G asked in Science & MathematicsPhysics · 2 years ago

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.



Sorry I meant Space CRAFT not station

5 Answers

  • 2 years ago
    Favourite answer

    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.

    • ...Show all comments
    • oldprof
      Lv 7
      2 years agoReport

      The velocity is also a function of the nozzle shape and the internal/external partial pressures of the exhaust.

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  • 2 years ago

    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.

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  • 2 years ago

    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/Tsiolkovsky_rocket_e... . Chapter 2.3.1 should have your answer.

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  • D g
    Lv 7
    2 years ago

    Yes the other person is correct if the object does have linear acceleration it is not a station

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  • Anonymous
    2 years ago

    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.

    • G2 years agoReport

      sorry I meant space craft

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