This answer on Worldbuilding (to "How to solve the old 'gun on a spaceship' problem?") posits that ISS can easily withstand a bullet fired from a handgun

Our current micrometeorite shielding can protect against the energies of bullets at point blank range, by several times. An aluminium slug of 7mm at 7kms can be stopped by whipple shields. That's a kinetic energy of ~36kJ. A handgun bullet at point blank range is more like ~8kJ.

Now, this is theoretical guess on an SE answer.

Since Space Shuttle (as well as most other space efforts through history) has been - at the very least - of interest - to the military, have any space vehicles or stations that were designed for actual launch been actually tested on how well their walls can withstand firearm-discharged bullets?

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    $\begingroup$ Something to consider regarding the muzzle energy of bullets. A 9mm is less than 700J. A 44 magnum is less than 4200J. A 338 lapua is less than 6700J. A 50 BMG is less than 21000J but is a pretty extreme outlier. The number of 8kJ muzzle energy for a pistol is well beyond any pistol cartridge I know of, even in the large and exotic category. All numbers are source from wikipedia and can be found by searching for a particular cartridge. $\endgroup$ Commented Oct 12, 2016 at 22:03
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    $\begingroup$ high velocity impacts have been tested against the shielding, but as for bullets... why would you need to? Of all the hazards in space that one seems to be one of the more remote. $\endgroup$
    – GdD
    Commented Oct 13, 2016 at 12:17
  • $\begingroup$ @GdD - shuttles, as well as space stations, and earlier space capsules, were all partly military programs. And/or had military variants. So the scenario/hazard isn't as useless to test it seems. $\endgroup$
    – DVK
    Commented Oct 13, 2016 at 12:48
  • $\begingroup$ I doubt that the Apollo LM would have passed such a test, had one been conducted. $\endgroup$
    – user
    Commented Oct 13, 2016 at 13:39
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    $\begingroup$ @GdD Gun on the Soyuz, Gun on the Almaz, Discussion of guns in space $\endgroup$
    – called2voyage
    Commented Oct 13, 2016 at 14:44

1 Answer 1


Not an actual bullet, but researchers fire plenty of things at spacecraft shields to test them. Here's one I found in full Performance of Whipple shields at impact velocities above 9 km/s. They fired aluminum balls in the range of .004g to .3g at 7 to 9.9km/s.

The shield failed against .3g at 7km/s. That's 7,350J or about three times what a typical battle rifle fires (see below).

It held up against .03g at 9km/s = 800J. That's a bit more than a typical pistol round (again, see below).

Before you go concluding a spacecraft is "bulletproof" against a handgun, there's a few caveats.

That's just for the shields they tested

This particular test was comparing some new designs against a baseline shield. Different spacecraft will have different shielding. The same spacecraft will have different shielding over different areas. This just gives a ballpark estimate.

Every hit damages the shield

enter image description here

"Bulletproof" armor isn't. Every piece of armor has a V0 and a V50 rating. These are it's design limits to stop a certain amount of energy 100% and 50% of the time over a certain amount of rounds. Eventually it will fail.

In addition, Whipple shields are designed to take damage to reduce the energy of the impact. A thin layer of metal vaporizes on impact dispersing some of the energy of the impact. This is known as ablative armor. Hit it again in the same spot and the armor will be weakened.

With the amount of damage these shields took from single impacts, I'll bet unloading a clip of 9mm will give you at least spalling if not full penetration.

For more detail on ballistic protection, see my attempt to make Captain America's Shield feasible.

Some of the energy numbers for bullets in the question and comments are WAY out of scale. It's worthwhile to review the energies of bullets. (Note: all numbers are approximate for the typical load. The same round has all sorts of variations in bullet weight and powder load. Barrel length also maters, a shorter barrel means lower velocity means less energy.)

First, typical pistol rounds.

  • 9x19mm Parabellum, the typical "nine" pistol round. 7.5 grams at .4km/s = 600J.
  • .45 ACP, the famous "Colt 45". 12g at .3km/s = 600J.

Now the most powerful pistol rounds.

  • .44 Magnum, "the most powerful handgun int he world" according to Dirty Harry. 16g at .4km/s = 1400J.
  • .50 Action Express made famous by the Desert Eagle. 19g at .45km/s = 2000J.

As you can see, that's no where near the 8000J claimed in the question. These are a full order of magnitude below the velocity of even the low end of an expected micrometeorite strike. That translates into 100 times less energy.

Next, modern battle rifles.

  • 5.56 NATO the standard western rifle round like from an M16, 4g at .9km/s = 1800J.
  • 7.62x39mm used by modern Russian rifles like the AKM, 8g at .7km/s = 1600J.
  • 7.62x54mmR used by the AK-47 (but not modern Russian rifles). 12g at .8km/s = 3600J.

That the 4g 5.56 NATO and 8g 7.62x39mm have nearly the same energy says a lot about how important a small boost in velocity is.

Now WWII era stuff you're not likely to see, but this is the most energetic a bullet gets without getting into specialty ammo and guns.

  • .30-06, the standard US round in WWII. 12g at .8km/s = 4000J.
  • .50 BMG, used by US WWII aircraft and still used today in vehicle mounted heavy machine guns. 45g at .9km/s = 19000J.

4,000J is the biggest bullet you're likely to see fired at a spacecraft... on the ground I guess, but you can go up to 20,000J. These are all moving at less than 1km/s.

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    $\begingroup$ I'd actually expect most bullets to have too little energy for a whipple shield to be effective. Whipple shields are designed, as you show here, to cope with much more energetic impacts, where vapourising part of the shield can occur. I believe they work by vapourising and disrupting the projectile, spreading the impact. I don't think a 9mm packs enough punch to vapourise very light aluminum sheet $\endgroup$
    – Leliel
    Commented Dec 15, 2016 at 22:44

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