What collateral damage would a mach 23 bullet cause?

Question

In my sci-fi story there is a man-portable railgun. The gun fires a 4mm-wide tungsten bullet with a muzzle velocity of Mach 23. The inventors of this weapon could have made the projectile faster but were concerned that doing so would cause it to go into orbit and didn't have enough time to fine-tune the velocity. What I'm most worried about is the effects on the nearby atmosphere that such a weapon would cause not only to the individual firing the weapon but also anyone he's fighting along side. Can anyone weigh in on this?

Answer 1

You had it right with the 2mm projectile before you edited.

A 20mm wide projectile would have far too much recoil for a man-portable weapon. Let's say it's 30mm long, since it needs to be longer than it is wide. That would give it a mass of 181 grams. By conservation of momentum, if 181 grams goes forward at mach 23 and you weigh 100 kg, you instantly go backwards at 50 km/h. Or in other words, your shoulder is meat pudding and you're going to have to go find where the gun went.

There's a reason why man-portable railguns are often portrayed in fiction as "needlers." If you shoot something much bigger than a needle, you couldn't handle the recoil.

So, 2mm. Say the projectile is roughly a cylinder, 10 cm long and 2 mm diameter. It weighs 6 grams. If 6 grams goes forward at mach 23 and you weigh 100 kg, you go backwards at 1.7 km/h - you'll be knocked a step back with every shot, but it's manageable.

First, would the 2mm projectile vaporize? The short answer is yes - but not before it could hit a target at 100m+ range.

Based on this article, the drag coefficient C_V settles around 0.6 for a properly shaped hypersonic object. Use the formula here to calculate the drag force: 1/2 * (density of air) * (mach 23)^2 * 0.6 * (frontal area of projectile). That gives us 22N of drag force.

Next we need to know how much energy it takes 6g of tungsten to melt. Tungsten melts at 3695 K, so from room temperature we'll say it has to heat up by 3400 K. But once it gets to 3695 K it doesn't instantly melt; there's something called the latent heat of fusion, which means you have to keep dumping more energy into it while it's at 3695 K before it melts. Adding up those effects, it will take 3847 J to melt the tungsten.

So, the drag force is 22N, and we want to know when it has done 3847 J of work on the projectile. Work = Force * Distance, so Distance = Work / Force. 3847 J / 22N = 175 meters. This means that by the time the projectile has gone 175 meters, it could be completely melted, because enough kinetic energy has been dissipated to melt it, if all the lost kinetic energy went into heating the projectile.

In fact, it won't be melted by 175 meters, because much of, perhaps almost all of, the lost kinetic energy will heat the air instead of the projectile. So the projectile could have a range much greater than 175m, especially if it was designed with ablative materials in front of the tungsten, like the Apollo re-entry module.

However, I will proceed on the assumption that it does melt in 175m, simply because it is very difficult to determine how much of the lost kinetic energy actually heats the projectile. It could go ten times that distance. But it will at least go that distance.

It will melt from the front towards the back, like a candle, over those 175 meters. (Presumably it has fins or something to stabilize it through this process.) We can assume that the melted tungsten will immediately stream away from the solid projectile by the force of the air. Once scattered in the air, the melted tungsten will have a much larger frontal area, so its air resistance will be much greater and the liquid will almost immediately come to a stop. As the liquid tungsten stops, it will deposit its energy into the air.

So that means that the high initial energy of the projectile will be dumped into the air at a linear rate, along the entire 175m path of the projectile.

The initial energy of the projectile is 184 kJ. So it will deposit about 1.05 kJ of energy per meter of its flight.

For comparison, an M80 firecracker has around 12 kJ of explosive energy in it. So the projectile will release 1/12 of an M80 firecracker for each meter it travels. That will be loud, but not too dangerous; it won't hurt the operator of the railgun.

Because it is continually melting, the damage the projectile does when it hits depends on the range. If you hit an enemy at close range, it will release most of the 184 kJ, which will do damage comparable to a hand grenade. By the very end of the 175m range, it will be more like a rifle bullet.

Answer 2

2mm projectile?

It's turned to plasma near instantaneously and unless the round is very long, the lack of mass means that it's unlikely to have enough energy from being turned into plasma to do any meaningful damage to anything nearby.

And yes, I do include downrange in that statement.

Speed is good, but mass is also important. I'd suggest increasing the round size to something more like 20mm - and using the friction and speed as a limiting factor: In a standard atmosphere it can travel X distance before it is totally consumed by vaporization.