Physics: energy to accelerate an object?

[Icewalker]

I'm working on some numbers for my science fiction book's plausibility, and I need some help because I don't know crap about physics currently.

I'm just trying to get some base numbers that I can fit in later, once I know things like how big a certain object will be, but until then I just want it in units like per ton.

I don't really care what kind of units you can calculate it in, because I can convert it.


I'm trying to figure out how much energy it would take to accelerate X weight to 1% the speed of light in a vacuum. Or 5%. or 10%.

For your use, the speed of light is 1,079,252,849 km/h, or 670,616,629.5087 mph

[thubby]

the formula is Ek = (1/2)mv^2

if i'm reading it right, thats energy (in joules)=(1/2) mass times vectorsquared

vector for most purposes is speed.

so getting a 1 gram item to 1% C would take

EK=(1/2)1*2 997 924.58^2

4,493,775,893,684.0882 joules

[UglyPanda]

Relativistic physics is annoying.

I think this is right, assuming no friction:
E = m(c^2)/sqrt(1- (v/c)^2) - mc^2

For 0.01 C:
E=.0001(c^2)m/sqrt(.9999)-mc^2
4.5e12 Joules/kg

For 0.05 C:
E=.0025(c^2)m/sqrt(.9975)-mc^2
1.31e14 Joules/kg

For 0.1 C:
E=.01(c^2)m/sqrt(.99)-mc^2
4.53e14 Joules/kg

Edit: Many stupid mistakes.
Edit 2: Atomic bomb calculations were all wrong. I don't feel like redoing them.
edit 3: I used the wrong number for C, it's 3e8, not 3e9.

[Sleet]

I'm pretty certain you can ignore relativistic effects for something as low as 1% c.

I'd just use energy = 1/2 * mass * velocity squared. Simple and easy.

( c is physics-speak for the speed of light, by the way. Use speed for velocity, because energy is a scalar quantity you don't need to worry about vectors.)

[Cobra_Ikari]

I don't remember. Although I used to know a way to get infinite constant theoretical acceleration...hmm.

...and a method with a speed on the order of 10^24 times the speed of light. Yay, numbers!

[Icewalker]

@sleet: I like how you gave me c, which I totally know, but then all this other stuff that I can't figure out in the least :smalleek:

Would velocity equate directly to speed, or does it involve some confusing exponential side-tracking into acceleration vs. speed?

If it does, then 1/2mv^2 sounds useful.

[Sleet]

Would velocity equate directly to speed, or does it involve some confusing exponential side-tracking into acceleration vs. speed?

Velocity is just speed with a description of direction. Speed is 35 mph. Velocity is 35 mph that -> way. In that formula, just use speed in meters per second and mass in kilograms, and you get energy in joules. It'll work. :)

[Icewalker]

Alright, that's what I thought, good.

[thubby]

Velocity is just speed with a description of direction. Speed is 35 mph. Velocity is 35 mph that -> way. In that formula, just use speed in meters per second and mass in kilograms, and you get energy in joules. It'll work. :)

wouldn't it just be grams, or you'd have to do speed in km/s?

[Sleet]

wouldn't it just be grams, or you'd have to do speed in km/s?

Nope - the base unit for mass in SI is kilograms, not grams. With the vast majority of other units, you'd be right, but mass is an exception.


Edit: Once you start getting up to 10% of C, Einstein's Special Theory or Relativity starts to matter and the energy to accelerate the object starts going up (to the point where it takes infinite energy to accelerate it to 100% c).

[AlisdairM]

I'm going to answer the question with another question ;¬)

Why are you concerned about some %age of the speed of light? As you start getting closer to this speed, you have to start worrying about all sort of other stuff too, including understanding the idea that speed on its own is a meaningless number - you need "speed relative to something", and different "somethings" start to give different answers.

At 1% of speed of light this should probably not be a concern, indeed it would only make a 1% difference at 10% of light-speed, but becomes significant very quickly after that.

The reason I ask is to make sure there are not other important questions you don't realise you should be asking!!

[SlightlyEvil]

I'm pretty certain you can ignore relativistic effects for something as low as 1% c.

Well, you can for most applications, but your result would be an estimate. Also, I think the OP wanted the energy for an arbitrary percentage, which is given by K = m(c^2)/sqrt(1- (v/c)^2) - mc^2.

[Sleet]

Well, you can for most applications, but your result would be an estimate.

No more so than using the relativistic equation, unless you know your mass and velocity exactly. Which you never do - there's always error in measurements.

A computation is only as accurate as the least accurately measured quantity.


Also, I think the OP wanted the energy for an arbitrary percentage...

This is true.

[AlisdairM]

Uh oh, let's not go dragging Heisenburg into this, or Shroedinger might come looking for his cats...

Ok, working on the assumption we are looking to provide energy for a literary device, rather than solid engineering, accuracy to an order of magnitude is probably good enough. Out of curiosity I did some rough figures for the kinetic energy of a 1Kg mass travelling at 0.9c. As above, I am going to ignore relativistic effects as a 1% error is neglible at this accuracy.

So kinetic energy is 1/2 m v^2 == 0.5 * 0.9 * 0.9 * c^2
== 0.4 c^2 (approx)

Using e = mc^2 for converting matter/antimatter reaction (most efficient fuel) we see approximately 40% of your mass is fuel accelerating to 90% light speed. This is very rough, as I have not factored fuel burning reducing mass during flight.

so we need 400g of antimatter per kilo to accelerate to 0.9c, assuming 100% efficient energy -> speed (not possible by 2nd law thermodynamics, but assume we can get close with our future sci-fi tech)

400g antimatter is ~ 2*10^16 joule, or roughly 5 Megatonne of TNT. If megatonne makes you wonder about nukes, that is roughly 400 Hiroshimas, or probably 40 modern nukes.

According to this site:
http://www.circlon-theory.com/HTML/joules.html
that convert to around 200 Kg or Uranium 235 for a fission reaction. I am struggling to find a good reference for how much hydrogen we would need for equivalent fusion reaction, but according to that site it looks like U235 fission releases around 3*energy of H2 fusion. Of course, the mass difference is 235/4 ~= 60, so we get roughly 20x energy from same mass of hydrogen fusion - I make that 10kg.

So per kg travelling at 0.9c, we need:
400g antimatter
10kg hydrogen fusion
200kg uranium fission
5 million tonnes coal
5 million tonnes TNT

coal equivalent from http://www.euronuclear.org/info/encyclopedia/coalequivalent.htm

Remember to multiply by mass of whatever you are accelerating, so getting a typical person to 0.9c using average mass of 68kg (common assumption in elevators!) uses most of a tonne of hydrogen for fusion fuel!

Take all calculations with a pinch of salt ;¬)

Oh, and if using nukes for fuel sounds odd, look up 'Project Orion' in your favourite search engine for some wacky 60s technology that thankfully was never tested.

[Sleet]

Uh oh, let's not go dragging Heisenburg into this, or Shroedinger might come looking for his cats...

Heisenberg's got nothing to do with it. It's a simple matter of significant figures. You just don't measure the mass of a spacecraft down to the milligram. (I've worked in aerospace. There's a lot more "close enough" than you'd think. :smallwink: )

I'm not trying to be pedantic about this, I'm just objecting to the implication that relativistic computations are exact. (Well, OK, I guess that's being pedantic. I'll shut up now. :smallsmile: )

[Ted_Stryker]

If v/c < 0.1, an object will certainly have kinetic energy of (1/2)*mv2, but the elephant in the room that hardly anyone is talking about is the propulsion technique you use to get up to that speed. If the ship carries all its fuel on board at the start of the flight, it's a rocket problem, and the question you really need to ask is how much fuel you need to carry.

Also, if you're going somewhere, it might be worthwhile to make provision for braking so that you don't whip past your destination, assuming it's basically at rest with respect to your launch point. Here's (http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html) a treatment of the relativistic rocket problem assuming a 100% efficient mass-conversion propulsion system.