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.
~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?
The mean orbital velocity needed to maintain a stable low Earth orbit is about 7.8 km/s (4.8 mi/s), which translates to 28,000 km/h (17,000 mph). However, this depends on the exact altitude of the orbit. Calculated for a circular orbit of 200 km (120 mi) the orbital velocity is 7.79 km/s (4.84 mi/s), but for a higher 1,500 km (930 mi) orbit the velocity is reduced to 7.12 km/s (4.42 mi/s).[10] The launch vehicle’s delta-v needed to achieve low Earth orbit starts around 9.4 km/s (5.8 mi/s).
Depending on T/W and other factors, delta V to LEO can vary. But once you’ve reached LEO, gravity loss and atmospheric drag are no longer factors. So I’ll give you some delta Vs going from a 300 km altitude circular orbit to higher altitude orbits.
300 km to 500 km - .11 km/s
300 km to 1000 km - .38 km/s
300 km to 2000 km - .83 km/s
300 km to 4000 km - 1.51 km/s
300 km to 8000 km - 2.37 km/s
300 km to 16000 km - 3.22 km/s
300 km to 32000 km - 3.83 km/s
300 km to 64000 km - 4.1 km/s
300 km to 128000 km - 4.13 km/s
300 km to 256000 km - 4.02 km/2
300 km to 512000 km - 3.87 km/s
300 km to 1024000 km 3.72 km/s
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?
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.
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 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.
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.
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.
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
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.