Yahoo Answers: Answers and Comments for I have a space station creating artificial gravity via linear acceleration. how would you use momentum to find the stations fuel consumption? [Physics]
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enIN
Mon, 10 Sep 2018 07:08:11 +0000
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Yahoo Answers: Answers and Comments for I have a space station creating artificial gravity via linear acceleration. how would you use momentum to find the stations fuel consumption? [Physics]
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From billrussell42: To maintain 1G of gravity, you have to acceler...
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Mon, 10 Sep 2018 12:16:59 +0000
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.

From YKhan: How do you expect a space station to create ar...
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Mon, 10 Sep 2018 07:19:50 +0000
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.

From oldprof: For impulse we have d(MV) = F dT; where F is t...
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Mon, 10 Sep 2018 18:53:46 +0000
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.

From D g: Yes the other person is correct if the object ...
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Mon, 10 Sep 2018 07:25:24 +0000
Yes the other person is correct if the object does have linear acceleration it is not a station

From Markus Imhof: You can't  at least not that easily. Ther...
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Mon, 10 Sep 2018 07:25:30 +0000
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_equation . Chapter 2.3.1 should have your answer.