Rocket with Variable Mass and Air Resistance.

  1. Problem:

    A rocket is launched vertically upward. The rocket has a mass Mr and carries Mo of fuel. The fuel burns at a constant rate (ß) and leaves the rocket at speed Ve relative to the rocket. Assume constant gravity (9.8m/sec^2). There is an air resistance given by F(a) = -kV.

    Where V is the velocity of the rocket.


    k = .1 n-sec/m
    ß = 100 kg/sec
    Mr = 1000 kg
    Mo = 10000 kg
    Ve = 3000 m/sec

    What is the terminal velocity of the rocket (the velocity when fuel runs out)?


    The general solution for this problem is an extension of conservation of momentum for a system of variable mass.

    mdv/dt = ∑F(ext) + Vrel(dM/dt)

    I have no problem when the external force is solely gravity, however, when this problem presents linear air resistance, I am running into a bit of trouble with the integration.

    My setup of the differential equation is as follows:

    Set dm/dt to constant. dM/dt = ß = 100 kg/sec.


    mdv/dt = 100Ve - mg - kV


    dv/dt + kV/m = (100Ve/m - g)

    I am attempting to integrate the velocity from V(0) to V(Terminal), and the time from t=0 to t=t(final)= Mo/ß = 10000/100 = 100 seconds.

    My question is regarding the velocity portion of the differential equation.

    I can't seem to get that V over to the left side of the integral by itself... and I'm wondering if I could get some sort of example on how to do that type of integration, or if I've gone astray somewhere.

    Does this integration qualify as a first order differential?


    dy/dx + P(x)y = Q(x) ?


    Okay, so I've set up the equation:

    m/k(100Ve/m-g) - (m/k(100Ve/m-g)e^(-kt/m))


    Last edited: Feb 19, 2007
  2. jcsd
  3. Yeah, it does. The solution's [tex]ye^{\int p(x)}=\int Q(x)e^{\int P(x)}dx[/tex]
  4. Tips for making this easier:

    Remember that ultimately the mass is constant (since the M comes from the force being applied) and don't try and trick yourself into thinking that it is a variable in the end.

    In a rocket formula with linear air resistance, it is helpful to compare your answer to something you'd expect without air resistance.
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