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Rocket Motion in interstellar space

  1. Nov 9, 2013 #1
    So suppose an astronaut in interstellar space has gas ejecting from her propulsion system.
    So the gas would cause her to move forward by some distance, d. Then, the F[itex]_{thrust}[/itex] acting on her must be constant (assuming the amount of gas ejected per unit time is constant, and the speed it is released at is constant). Then, since the mass of the system (astronaut+equipment) is decreasing the acceleration of the system must be increasing to enable the F[itex]_{thrust}[/itex] to remain constant. Please correct me if my reasoning is wrong.

    (1) So naturally it wouldn't be valid to calculate the distance she travels by finding the acceleration from Newtons second law, and using that along with a kinematical equation to solve for the distance d (since the acceleration isn't constant).
    (2) So instead I would use [itex]v = v_{g}ln \frac{m_{i}}{m_{f}}[/itex], where [itex]v_{g}[/itex] is the speed of the gas being ejected.
    Then integrate for the position , [itex] x(t) = \int_o^t {v_{g}ln \frac{m_{i}}{m_{f}}}dt[/itex],
    where [itex]m_{f}=m_{i} - \frac{dm}{dt}t[/itex], where [itex]\frac{dm}{dt}[/itex] is the rate at which gas is ejected.

    But then for a question I had to do, both methods (1 and 2) produced the same final answer which I found to be very odd. Any possible explanations?
  2. jcsd
  3. Nov 9, 2013 #2
    How did you figure an acceleration in order to solve the problem by the suspicious method 1?
  4. Nov 9, 2013 #3
    From [itex]F_{thrust}=\frac{dp}{dt} = \frac{d(mv_{g})}{dt} = v_{g}\frac{dm}{dt}[/itex]
    Then, [itex]a=\frac{F_{thrust}}{m_{i}}[/itex]
    And then [itex] x(t) = \frac{1}{2}at^{2}[/itex]

    Now this assumes the acceleration is constant, yet produces the same answer.
    Last edited: Nov 9, 2013
  5. Nov 9, 2013 #4
    If you used the initial mass than the two methods should give different results. You would have to use some kind of average mass in order for both methods to give the same solution. Why don't you post your calculations for the second method so that we may figure what's up?
  6. Nov 9, 2013 #5
    Paying closer attention to your equations I noticed what seems to be a logarithm. Where did that come from?
  7. Nov 9, 2013 #6
    Well let [itex]m_{i}=115 kg[/itex] ,[itex]\frac{dm}{dt} = 0.007 kg/s, [/itex], [itex]v_{g}=800 m/s[/itex], [itex]t= 6 s .[/itex]
    Then, by method 1 [itex] F=5.6 N \Rightarrow a=0.0487 m/s^{2}\Rightarrow x= 0.877 m [/itex]
    , which is the same answer obtained by method 2.
  8. Nov 9, 2013 #7
    By solving the differential equation [itex]\frac{dv}{dm}= -\frac{v_{g}}{m}[/itex].
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