An Impact at a Relativistic Speed

[Corkeyno2]

I am making a video about the dangers of travelling through space at high velocity. Say there was a spaceship travelling towards a stationary rock of a few kilograms, and they impacted, I want to calculate the energy released. Does anyone know what formula I could use? Any advice would be greatly appreciated.

Thanks,

Corkey

[Thalestris24]

Here you go:

https://www.boundless.com/physics/textbooks/boundless-physics-textbook/special-relativity-27/relativistic-quantities-180/relativistic-kinetic-energy-662-6210/

Louise

[Corkeyno2]

2 minutes ago, Thalestris24 said:

Here you go:

https://www.boundless.com/physics/textbooks/boundless-physics-textbook/special-relativity-27/relativistic-quantities-180/relativistic-kinetic-energy-662-6210/

Louise

Thanks. Is this the formula for energy release during an impact?

[Thalestris24]

Just now, Corkeyno2 said:

Thanks. Is this the formula for energy release during an impact?

Yep - it's the relativistic for of KE = 1/2(mv2) (where the 2 is supposed to be an index i.e. = squared but can't do it on here!)

[Corkeyno2]

3 minutes ago, Thalestris24 said:

Yep - it's the relativistic for of KE = 1/2(mv2) (where the 2 is supposed to be an index i.e. = squared but can't do it on here!)

cheers

[Corkeyno2]

5 minutes ago, Thalestris24 said:

Yep - it's the relativistic for of KE = 1/2(mv2) (where the 2 is supposed to be an index i.e. = squared but can't do it on here!)

sorry for being a noob at this - so it calculates how much energy is released when two objects slam into each other?

[Corkeyno2]

9 minutes ago, Thalestris24 said:

Yep - it's the relativistic for of KE = 1/2(mv2) (where the 2 is supposed to be an index i.e. = squared but can't do it on here!)

so if the rock had a mass of 10 kg, and the vehicle had a mass of 3 million kg, and the vehicle hit the rock at 0.999 c, how much energy would be released?

[Thalestris24]

It's a bit more complicated and depends on whether the collisions are elastic or inelastic in classical physics. You can google it for relativistic collisions - sorry, I'm not working it out for you!

[Corkeyno2]

12 minutes ago, Thalestris24 said:

It's a bit more complicated and depends on whether the collisions are elastic or inelastic in classical physics. You can google it for relativistic collisions - sorry, I'm not working it out for you!

nevermind... I can't imagine the result would be very pleasant anyway

i mean like you would blow up

[Daz69]

I'd hazard a guess that Newton's laws don't really work in space? Didn't do physics, wished I had though.

[Daz69]

Corkey, would the rock be stationary in space? Surely it would also be moving, not necessarily in the same plane of course, but it must be moving?  

[Corkeyno2]

Just now, Daz69 said:

Corkey, would the rock be stationary in space? Surely it would also be moving, not necessarily in the same plane of course, but it must be moving?  

for my purposes, the rock is stationary

[Corkeyno2]

1 minute ago, Daz69 said:

Corkey, would the rock be stationary in space? Surely it would also be moving, not necessarily in the same plane of course, but it must be moving?  

for my purposes, the rock would be stationary

[Thalestris24]

Corkeyno2, you're welcome to plough through this:

https://arxiv.org/ftp/arxiv/papers/1011/1011.1943.pdf

[Daz69]

K.I.S.S. theory then, like it  

[Thalestris24]

Hi again

ok... It's probably best to use the spaceship as the frame of reference. That means you're standing still at the helm of your starship  All of a sudden a 10kg rock heads towards you at 1/3 the speed of light. That's the same as your spaceship travelling at that speed and hitting a stationary object. However, it's much simpler to do it the other way round. That means that you can consider all of the kinetic energy in the rock to be released. 1/3 the speed of light isn't possible by any technology we currently have but we can be hypothetical  The amount of energy released will be 5.45 x 10^16 Joules which converts to just over 13 megatons of tnt. So you probably wouldn't know what hit you, lol. There is a relativistic energy calculator here and a Joules to megatons one here

Louise 

[goodricke1]

Another offshoot of the Fermi Paradox is to ponder why we have not already been struck by a relativistic missile designed to wipe out the competition.

[ollypenrice]

Why 0.999C? That's a strange number!

Olly

[M106]

20 hours ago, Daz69 said:

I'd hazard a guess that Newton's laws don't really work in space? Didn't do physics, wished I had though.

They apply to the entire known universe; although its possible that strange things may occour at the event horizon of a black hole which Newtons laws of motion cant deal with

[SilverAstro]

7 hours ago, goodricke1 said:

the Fermi Paradox    to wipe out the competition.

We already have ! our ancestors did it to them, that is why we are alone. It is a resolution to the paradox that i have not seen. Our  ancestors were a bit worried that if they did not achieve a 100% first strike then the remaining Galactic Police would come to call,,, consequently they covered their tracks so well that even we now dont know that it was done. And Von Daniken got it wrong way round !!

Isnt the kinetic problem the reason why a Bussard ram wouldnt work, hitting all those (mayby stationary) hydrogen atoms at near light speed would be exceeding unhealthy ??

 

[goodricke1]

14 hours ago, SilverAstro said:

Isnt the kinetic problem the reason why a Bussard ram wouldnt work, hitting all those (mayby stationary) hydrogen atoms at near light speed would be exceeding unhealthy ??

I'm sure an advanced species would find some way of sweeping up those nasty H atoms... a Dyson Hoover maybe?

[Corkeyno2]

On 1/6/2017 at 19:35, ollypenrice said:

Why 0.999C? That's a strange number!

Olly

c represents the speed of light, or 299,792,458 m/s

[ollypenrice]

Yes, but since no massive body can get anywhere near 0.999C the question is rather academic!

Olly

[SilverAstro]

On 06/01/2017 at 20:29, M106 said:

 which Newtons laws of motion cant deal with

One does not need to find a tame black hole for that, Newton cannot deal with the orbit of Mercury. It puzzled Newton greatly, he knew something was wrong as well but he was not the type to admit it, so he left everyone else to look for Vulcan and things well knowing that it was not a Vulcan type problem.

 

27 minutes ago, Corkeyno2 said:

c represents the speed of light, or 299,792,458 m/s

You would think that any intelligent species on seeing that number would have the gumption to file a bit off their m measuring stick to make a nice round 300.10^6 ? Proper engineers use 300 anyway !

 

[Alien 13]

The online calculators I saw wont let you input C the best I could get was 0.9999999999999999 C, the energy released compared to 0.99999999999999 C was huge.

Alan