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May 3

relativistic projectile

I'm reading a story in which (among other things) someone is trying to build a gun that shoots 1.5kg iron slugs (5cm diameter) at 60% of the speed of light, for use in space combat.

1) How do I calculate the kinetic energy of the slug? Do I just use ${\displaystyle E/{\sqrt {1-0.6^{2}}}=E/0.8=1.25E}$  where E is the Newtonian approximation? I think that is about as much energy as a 7MT nuclear bomb, if that matters.

2) What happens if the slug actually hits a spaceship? Would it most likely just punch a hole all the way through, without slowing down much? Assuming a large enough ship to self-seal around the holes, is that all that effective a weapon? I.e. the ship is USS Enterprise size or larger.

Thanks. 2601:644:8581:75B0:3DAF:465A:7AA1:65A0 (talk) 04:04, 3 May 2025 (UTC)[reply]

The article Railgun may be helpful.-Gadfium (talk) 04:30, 3 May 2025 (UTC)[reply]

The total energy is ${\displaystyle {\frac {mc^{2}}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}$ . Subtract from it the rest energy ${\displaystyle mc^{2}}$ . See kinetic energy (there is a section on the relativistic generalization). Icek~enwiki (talk) 06:12, 3 May 2025 (UTC)[reply]

Using the Lorentz factor

${\displaystyle \gamma ={\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}},}$ 

this difference can be written compactly as ${\displaystyle E_{\text{kin}}=(\gamma -1)\,mc^{2}.}$ 

When ${\displaystyle v=0.6\,c,}$  we have ${\displaystyle \gamma -1=0.25.}$  The energy required to get the projectile up to speed is at least equal to the kinetic energy it gains, about 33.7 PJ. For comparison, the energy released by the Trinity nuclear test was about 0.1 PJ. BTW, the material composition of the slug is immaterial; it might as well be a canister of elderberry preserve.  ​‑‑Lambiam 08:39, 3 May 2025 (UTC)[reply]

That would depend on the spaceship. Is the hull made of unobtanium? Are there shields of some sort? Clarityfiend (talk) 11:07, 3 May 2025 (UTC)[reply]

At this speed each iron nucleus will have energy of about 15 GeV, which will absorbed by the material the hull is made of. This with result into fireball of high temperature plasma exploding inside the ship. Ruslik_Zero 20:07, 4 May 2025 (UTC)[reply]

Hmm, will it, though? This isn't a single nucleus; it's a big ol' hunk o' metal. The spaceship atoms in the way of the slug atoms are going to be getting out of the way in a hurry. How much they interact with the rear portion of the slug strikes me as a fairly difficult simulation problem that depends on a lot of details that haven't been specified, but I can imagine a fair portion of the energy being carried out the other side of the spaceship, still as kinetic energy.

I certainly agree that it isn't going to be good for the spaceship, but if the question is whether we're wasting energy that isn't going into the kill, I don't think we can answer that with the information given. --Trovatore (talk) 20:27, 4 May 2025 (UTC)[reply]

Thanks everyone. Yes the ships have deflector shields that use antigravity generators of some kind, which is why such fast projectiles are needed. The shields can handle the impacts at 0.1c which is why they are working on getting the speed up to 0.6c. If the defector shield means the slug's energy is transferred to the shield though, that's probably worse than just punching a hole in the ship and coming out the other side. 2601:644:8581:75B0:C710:F116:861:28C5 (talk) 09:30, 5 May 2025 (UTC)[reply]

Hmm, I wonder whether the articles Stopping power and Terminal ballistics might be of any use to answer your question? 2601:646:8082:BA0:C887:6F01:C269:367F (talk) 13:07, 5 May 2025 (UTC)[reply]

You'll also have fun with the momentum. Typically, naval guns are about 100 times heavier than their projectiles. Assuming this holds for this relativistic gun, it would have to move back at 0.6%c, or 1800km/s. That's about Mach 5000 - talk about hypersonic weapons ;-). --Stephan Schulz (talk) 22:49, 16 May 2025 (UTC)[reply]

Why is GP-A not on File:Time_Dilation_vs_Orbital_Height.png total time dilation curve?

(CC:@Prokaryotic Caspase Homolog:)

 

Time dilation as a function of orbital height relative to a stationary observer on Earth

In the attached thumbnail, GP-A which is presumably Gravity Probe A is placed on the gravitaional time dilation graph instead of the total time dilation one like the others. Is there a reason for it? Thanks, cmɢʟee⎆τaʟκ 22:33, 3 May 2025 (UTC)[reply]

GP-A did not orbit. It was launched nearly vertically, reached 10,000 km, and came back down. —Amble (talk) 02:54, 4 May 2025 (UTC)[reply]

Thanks. That explains it. Cheers, cmɢʟee⎆τaʟκ 19:13, 4 May 2025 (UTC)[reply]