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/sci/ - Science & Math

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2364658 No.2364658 [Reply] [Original]

How to compute how much energy is needed to get a 5 kg object to a speed of 10000 m/s in outer space?

>> No.2364661

From what starting point?

>> No.2364662

>>2364661

>> No.2364674
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2364674

>>2364661
>>2364662

>> No.2364697

Orbital speed or surface relative speed?

>> No.2364726

>>2364697
And starting from stationary on Earth surface and going to the nearest circular orbit, or starting in stable circular orbit and going to a radial velocity of 10,000 m/s away from Earth's surface?

>> No.2364730

That really does depend on the fields present. If it's in completely empty space, and no other objects exist to exert forces on it, then it's a very simple calculation.

But if it's heading directly toward a planet, or directly away from a planet, obviously the energy would change as your force acts with or against the local fields.

To give an exact answer, you would need at least the solution to the n-body problem.

>> No.2364732

>>2364726
Space to space

>> No.2364733

>this thread
relativitymind

>> No.2364772

what does m/s

>> No.2364786

Kinetic energy in flat (negligible gravity) space is:

KE = 1/2*m*v^2. So, 0.5*(5 kg)*(10,000 m/s)^2 =
250,000,000 J. So, 250 megajoules.

>> No.2364787

>>2364772

what does m/s mean**

>> No.2364791

>>2364787
meters per second. It's a speed.

>> No.2364793

>>2364786
So how many joules does it take to beat light speed? Lrn2relativity.

>> No.2364800

>>2364786
You really, really mad.

Question in OP, in what amount of time ? I assume that your 'get to a speed' means 'from 0 to '.

>> No.2364809

>>2364793
The relativistic calculation is not needed. Learn to be a physicist - you use the simplest model that gives you acceptably accurate results. Always.

>> No.2364911

>>2364809
> implying OP specified 'acceptably accurate'

I disagree with that, even though I'm sure more people are on your side. I always head for the most accurate answer I can give, even if it takes me a few more seconds filling out variables in a more complicated equation.

>> No.2364923

>>2364786
More than that, I expect. I assume he's getting the 5kg object up to speed via thrust. Which means he's ejecting mass with kinetic energy of it's own.

>> No.2364941

>>2364911
10000m/s <<<<<< 180,000m/s

Relativistic calculations are not necessary.

>> No.2364951

>>2364941
Wow, I was dumb for a second. Should be 3 X 10^8 m/s instead of 180,000.
Was thinking of miles/second I think

>> No.2364956 [DELETED] 

>>2364941
That's even for mi/s. In m/s, it's 10,000 << 300,000,000.

>> No.2364966

>doesn't know what m/s is
>confuses standard units with medieval units

americans, americans everywhere

Also, the relativistic change factor at 10 000 m/s is 1.0000000005563252. That is, the object's mass is changed by a factor of less than 5.5*10^-8 percent. That's as close to "insignificant" as it gets.

>> No.2364975

>>2364966

This. You'd have a point if the OP specified a velocity 10 or 100 times higher, but this is just retarded nit-picking.

>> No.2364977

you need to figure out how efficient are railguns and apply that knowledge

also, there's no air resistance or gravity in outer space so that's a huge win

>> No.2364993

>>2364786
>herpaderp assuming 100% efficiency

1-10% would be a better guess