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Rocket Motion

  1. Oct 15, 2004 #1
    This question was posed in class the other day for extra credit:

    A rocket with initial mass 70000kg burns fuel at a rate of 250kg/s; it has an exhaust velocity of 2500m/s. If the rocket is at rest, how long after the engines fire will the rocket lift off?

    I've been trying to solve it using some variation of the rocket equation:
    v=v(exhaust) * ln(M(0)/M(t)) - gt
    but to no success.
    Any hints or help would be greatly appreciated
  2. jcsd
  3. Oct 15, 2004 #2
    There is a problem with the equation you are using
    it is r (rate of burning the fuel) for g you used, also it is dimensionally incorrect we can have log of only natural numbers.
    If it works thats Ok, otherwise try by equating force by the ejected gass to mass of rocekt at the moment.
  4. Oct 16, 2004 #3


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    Be careful. The equation of your rocket is:

    [tex] v=2500\cdot ln\Big(\frac{70000}{70000-250t}\Big)-9.8t[/tex]

    There is an interval of velocities 0<t<47.9 s in which v<0. The rocket will start to lifting off when v>0 or t>47.9 s. (solve numerically the equation v=0, you will obtain t=0 and t=47.9 s).
  5. Oct 16, 2004 #4
    Thanks both of you for your help, very much appreciated.
  6. Oct 17, 2004 #5
    I think you're all making this much too complicated.
    Just calculate the force the rocket produces using the equation force = change in momentum / time, with knowledge of the fact that momentum is mass times velocity. The weight of the rocket is given by acceleration due to gravity x (original mass of rocket - (rate at which the rocket loses mass x time). Set the weight of the rocket to equal the its force of propulsion (which you already calculated) and solve for time. At that time, the force of thrust will balance the rocket's weight and immediately afterwards it will start lifting off.
    Last edited: Oct 17, 2004
  7. Oct 17, 2004 #6
    I get about 25 seconds.
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