up, this is impossible to predict.

So maybe we'll be okay...

Well, "No". The numbers are so high, even if impacted zone is small, and the vast-vast majority of the impact is safely left behind, its almost inconceivable that the tiny remaining factor is not enough to cause rapid unscheduled conversion into exotic forms of plasma.

$\endgroup$
  • $\begingroup$ And the nuke-scale energy is just considering the kinetic energy of the motion, not the energy released by the actual nuclear (fusion) reaction that would result. $\endgroup$ – reirab Feb 27 at 17:51
  • $\begingroup$ @reirab, Yes, and that the small rock is really a grain of sand... Let's just say I wouldn't be signing up for that trip anytime soon. $\endgroup$ – ANone Feb 28 at 9:42
1
$\begingroup$

N-N, O-O, and N-O fusion are all endothermic. The thermonuclear fusion happening at the front of the ball will be sucking in energy at a stupendous rate to make heavier elements. This endothermic process is why some stars actually collapse when they reach end of life. I am not sure how this goes into modeling things here.

$\endgroup$
  • $\begingroup$ The KE of the rock is almost as much as its total mass energy. No nuclear reaction is going to make a significant difference. If the rock was made of anti-matter the bang would be a few times bigger, but that's all. $\endgroup$ – Steve Linton Mar 3 at 0:08
1
$\begingroup$

I'd like to add onto Knudsen's answer, by stating that any such consideration of a fusion reaction has to factor in the the composition of both the rock and the spaceship (both its air and structural components). The energy that results from any nuclear reaction is a result of the difference of the output mass from the input mass, where E=MC^2 gives us the energy from that difference. This mass difference is itself a result of a difference in the binding energy between the nucleons of the inputs and outputs. Higher binding energies per nucleon mean lower mass per nucleon.

If your space rock was made primarily (70%-90%) of iron, all fusion reactions involving the iron would be endothermic regardless of the other input product, as iron has the highest nucleon binding energy per nucleon. If we presume then that most of the reactions observed will involve the iron as it is the most abundant element in the rock, the net reaction will be endothermic as well. An endothermic reaction can't cause a chain reaction.

As Knudsen pointed out, there are endothermic fusion reactions that can occur with inputs less massive than iron. I would wager that the composition of most space rocks would not be conducive in general to any sort of exothermic reaction, unless it was primarily hydrogen by mass. Therefore I don't think you need to consider the fusion reaction as a factor in the destruction to your ship (unless the endomthermic reactions slow down the rock enough for it to impart more of its momentum to the ship, but that's beyond the scope of my answer).

$\endgroup$
-2
$\begingroup$

At 0.9c any type of collision would certainly be catastrophic, football sized or otherwise. In a Newtonian model contact/collision is assured with incomprehensible energy released in many forms. An atomic reaction would be even more incomprehensible. But at that speed, the interaction of the atoms in the football would be relative to the football, and likewise for the ship. The relative difference is so high that the atoms of one object would appear frozen to the other. The atoms of one object would appear as open space to the other. Only the subatomic particles of one object would need to pass thru the other - allowing for the possibility that the football passes thru the ship without any damage or energy transfer at all - assuming of course that there is not a reaction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.