this post was submitted on 25 Oct 2024
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[–] [email protected] 16 points 6 days ago* (last edited 6 days ago) (6 children)

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

Air has a density of approximately 1.225 kg/m 3 (0.0765 lb/cu ft)

https://oow-govmil.com/firearms/50-m2hb-qcb-2/

Barrel Length – 45 inches (114.3 cm);

https://en.wikipedia.org/wiki/.50_BMG

Bullet diameter: 12.98 mm (0.511 in)

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/

Bullet Weight: 661 gr

~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

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).

https://www.gd-ots.com/wp-content/uploads/2017/11/M2HB-50-Caliber-Heavy-Machine-Gun.pdf

Muzzle velocity: (M33) 3,050 feet per second (930 meters per second)

Hmm.

https://space.stackexchange.com/questions/15349/how-can-the-delta-v-to-a-specific-altitude-in-earth-orbit-be-calculated

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?

https://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html

Escape velocity: 2.38 km/s

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.

[–] [email protected] 2 points 4 days ago

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.

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