I would like to better understand intragalactic ship design (aka, "spaceships").

  • It's travelling at a barely tolerable 0.05c (86 years to Alpha Centauri).

  • It happens to impact with a baseball (145 grams). (In reality it would impact with rocks. We believe there's nothing out there other than inconsequential dust, but the reality is we don't know because we can't detect anything this small, so assume I'm right, please.)

  • Maximum material thickness: 100 cm.

  • For the purposes of this question, whether or not the material will be dented is ignored, so long as it is not pierced or burned away.

I believe the basic equation is Joules = 0.5 * mass(Kg) * velocity(m/s)2. In which case, the baseball hit my ship with 16 TJ of energy.

Question Is there a material today that could act as the forward plating of my ship that could withstand such an impact?

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    $\begingroup$ At that speed, I believe relativity must be taken into account. The equation for energy is right but mass must be adjusted. $\endgroup$ – Renan Apr 20 '18 at 16:54
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    $\begingroup$ You're correctly calculating the kinetic energy of the baseball (relative to the ship). But incorrect to assume all the energy is going to be released into the ship. You may see this if you flip the frame of reference and ask if a baseball will survive being hit by a ship of that mass and speed. A partially inelastic collision will occur and it will likely be at least two dimensional. $\endgroup$ – Samuel Apr 20 '18 at 17:10
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    $\begingroup$ No. Proposed relativistic spaceship designs include some form of a "shield" that travels in front of the main ship and absorbs impacts. $\endgroup$ – Alexander Apr 20 '18 at 18:14
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    $\begingroup$ related XKCD: what-if.xkcd.com/1 $\endgroup$ – CaM Apr 20 '18 at 18:34
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    $\begingroup$ It's worth noting that interstellar space is in general extremely empty and hitting a baseball may be a very rare calamity rather than a regular occurrence. Like a meteorite hitting an airplane or something. (I'd put this in an answer if I had numbers instead of a gut feeling) $\endgroup$ – Elukka Apr 20 '18 at 20:32

10 Answers 10


 Is there a material today that could act as the forward plating of my ship that could withstand such an impact?

The problem is the same with the rock hitting the ship while the former is traveling at .05c and the latter is standing still. We may use Newton's impact depth equation giving penetration depth of a projectile P in a shield S, $depth = length_P \frac{density_P}{density_S}$: at that speed, the barrier will act as a liquid and the rock, having a length of 5 cm and a typical density of 3.5 g/cm^3, will penetrate to a depth of about 17.5 cm in water (this excludes penetration-optimized shapes). At that point it will have ceded all its considerable momentum to the surrounding material, converting a significant fraction of it into compressive and friction heat, and it will explode.

In the volume of impact a plasma jet will form, still possessing a considerable momentum, and will start penetrating inwards; since the elementary momentum is given by the product of density by speed, and it is a finite quantity, the greater the density the lesser the speed.

So you want to have the densest possible material (which would be osmium, density of around 22) for the first 20-25 cm, then you need to survive the explosion of ~240 tons of TNT and temperatures briefly in excess of several hundred thousand K, plus the jet of osmium plasma that has absorbed the momentum of the impactor. This calls for some superrefrigerated phase-changing metamaterial (mostly ice-XI) and some way of distributing the impact laterally as quickly as possible, which calls for an enormous Young's modulus - basically a carbyne layer.

Finally you need to consider spallation. The shock wave will travel through the armor, and blast the opposite side even if the projectile doesn't push all the way through. You need a further layer of high density, high tensile material to block that.

I'm not too sure that all of that is going to fit in a 100cm thickness...

We overlooked something!

The thickness of 100cm refers to an impactor hitting head on. But if we build the shield as a sloped, conical glacis - a vacuodynamic shape - we reap some very important benefits:

  • the collision will be at an angle, thereby wasting a large part of its energy into a shower of fragments taking away most of the momentum harmlessly.
  • the penetration path will be increased by the inverse sine of the slope angle; an angle of 30° will immediately double the thickness of the material as seen by the impactor.

I think we can do this! :-)

Charged Whipple shield

A standard Whipple shield will probably not fare well against solid objects in the hundred-gram range.

But we can imagine a cubic lattice of osmium pellets connected by very strong insulating threads (e.g. Kevlar) no more than three or four centimeters long horizontally, and a dozen meters vertically; the lattice itself is as wide as the ship's front.

At takeoff, the lattice is folded and is only some centimeters thick. Then we start pumping electric charge into it (somehow). Coulomb repulsion starts driving the pellets away one from the other, until they form several layers of four-centimeter square mesh, separated by a dozen meters of empty space. When the charge is high enough, the lattice becomes increasingly rigid.

Now a 5-cm rock comes in at .05c relative speed. It impacts on one, possibly two pellets of the first layer, and explodes, forming a cone of debris still traveling at .05c. It has also absorbed a lot of electric charge, and therefore each debris particle is strongly repelled by all the others - which contributes to the cone's expansion - and by the incoming subsequent layers, which both expands the cone and slows it down. We can't pack 16 TJ of Coulombian potential in one hundred twenty meters' worth of lattice (or can we?), as the lattice would start discharging by emitting charges into space faster than we could replace them, but sure we can make it behave like a sort of electric reactive armor.

In the end, the final layer of the shield only needs to be able to deal with small-size buckshot; a layered sandwich of high-density material to absorb momentum, high-tensile material to diffuse the shock and vacuum to stop P-waves will suffice.

When slowing down at arrival, the lattice is powered down and folded back.

It's true that the total installed thickness is two orders of magnitude greater than your requirement of 100cm, but its equivalent thickness might well fall under that.

Protecting against impactors

Protecting against impact will resemble that game called Missile Defense, with the impactors arriving at a relative .05c. enter image description here

But you cannot use missiles. What you do is saturate the space in front of the ship with millimetric radar, which will give you a low-noise estimate of the incoming impactors and something about their nature. It's reasonable to expect detection at about 500-600 kilometers, maybe more (high vacuum, few disturbances). You will use several radars to immediately gauge position and speed of incoming projectiles through parallax and Doppler shift (and also for redundancy). At 500 km distance traveling at .05c you have a warning time of about 30 milliseconds.

You can't safely swing a weapon mount in that time. So you use a massive phased laser array instead, to direct the equivalent of a focused megawatt of power from a supercapacitor bank into the rock, which can be expected to shatter.

This robs the inbound projectile of perhaps one percent of its energy, but importantly it reduces its size, and penetration is proportional to that. It also weakens its structure, increasing the chances that a glancing impact will remain just that - a glancing impact.

At the same time, the relative speed imparted to the impactor will be directed towards making the impact angle shallower, further reducing the damage.

If there's enough time, supplemental strikes could further reduce the damage by pulverizing the most threatening fragments.

  • $\begingroup$ Wikipedia seems to say that Newton's impact formule only holds if the projectile holds together. Thats good for you, since a projectile already disintegrating can give it's momentum to more shield material. $\endgroup$ – b.Lorenz Apr 21 '18 at 14:38
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    $\begingroup$ But your angled armor solution is cheating: It maintains 100cm thichness, but it multiplies the "kg of armor needed to cover 1 mm2 of usefull cross section" metric just like a non-angled, thicker armor would. (And you can not count on boucing it aside at this speed) $\endgroup$ – b.Lorenz Apr 21 '18 at 14:41
  • $\begingroup$ @b.Lorenz "if you don't cheat, you're not trying hard enough" :-) - but yes, with enough density differential the projectile sort of slides aside. It will gouge deep furrows, of course. And yes, the weight is probably an issue. $\endgroup$ – LSerni Apr 21 '18 at 16:53
  • $\begingroup$ @cmaster that wouldn't be a small nitpick - the whole system would never work! I was thinking of the beads as connected by insulating carbon nanotubes, charged one by one when detaching from the ship; but I've just google-checked, and apparently this is something that needs extra care: arstechnica.com/science/2009/03/… . Of course, metal-like nanotubes would not work. Amended answer. $\endgroup$ – LSerni Apr 22 '18 at 21:42
  • $\begingroup$ @cmaster done. I was under the impression that you could have insulating nanotubes, but you're right - it appears that while you can, the nanotubes you get lose most of their tensile strength (and the tubes aren't really good insulators). At that point Kevlar is too much better. $\endgroup$ – LSerni Apr 23 '18 at 6:23

You're going at this wrong. Your defense material is aluminum foil or something of the sort.

Once your ship is up to your .05c you spread out a sheet of aluminum foil and push it ahead of you. Periodically launch another one.

You're no doubt saying the the baseball will go right through and all but ignore it. Here on Earth that would be true, but look back at the energy of your impact. It's not 16TJ because for practical purposes it's the aluminum foil hitting the baseball. It's still a lot of energy, though. Yes, the baseball is hardly slowed by this--but it is vaporized. You now have a very rapidly expanding ball of plasma. That's much easier to stop and very well might be dealt with by subsequent sheets and certainly can be stopped by your hull--it's just a dense pocket of pretty low energy radiation at that point.

I've also seen dust suggested for a similar use but it's much harder for the engineering department to verify the protection offered by their shield that way.

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    $\begingroup$ This is... something I've never heard before. Do you have any citations regarding this technique? I would like to learn more about it. It seems really interesting! $\endgroup$ – T. Sar Apr 21 '18 at 8:42
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    $\begingroup$ It's just an extreme version of what the Space Shuttle did: two thin layers separated by an empty gap. Except here the energies are greater, so more layers used to disperse the impactors. @T.Sar $\endgroup$ – Nij Apr 21 '18 at 9:09
  • $\begingroup$ @Nij What the space shuttle did is what is explained on Arcanist's answer a bit down below. This is a bit different - this isn't a passive shielding, this idea is pushing forward a Whiple Shield of sorts ahead of the vessel. That said, the idea postulated on this answer is pretty much useless for slower-moving projectiles. $\endgroup$ – T. Sar Apr 21 '18 at 11:15
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    $\begingroup$ I think the idea of "slower-moving" is perhaps inapplicable when discussing nonnegligible fractions of c. As I said, an extreme version of it - silly C20th ships limiting themselves to a single enclosed hull... $\endgroup$ – Nij Apr 21 '18 at 11:23
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    $\begingroup$ its called a whipple shield en.wikipedia.org/wiki/Whipple_shield $\endgroup$ – John Apr 21 '18 at 12:17

Unlikely, but it doesn't necessarily need to

Existing spacecraft don't need to deal with relativistic speeds, but they still move incredibly fast. And while satellite-destroying meteors are rare, micrometeors are pretty common. A meter of shielding is far beyond the budgets of any spacecraft we currently build.

Instead of trying to use thick solid shields, modern spacecraft rely on Whipple Shielding. A Whipple Shield is a spaced shield - it has two layers with a gap between them. The first layer intercepts obstacles, but it isn't intended to stop them - its purpose is to disperse the energy over a larger area.

Image from Hypervelocity impact performance of open cell foam core sandwich panel structures by Shannon Ryan et al.

Even as the first layer gets perforated, the holes are small and don't significantly compromise the layer's structural integrity. The second layer isn't penetrated by the more diffuse impact, and maintains an airtight seal.

There are, of course, many variations on the design that Fred Whipple proposed in 1946. You can use more than just two layers - that way if the energy is not sufficiently dispersed by the first impact, it can be dispersed several more times before risking your final layer. You can use flexible materials such as Kevlar instead of rigid aluminum. But the basic principles remain the same. By using multiple layers separated by gaps you disperse the impacting energy and make it possible to protect your ship with a great deal less shielding than you would need with a solid shield.

  • $\begingroup$ This reminds me of battle tank armor. $\endgroup$ – RonJohn Aug 28 '18 at 0:47

Your goal is to let this hypothetical material dissipate 16 TJ with just 1 cubic meter of material per square meter of impacted surface.

That amount of energy is comparable to 1 kiloton of TNT: smaller than Little Boy, but still quite a lot.

If you want to stay compact (meaning few square meters of shield surface) I think there is no material which can fit your purpose.

P.s. the formula for the relativistic kinetic energy is

$$E_k = m\gamma c^2 - mc^2 = \frac{mc^2}{\sqrt{1 - v^2/c^2}} - mc^2$$

  • $\begingroup$ Why assume a perfectly inelastic collision would occur? $\endgroup$ – Samuel Apr 20 '18 at 18:13
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    $\begingroup$ @Samuel, the shock wave of the impact would destroy the impactor and the impacted surface way before any shock wave could propagate into them. $\endgroup$ – L.Dutch Apr 20 '18 at 18:52
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    $\begingroup$ As a rule of thumb relativistic effects become meaningful after about 0.5c. For 0.05c in question β = 1.0012 making Newton good enough to about 3 s.f. $\endgroup$ – Maciej Piechotka Apr 22 '18 at 19:26

There isn't. The energy will make any solid act like a viscous liquid - doesn't matter what the material is.

The practical way of passive defense is Whipple shields - multiple layers of material with a lot of spacing.

The layers explode the impactor plus the shield into pellets, spacing lets the pellets spread out, each successive layer of material spreads them into smaller pellets. So there are optimal layer and spacing thicknesses.

Any Whipple shield is inferior to a solid block per unit depth, but superior per unit weight. Very roughly, a single shield layer with mass loss matching the penetrator's mass should halve its effective energy. So, at 0.15c, and at 1 km/s as tolerable impact for the final armor, you should be looking at 31 layers.

Given your geometric parameters, ~0.02 m^2 initial impact area can be counted on, so you need a 7.2 kg/m^2 density per layer. In other words, 3 mm aluminum. Add 200+ mm behind each layer and stack 31+ such layers before your main armor, and you should be covered.

I's more than 1m geometric thickness, but it's just 100mm of total material thickness, plus whatever it takes for the final armor to stop 1 km/s impact. The main armor is subject to conventional mechanics and just another 100mm would do. But 1m of ceramic and metal would not be unreasonable for the front of a large colony ship.

  • $\begingroup$ The total armor thickness (including the spacing) is 6 m, isn’t it? $\endgroup$ – prl Apr 22 '18 at 1:39
  • $\begingroup$ A little over 6m, yes. I don't see it doable in 1m, without active defenses (which should definitely be there, but better to have two ways to survive). It's a bare minimum and the impact would leave a large crater, but that's the consequence of maximum efficiency (not overengineered) design. $\endgroup$ – Therac Apr 22 '18 at 3:39
  • $\begingroup$ "Any Whipple shield is inferior to a solid block per unit depth" - I'm not quite convinced of that. Do you have a source? $\endgroup$ – JimmyB Apr 23 '18 at 11:43
  • $\begingroup$ Not that I could cite. It's just that in any calculation for the shield, thicker layers are better, just not as good as more layers. Although, on second thought, the above shield is insufficient; I'm sure nothing less will work, but it ignores some of the more complex effects. $\endgroup$ – Therac Apr 23 '18 at 16:48

The easiest seems to be to destroy the rock before it hits the ship.

Once the ship has reached cruise speed, the engines are stopped, and it now travels at constant speed. At this point you need to spray a fine mist of water, or perhaps dust, in front of the ship, and let it expand until it fills a sufficient volume.

It can't be a gas, as it would expand very quickly (unless the ship is heavy enough to have its own atmosphere). I'm not sure water droplets would stay solid and not sublimate, I'd rather use dust, ie small solid particles. Feel free to pick the material you want. Perhaps the dust cloud can be held into a useful shape by using magnetic and electrostatic fields, if the ship has its own magnetic field, a bit like the Van Allen belt...

Now, if your baseball sized rock enters a cloud of dust (or gas) at relativistic speed it will be vaporized almost instantly, and it will turn into expanding plasma. The advantage a cloud has over an aluminium foil as suggested above is that the more the plasma expands, the more material it will encounter, which will spread it even more. Hopefully it spreads over a wide enough area to bring the energy density per square meter down to a manageable value for the hull of your ship. Also a significant proportion of the kinetic energy would be turned into radiation (ie, light) and radiated in all directions (thus only a very small proportion would hit the ship).

Another advantage relative to the aluminium foil is that the foil isn't see-through.


Dark matter.

Dark matter might not be interactive after all

Dark matter interacts with other matter only via gravity and with electromagnetic radiation not at all. So far no-one has been able to detect an impact from a particle of dark matter on a particle of normal matter. It may be phantomlike, passing through normal matter without a touch.

But for slowing a fast moving object, gravity works great. Your baseball will pass through your front shield of dark matter and be slowed gravitationally after it passes the halfway point, where the net gravitational pull is the opposite of the moving ball's momentum. Eventually it will be slowed to a stop. This will require a tremendous amount of dark matter and it will be tricky to contain any amount of a substance interested only in gravity - hopefully your future tech includes artificial gravity. Fortunately there is apparently a tremendous amount of dark matter for the taking, and the OP only requires "withstand"; once in place your planet-sized chunk of dark matter will be impervious to impacts of any sort and good for the duration.

A big question and one I cannot find an answer for - what happens to the energy of the baseball decelerating into dark matter? If normal matter falls towards a black hole, gravitational potential energy is converted to heat and shed as radiation. If dark matter can fall under gravity but is unable to shed energy as radiation, what happens to that potential energy? Does dark matter get irreversibly hotter?

For your question it is a little easier because the baseball can get hotter and certainly will, dissolving into plasma long before its momentum has ceased.

  • $\begingroup$ Dang! Dark horse contender gets the bounty! Thanks @JBH! $\endgroup$ – Willk Sep 4 '18 at 21:21

In one of the Star Trek books, the author describes the crew trying to figure out how to destroy a planet-sized machine, at all cost. One idea is to accelerate the ship to close to light speed and ram. Computer analysis determines (according to the author) that the outcome would be a Startrek sized hole going straight through the planet, with not much actual impact.

Maybe the same would happen with a baseball or rock at 0.05c: If you don't try any shielding, it might just go straight through the ship and leave a baseball sized hole. You might just accept the possibility, make sure that you can fix any air leaks very, very quickly and hope you can repair the damage.

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    $\begingroup$ I'm not sure about this one. Upon impact with the planet the ship would immediately cause the planetary material to convert to plasma, that plasma has to go somewhere, and you're converting more as you go through the planet. Methinks the result would be a new asteroid belt. The obligatory XKCD listed above and linked here is more believable, and that's just the impact with atmosphere. I don't think the author of the star trek book did the math. $\endgroup$ – JBH Apr 23 '18 at 0:22

Maybe Neutronium? It's 4 * 10^17 kg/m^3... 13 magnitudes denser than anything else.

Only problem is keeping it that dense.

Or maybe use black holes as shields, to just absorb the impactor, but a baseball-sized black hole, is already too big to be useful, as its gravitational effect affects the movement of the ship too much to compensate for.

And both of those methods are out of reach of current science, while possibly being the only things that could withstand the impact you're subjecting the spaceship to.

  • $\begingroup$ We don't have the ability to make either neutronium or black holes today. $\endgroup$ – Mark Apr 21 '18 at 18:51
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    $\begingroup$ You can't use a black hole to shield off anything: 1. They are really small for their weight (good luck trying to hide a spaceship behind a planet-weight black hole...), 2. they might absorb some stuff, but they'll also attract any stuff passing closely, leading to a zone of increased danger right behind them, 3. they are a bit hard to manipulate (you can only use gravity, and your ship is behind the black hole, so into which direction does it accelerate? Bummer...) $\endgroup$ – cmaster Apr 22 '18 at 21:16

In extremely high energy cases like this, the Newton penetration equation applies. The amount of energy spent accelerating the target material out of the way will greatly overshadow any contribution by strength.

Newton's penetration equation is - penetration depth (s) = (density of projectile / density of target material) * length of projectile.

For a 400 kg/m^3 leather baseball about 0.15 m in diameter impacting some 8000 kg/m^3 steel, the depth of the impact crater in the target material would be 0.008 meters. For a 1000 kg/m^3 ice cube also 0.15 m long, the impact crater would be about 0.02 m deep.

i think the takeaway is that anything not close in size and mass to your ship will be annihalated. However, your hull will get eaten away over time by these collisions.

Although its a popular device, force fields would have to be terrifically powerful, as you have already calculated, to produce the same benefit as a sheet of lead or steel.

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    $\begingroup$ Why do you think Newtonian equation applies in extremely high energy cases? Newtonian equation only tells where the momentum would stop, but not what damage the participants would take. $\endgroup$ – Alexander Apr 20 '18 at 18:09
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    $\begingroup$ See @LSerni's answer above. Newton's approximation is all about momentum, it says nothing about the energy. What you have is an enormous amount of energy released as heat in a very small volume. $\endgroup$ – AlexP Apr 20 '18 at 19:55
  • $\begingroup$ The Newton equation has been tested for estimating the depth of meteor impact craters. I do not have a reference handy, but I believe wikipedia has an article of Newtons penetration equation that discusses using it to estimate meteor crater depth. $\endgroup$ – James McLellan Apr 20 '18 at 19:57
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    $\begingroup$ Two other points - you can formulate Newtons equation as an energy equation. second, at these velocities energy accounting will be horribly messy. Phase transitionss - solid to liquid - liquid to gass - in both the impactor and the target material will take a lot of energy out of the system. $\endgroup$ – James McLellan Apr 20 '18 at 21:36
  • $\begingroup$ The Newton penetration equation doesn't apply here. You'd be better off using Lanz-Odermatt, though it's also horribly misapplied. $\endgroup$ – Therac Apr 21 '18 at 13:35

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