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Author Topic: Collisons at kinetic energy > binding energy  (Read 1730 times)

Only2ndplace

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Collisons at kinetic energy > binding energy
« on: February 05, 2018, 09:06:49 AM »
I've been testing around a bit with shooting the Great Pyramid of Gizah at earth at varying speeds and then had the idea to shoot it with a kinetic energy greater than earth's gravitational binding energy to see how much it would fragment (if the energy is distributed equally among all pieces of the planet it has to be ripped apart).

The binding energy of a body can be calculated via:
E = 3/5*G*M^2/R, where G is the gravitational mass and M and R are the mass and radius of the object respectively. Comparing this with the kinetic energy of a projectile of mass m, it must at least have a velocity of:
v = sqrt(6/5*G*m*M^2/R)

If I didn't put in the numbers wrong, this leads to a speed of 5.7*10^13 lightspeeds for the pyramid (non-relativistically, obviously).

Since the lower limit of time steps in US^2 seems to be 0.01 ms/sec, you can't really catch what's going on in collisions of these speeds at the point of impact, but so far I've had two outcomes:

Once, one side of the earth heated up and it shot of at a few percent lightspeed, with no visible fragments (you can check for fragments with STRG+A).
Two other times, earth was not affected visually at all, but lost exactly 10% of its mass, which was distributed over 10 fragments, that shot of at 10^12 lightspeeds.

I'm not firm in collision or penetration physics, but I think Newton's law of penetration states, that an objects penetration depth does not depend on its speed, but only its density and the density of the target material. I think for equal densities (which is pretty much the case here) the penetration depth is just the bullet length.

In that case, the second outcome seems somewhat plausible, at least in the way, that not the entire planet is destroyed, as only a small percentage of it would receive almost all of the energy.
On the other hand, there should obviously be some sort of crater and probably more fragments, but I think there's in general very different physics between such insane high speed collsions and smaller ones, which can be described with the mentioned penetration law. (That, and there's also not really a lot of data on planet destroying pyramids I guess)

Anyway, no screenshots, cause there was not a lot to see, but maybe you can try out for youself. Maybe even with some other planets/stars and overcoming their binding energy. Let me know if there's anything cool you find out.

Physics_Hacker

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Re: Collisons at kinetic energy > binding energy
« Reply #1 on: February 05, 2018, 03:46:05 PM »
0.01 ms is definitely not the lowest timestep you can go to, I've slowed it down so much before that I was able to put something going probably more than billions of times c and was still able to watch just fine. Also, the weird effects you're getting might be because you've only had it set to 0.01 ms which is huge when considering things going so much faster than c.

Only2ndplace

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Re: Collisons at kinetic energy > binding energy
« Reply #2 on: February 06, 2018, 07:01:14 AM »
Definitely a good point about the time step and the accuracy of the simulation, though I still don't think it would simulate the collision accurately either way. However, it could explain why there were two different outcomes.

Okay, so I fiddled around a bit, and I can't get it to go much slower. Putting in time steps manually does stop at 0.01 ms, at least in the current build. Curiously, one can get down to about 10^-5 ms by using the slider, but that's still way to big as the pyramid crosses an earth radius in times of about 10^-13 ms. I also tried using the timer function, but when you put in such low numbers it just jumps to 0 as soon as you press start and doesn't advance the simulation at all.

That's why I just tried the simulation with the lower limit of 10^-5 ms, which reproduced the first result, where the earth retains its mass completely and there are no fragments. I took note of the exact speed this time: 6% c, which checks out with conservation of momentum. Also, advancing the time steps to 10 s I could this time also see the slow spreading of the shock wave and melted surface and noticed, that earth's rotational period has sped up to about 10 s, so I'm assuming angular momentum was conserved as well.

As a bonus, there now appears to be an incorporeal shadow hanging over earth, even though there are no other objects in the simulation (see GIF).

Physics_Hacker

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Re: Collisons at kinetic energy > binding energy
« Reply #3 on: February 06, 2018, 01:42:10 PM »
It seems very bugged out with the shadow thing, maybe it's due to your PC, I don't know.

Okay, so I slowed the simulation down to 6.91 10-9 seconds before it jumped to 0, put in the number of times c you gave as the velocity (5.7 1013) and I got a pretty similar result except Earth is still visible, half the planet is molten, and after a certain point the timestep indicator goes completely red and the simulation just stops.

It might be best to stick to planets slamming into planets with more than their binding energy.