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Rocket motion problem

  1. Jan 15, 2009 #1
    1. The problem statement, all variables and given/known data

    A rocket of initial mas M, of which M-m is fuel, burns its fuel at a constant rate in time tau and ejects the exhausts gases with constant speed u. The rocket starts from rest and moves vertically under uniform gravity . Show that the maximum speed achieved by the rocket is u ln([tex]\gamma[/tex])-g[tex]\tau[/tex] and that its height at burnout is

    u[tex]\tau[/tex](1-ln([tex]\gamma[/tex])/([tex]\gamma[/tex]-1) where [tex]\gamma[/tex]=M/m[assume that the thrust is such that the rocket takes off immediately.)

    2. Relevant equations

    3. The attempt at a solution

    I had no trouble finding v, I had trouble integrating v to obtain the height. v=u ln (gamma)-g*tau . h=[tex]\int[/tex]v dt= [tex]\int[/tex]u*ln(m0/m(t))-.5*gt^2

    u is treated as a constant I think since I am integrating v with respect to dt. [tex]\int[/tex]ln([tex]\gamma[/tex])=[tex]\gamma[/tex]*ln([tex]\gamma[/tex])-[tex]\gamma[/tex]. Now I am stuck on this part of the solution.
  2. jcsd
  3. Jan 15, 2009 #2
    You are integrating a ln function. The integral is

    int[ln(ax)] = xln(ax) - x.

    The exhaust velocity, u, is treated as a constant. This integral is obtained by integrating by parts. An introductory calculus text will have the derivation for the intergral of ln(ax). Apply the limits of integration to each part of the solution.
  4. Jan 15, 2009 #3
    Sorry about the previous post. Your equation for v is correct. Since v = dx/dt, separate the variables such that dx is on the left side and dt is on the right side. Recall, you are finding the maximum height so integrate dx from x = 0 to x = h and integrate (u ln (gamma) - gt)dt from t = 0 to t = tau. Then the remainder of the rocket flight is only under the force of gravity and becomes a vertically fired projectile with an initial velocity. Use v2 = vo2 - 2g(x - xo) to find the additional height by setting v2 = 0. Add this height, x - xo to the height at burnout.
  5. Jan 15, 2009 #4
    so I should still integrate (u ln (gamma)-gt)dt I am okay with that part of the problem and I am okay with integrating ln (gamma) because I can integrate ln(gamma) using integration by parts. How would the equation v2 = vo2 - 2g(x - xo) assist me in helping me find the height? Why do I even need the equation for v^2 for this problem?
  6. Jan 15, 2009 #5
    The additional equation v2 = ... is used because the rocket doesn't suddenly stop once the fuel is gone.
  7. Jan 15, 2009 #6
    I am integrating u*ln (gamma)-gt)dt to obtain the height. I don't see any other use for the additional equation. couldn't I just plug in my initial conditions in u*ln (gamma)-gt)dt to find the maximum height?
  8. Jan 15, 2009 #7
    Sorry, you do not need the additional equation. I missed the part "max height at burnout". Yes, just evaluate the integral using the initial conditions for height and time and the final time tau.
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