Impacts of high speed objects on various targets

[jseah]

Basically, what happens when a very high speed object hits another thing?
Example scenario is a 1kg iron sphere impacting on a flat piece of metal armour of 1cm, 5cm and 1 meter thickness at relative velocity of ~4000km/s. (KE is roughly 2 kilotons of TNT)
- Does the projectile just explode on contact? Or will it punch through cookie-cutter style?
- How does the damage scale with respect to relative speed and projectile mass?
There was alot of argument about the speed of sound in the armour.

A related question is how a similar object affect a planet like earth?
A high-altitude nuke-like explosion seems most likely for the small 1kg objects, but for the planet case, considering both small (1kg to 0.5tons) and large objects around 5 to 20 tons at the same speed might be relevant as well. (1ton objects striking starships is kinda a moot point, everything is dead)

[Tass]

I don't think this actually needs to be in fictional. It is a real science question.

My guts say at this speed it will behave pretty cookie cutter like. I don't think the ball of plasma formed will have time to spread that much at that speed. Of course the spreading scales with temperature which scales with speed, so maybe.

I still think you will get a roughly cone shaped hole through the armor.

[Soralin]

Well I have this: http://www.youtube.com/watch?v=QfDoQwIAaXg

Although it's a few orders of magnitude slower than what you're talking about.

[WarDaft]

I don't think this actually needs to be in fictional. It is a real science question.

My guts say at this speed it will behave pretty cookie cutter like. I don't think the ball of plasma formed will have time to spread that much at that speed. Of course the spreading scales with temperature which scales with speed, so maybe.

I still think you will get a roughly cone shaped hole through the armor.

I don't think it will be cone shaped exactly.

If the sphere struck an identical and relatively stationary sphere, it would definitely produce a cone-shaped trajectory of plasma. But for a meter thick armor plate, the plasma will encounter much more momentum obstructing it, slowing it and broadening the explosion substantially (which further slows and broadens the explosion which-). If we impact a 1kg iron sphere on a 1 meter cubed block of iron at 4000 km/s, I would be surprised if there were fragments of any significant size left. A cubic meter of iron weights, after all, nearly 8000 kg. A simple cone through it could easily weigh hundreds to over 1000 times as much as the impactor, slowing the exit of the plasma substantially, allowing it to expand far more than you might otherwise expect.

...

There's also the neat question of how fast you would have to throw the sphere to trigger fusion just from the pressure and heat created in the impact.

[jseah]

The reason why I put this in fictional science is that this question only really comes up in starship combat using mass drivers / ramming missiles. At the closing speeds stated, this is going to be fictional for a long time or forever.

While a 1kg iron ball hitting a 1cm thick piece of armour at 4kkm/s relative would generate a cone of plasma behind it, the armour itself would absorb some of that energy. And at 8TJ or so (can someone check my math?), even 0.1% of the energy absorbed would generate an explosion powerful enough to destroy the entire armour plate.

The problem of course, is deciding how much of the energy would go into the armour itself.

What about the planetary impact case? (kinetic missile vs planet; better to fragment into lots of 1kg pellets or stay as a 5ton block?)

[WarDaft]

What about the planetary impact case? (kinetic missile vs planet; better to fragment into lots of 1kg pellets or stay as a 5ton block?)

That one's easier. Does the planet have an atmosphere? If so, do you actually want it to hit the ground? If so, 5 ton block. If not, do whatever. If not, then do you want widespread damage, or to take out a particularly tough target? If the former, pellets. If the latter, a block.

For planets with an atmosphere... people tend to underestimate just how much air weighs, it being completely invisible. The column of air between a projectile and the ground can actually weigh a lot, and no matter how fast you throw the object, it still has to push all of that out of the way. And that's going to make it diffuse and spread, putting more air in its path. If you want a 1kg object to strike the ground, you can't throw it very fast. As in, if you release it stationary at mid earth orbit, it will probably burn up just from the speed it picks up falling unless it's made of heat shield materials.

[jseah]

I have heard it somewhere at 1cm^2 of ground has 1kg of air above it.
Thus a 5ton cylinder with any reasonable cross-section will never reach the ground simply because having a few kilograms of anything hitting it at 40-100kkm/s will destroy it.

Now the question becomes: how powerful does a high altitude airburst have to be to damage things on the ground severely? (by severely, I mean like a nuke)

5tons at 60kkm/s = 18E18 J (can someone check this?)
Is that enough?

[Soralin]

18x10¹⁸J, Well, we can check on the Boom Table, and see that it's somewhere over 3Gtons TNT equivalent, which puts it at more than the energy of a 9.5 earthquake, or a bit over 85x more powerful than the biggest nuke ever detonated, or more than 285000x more powerful than the nuke at Hiroshima.

And, this isn't exploding outward in all directions, it's all from kinetic energy, so all of that energy is going to all still going to be directed largely downwards.

[WarDaft]

It still has the the problem that due to air in the way, it's going to disperse some independent of its speed, and as it disperses it's going to have more air in its path to disperse it even more. It's still going to slam into the ground, but it's going to fan out quite a bit before it does. I still do not have the slightest idea how to actually calculate how much it might actually disperse.

[starslayer]

The energy's high by a factor of ~2 (6E4 km/s is 1/5 c, so relativistic effects are starting to creep in; in Newtonian mechanics, it's by a factor of exactly 2), but that's still ~2 Gt. That's waaaay more than enough to do some serious damage to an extremely wide area, even released very high in the atmosphere. For comparison, the Tunguska blast was ~10 Mt, and that blew down trees over an area of about 800 square miles. That asteroid was in the 10 meter size class and blew up 5-10 km above the ground. This 5 ton cylinder is going to disintegrate much higher, but its momentum is so high, and it's going so fast on impact, that the pieces (such as they are; it'll completely vaporize) would, I think, be scattered over a relatively small area. The explosion when the vaporized remains of the sphere hit the ground would be enormous, though.

[WarDaft]

I guess we can approximate. I'm going to do my best to underestimate at each step, and so give a minimum radius for the impact with the ground.

Suppose it's 5 Mg, 1 meter squared cross section. At 50 km, it's going to encounter 1 gram of air (a bit more really, but close enough) every meter it falls. Lets get a lower bound by assuming the air particles slam into it so hard they get stuck, and just add to the mass (and so keeping as much energy as possible in the impactor.) So over 1 meter, its mass increases by 1 gram, and (using Newton to keep this simple) it's speed drops by just about 1.2 m/s, which represents the transfer of 3.6 TJ. This is, give or take, about a million times as much energy as it would take to vaporize 5 Mg of iron. So we know it's exploded by this point. (We could now make things totally circular and point out that a super high momentum plasma object would not have any elastic response to air and so simply soak it up as it went along, to justify our first assumption that the air molecules just get stuck in it.) Let's assume that it makes it this far before vaporizing however, so we have a 1 meter cross section high density plasma cloud, heading straight down at 60 Mm/s, 50 Km above the Earth's surface.

Let's assume a constant density of 1 gram/m³ for air for the next 10 Km. That means it will pick up 10 Kg of mass, and so slow to about 58,823 Km/s and have a total energy transfer of at least 3.495*10¹⁶ J from the original mass's kinetic energy. There's really no where else for that energy to go but to heat up the object even more, so should have a temperature of 1.56*10¹⁰ and correspondingly should have an internal pressure of 1.8*10¹¹ ATMs. I'm simply not sure how to relate that to its rate of expansion, so here's our biggest approximation yet: I'm going to assume the density of the plasma object is uniform, then divide the pressure by 100 and assume the edge of the plasma cloud is accelerating as if acted on by that much force over 1 decimeter. Reversing the energy calculations suggests that it is now expanding outwards at 19 km/s, roughly 1/3000th of its current speed. A 1 meter cross section cylinder has a radius of 0.564 m. The next 3 km are therefore going to see an increase of at least 1 meter in radius, giving us a new cross section of at least 7.68 m². I cannot see any reason why the expansion of a uniform object under pressure in a relative vacuum (a ratio of 10¹¹ makes the Earth's atmosphere effectively a vacuum by comparison) could not be done by integrating the internal pressure as an acceleration on the outermost material and maintained uniformity, but IANAP so I divided by 100 and applied it only over a small portion of the distance I was hoping for.

Ignoring all the energy picked up between 37 and 40 Km up, we can take our new cross section of 7.68 m² and a new density of 6.3 g/m³. Assuming constant size (to maintain underestimation) the next mere kilometer will result in soaking up another 48 Kg of air, or 144 kg over the next 3 km. The radius should have been at least 2.56 now with a volume of up to 70 m³. Updating our cross section to 20.6 m² and the density of air to 9.9 g/m³, the next 3 Km of air promises to give us at least 600kg more mass. So the mass is now at least 5754 Kg with a downward velocity of 52 Mm/s and so the whole mess has transferred at least 1.18 * 10¹⁸ J from the initial objects kinetic energy into other forms of energy. That's more than 1/9th of it, and we're still up in the trace atmosphere. At this point the volume is at most 70 times our initial object's volume, and we've got at least 50 times as much energy in it, so it should still surely have at least 1/2 the pressure it had before, but with a larger radius, we can consider acceleration over a longer distance. Let's make it 6 times longer (since our radius is more than 5 times what we had before, this should be an even more accurate approximation) so we double the rate of expansion to 1 meter per 1 Km fallen. 31 Km up, the air has a density of 15.8 g/m³ and we have a cross section of 39.8 m².

The next 5 km promise to give us at least 3145 Kg of material. Our net pick up is 3899 Kg. Our current speed at 26 Km up is at most 33711 Km/s, so at least 3.94*10¹⁸ J has bled off from the initial kinetic energy. Our radius is now at least 8.56 m, our cross section is at least 230 m², and the density of air is 34.25 g/m³. We are also no longer falling even close to 60 Mm/s, so we don't even need to consider acceleration due to pressure to increase the rate of expansion with respect to height, we're just not falling as fast. We'll assume the pressure compensates for the increase in material picked which would otherwise slow down the expansion (though it should do much more.) We'll round down a bit and say we expand 10m/6km. Why 6 km? Because we're going down to 20km next. The next 6 km should give us at least 33,465 kg of mass. Our speed is down to 7080 km/s, barely more than 10% of what we started with, and we're only a little more than half way, and still not into the actually dense portion of the atmosphere. At least 8/9ths of the kinetic energy we started with has been bled off. Our new radius is 18.56 meters, our new cross section is 1058 m², our new density of air is 88.9 g/m³, and our new rate of expansion is almost 8 meters per km fallen.

The next 5 Km, we will pick up at least 480,949 Kg of mass. We're down to below 600 Km/s. Our rate of expansion is more than 10 times higher. At least 99% of the initial kinetic energy has been converted to other forms. Our radius is at least 58 m, and cross section at least 10,000 m². The density of air at this height has more than doubled to 194 g/m³. We cannot assume the mass is a sphere at this point, but if it were, it would be no more than 3 times the density of the air at this atmosphere and half the density of air at sea level.

Given all that, it really looks like it's going to be an air burst, but if someone who was more confident of the mechanics could through a bit of math, that would be a lot better.

[jseah]

=| I seriously doubt anyone wants to be near that kind of airburst.

Then again, at 1/5c, would incoming air particles slamming into the 5ton block generate some kind of radiation? I mean something like scattered x-rays or something. Might even generate a small emp.

[WarDaft]

I doubt there would be anything small about it, but I can't even extrapolate to calculate how powerful it would be.

[Tass]

I have heard it somewhere at 1cm^2 of ground has 1kg of air above it.

That is true. It is the same as ten tons per square meter, or a ten meter shield if it had density like water (which makes good sense, it is a well known fact that about ten meters of water increases pressure by an atmosphere).

That is also why it makes a pretty good radiation shield. It is actually impressive how transparent it is to visible light.