# I have a space station creating artificial gravity via linear acceleration. how would you use momentum to find the stations fuel consumption?

Im pretty sure the calculations involve energy, but Im not too sure.

You can make up any values that are needed. e.g mass, velocity etc. It just has to show how much fuel is consumed in accelerating the space craft such that I can calculate the cost.

Thankyou!

You can make up any values that are needed. e.g mass, velocity etc. It just has to show how much fuel is consumed in accelerating the space craft such that I can calculate the cost.

Thankyou!

## 6 Answers

- How do you expect a space station to create artificial gravity via linear acceleration? They are usually modelled as centrifuges, which use centripetal acceleration to create artificial gravity. If you wanted to create gravity through linear acceleration then you would have to send it through space accelerating at a steady state (presumably at 1.0 g) in a straight line. If it's accelerating in a straight line, then it's not a space station, it is a space ship.10
- Yes the other person is correct if the object does have linear acceleration it is not a station00
- You can't - at least not that easily. There are several paths to the result; I'd suggest you check out the Wikipedia entry on the Tsiolkwski rocet equation https://en.wikipedia.org/wiki/Tsiolkovsk... . Chapter 2.3.1 should have your answer.01
- The craft would reach speeds close to speed of light in about a year (as the other answer has stated already that it would reach c/2 in six months).

For hydrogen and oxygen combustion, you would have to take the work done to equate it with the energy released on combustion of the fuel.00 - For impulse we have d(MV) = F dT; where F is the thrust of the space station with mass M and velocity V. dT is the period over which the thrust is maintained.

So we have M dV/dT + dM/dT V = F and then dM/dT V = F - M dV/dT; so that dM/dT = (F - MA)/V = (F - Mg)/V is the rate at which the mass of the station is changing. And that means the rate at which the fuel mass is being expended. ANS.00 - To maintain 1G of gravity, you have to accelerate at that rate.

if the mass of the craft is m, then you need to apply a force F = ma where a = 10 m/s²

if you continue that for t seconds, d = ½at²

d = ½10•t² = 5t² meters traveled

work = energy needed = Fd = ma(5t²) = 10m(5t²) = 50mt²

that tells you the energy needed.

velocity = at = 10t

momentum = mV = 10mt

If you are just throwing reaction mass out the back at velocity V₀, then the momentum of the fuel thrown out the back is m₀V₀

m₀V₀ = 10mt

so the mass of fuel used is

m₀ = 10mt/V₀

To try some numbers....

if you have a craft mass of 10000 kg, a time of 3600 seconds (1 hour), fuel velocity of 1e8 m/s (1/3 of light, pretty extreme)

then mass of fuel would be

m₀ = 10mt/V₀ = 10•10000•3600/1e8 = 3.6 kg

this is for every hour.

at some point, the mass thrown away would become a significant portion of the total mass, and the equations would change.

or course you would reach relativistic speeds at some point so the equations would change. v = at.

to reach c/2, t = 1.5e8/10 = 1.5e7 seconds or about 6 months.30

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