[Source for the Art](The source for the art: https://twitter.com/konwashi_2/status/1849000087248376017?s=46)
this post was submitted on 25 Oct 2024
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Once fired, the bullets would just retain their muzzle velocity.
I don't know what kind of muzzle velocity increase would happen -- no air means that they'd be moving somewhat-faster. I'd think that that'd depend on a number of things, probably fluid dynamics and stuff. But maybe you can back-of-the-napkin it by figuring that any acceleration is roughly bounded by the energy required to accelerate the mass of air involved to muzzle velocity. I don't know exactly how much air that is. Certainly the air inside the barrel, but also some of the air outside the muzzle.
https://en.wikipedia.org/wiki/Density_of_air
https://oow-govmil.com/firearms/50-m2hb-qcb-2/
https://en.wikipedia.org/wiki/.50_BMG
So that's about
(.511/2)^2*3.1415*45=9.22850 in^3, or 0.00015122811 m^3, so .00018525 kg, so 0.1g of mass of air in the barrel.https://barrett.net/products/accessories/ammunition/50bmg/
~15.4g per grain, so ~43 grams. So I figure that the mass of the air in the barrel probably isn't a huge factor, and I don't know how to compute the effective amount of air that needs to be accelerated outside the barrel and how much...that's probably a fluid dynamics question .
Let's just say that it's three times that amount of air. Even if so, that's a pretty miniscule factor compared to the mass of the bullet, like under half a gram. So I figure that the muzzle velocity probably isn't all that much higher in space.
Are we moving fast enough to do much in terms of orbit change?
https://en.wikipedia.org/wiki/Low_Earth_orbit
https://www.gd-ots.com/wp-content/uploads/2017/11/M2HB-50-Caliber-Heavy-Machine-Gun.pdf
Hmm.
https://space.stackexchange.com/questions/15349/how-can-the-delta-v-to-a-specific-altitude-in-earth-orbit-be-calculated
So, the International Space Station is at 340 km. So figure that our astronaut is acting as gunner on a vehicle in LEO at 300 km, can use the above table.
The upper end of LEO is 2000 km.
So you've got about 1 km/s in terms of delta V to work with there. So if our astronaut is acting as roof gunner on a Humvee in LEO, that's actually quite a lot of ability to reach. Given sufficiently-accurate aim, and maybe willingness for a bullet to do a sufficient number of orbits before it collides with a target, he's got the range to hit anything in low earth orbit.
He can't hit GPS/Galileo/GLONASS satellites (19k km to 23k km). And he can't hit escape velocity, so that Model S that Elon Musk launched into an eccentric path wandering the solar system is probably safe. But he can hit a lot of stuff.
Okay, the image is of the Moon. How about acting as a gunner there?
https://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html
No. From the surface of the Moon, his bullets are going to come back to the Moon. So he can fight lunar battles, but he can't be engaging targets on other celestial bodies or in their orbit, like the Earth.
Major concern here for LEO engagements is that any shots that miss are a liability coming back to hit the gunner.
Basically once a bullet's fired, a new orbit is defined for that bullet, a new elipse can be drawn. That now elipse is constrained by the position and direction of that bullet the moment it's fired. Unfortunately that means that one bullet orbit later the bullet is going to be in the exact same position with the exact same velocity. The gunner had better hope that orbit phases are misaligned.
Shooting at targets in the retrograde direction might be safest, they're more likely to dip into the edge of the atmosphere and start to lose a bit of velocity ensuring they never come back.
Assuming the M2 has some kind of absurdly precise, even more advanced than a CIWS mechanical aiming mount and traverse system , connected to some kind of orbital trajectory calculator computer system, as well as a system that knows its own precise location, and trajectory and velocity, and the exact location and velocity and trajectory of his target.... (you know potentially on the other side of the fucking planet, moving at about Mach 28 or 30)...
...suuure.
Otherwise, no, the M2 will only be hitting things within visual distance, which are also travelling at a velocity and trajectory pretty darn similar to his own.
Did you compute for air in the barrel after calling out there was no air in space? Just curious about that, s'all.
With that, there would be a hell of a suction on the bullet after the cartridge was fired. Also, the detonation doesn't happen all in one go and continues as the bullet moves to the muzzle. (I did quite a bit of experimenting with that to reduce muzzle flash, actually.) So, the bullet is accelerating until the pressure is released when the bullet passes the muzzle.
While air in the barrel isn't really a factor on earth, surrounding air pressure absolutely is. It affects burn rate most but how it affects burn rate is a characteristic of the powder itself. (In zero-G, I would speculate a slower burn because the powder would be more prone to floating if it wasn't a compressed load.)
When I plan to go to lower altitudes, I typically use lighter loads or I risk over-pressure conditions. (I'll basically just compute for a couple hundred pounds less pressure chamber.)
After all that, I have no fucking clue what would happen in space because the conditions are so wonky.
Yes. What I'm looking for is to try to figure out what the increase in muzzle velocity in space is based on relative to a known muzzle velocity on Earth in air.
I admire your dedication to the math but isn't that all kind of irrelevant? No oxygen means no explosion when the firing pin strikes the cartridge which means your projectile remains stationary.
Edit: I just saw that other guys comment on this very topic. I don't know if he's correct but it's at least plausible enough to make the topic worth exploring. Please carry on.
He's correct. You can fire firearms underwater for the same reason (though you can also destroy weapons doing so, not something to casually do).
https://www.youtube.com/watch?v=OubvTOHWTms