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Thrust force of a rocket ejecting mass

  1. Sep 29, 2013 #1
    1. The problem statement, all variables and given/known data

    OK, this seems simple but I want to make sure I am not doing something totally wrong. The problem says: use the conservation of mass of a system of many particles to shoe that the thrust force of a rocket that ejects mass at rate [itex]\frac{dm}{dt}[/itex] is equal to [tex]F=-v_e \frac{dm}{dt}[/tex] where [itex]v_e[/itex] is the velocity of the mass ejected.


    3. The attempt at a solution

    I looked at it this way: momentum is conserved so if we start with mass m of the rocket, mv = k (where k is a constant).

    SInce we have a simple differential equation [itex]F=-v_e \frac{dm}{dt}[/itex] it can be integrated as [itex]-m v_e = Ft[/itex]. Taking the derivative w/r/t time we get [itex]-m \frac {d v_e}{dt} = F[/itex]

    That gets us the F=ma part of the equation, showing that that works. But I notice that if momentum is a k (constant) then [itex]-m v_e = k = Ft[/itex].

    There's a step I am missing here I think. I feel like I am almost there. Any hep -- and anyone telling me I have approached this in entirely the wrong way -- would be appreciated.
     
  2. jcsd
  3. Sep 29, 2013 #2

    D H

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    The momentum of the rocket is *not* a conserved quantity. What you are missing is that to look at things from the perspective of conservation of momentum you need to look at the rocket+exhaust cloud system.

    Another issue is how you look at force. There is a direct connection with the conservation laws if you use ##\vec F=d(m\vec v)/dt##. However, this means force is a frame-dependent quantity if mass is not constant. Force becomes frame invariant if you use ##\vec F = m\vec a##, but now the immediate connection with the conservation laws is lost.

    It might help if you look at things from the perspective of an inertial frame that is co-moving with the rocket (i.e., an inertial frame in which the rocket's instantaneous velocity is zero). The F=dp/dt versus F=ma imbroglio vanishes with this choice.
     
  4. Sep 29, 2013 #3

    arildno

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    A very simple version, alon DH's reasoning, is the following:

    Let us move along with the rocket through a tiny time Dt, in which a little mass Dm has been ejected with a velocity, relative to the rocket, v_e.

    That little packet of mass has expreniced a momentum change, what we call an impulse, of Dmv_e

    Impulse equals Dt*F, so the little packet of mass experienced the force:

    F=v_e*Dm/Dt.

    Now, by Newton's 3 law, you'll get the result for the thrust force for the rocket!
    ---
     
  5. Sep 29, 2013 #4
    Thanks to you both. I felt like this was a much simpler problem than it looked...
     
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