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Homework Help: Rocket Propulsion Diff Eq

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Consider the case of a rocket taking off vertically from rest in a gravitational field g. The differential equation is given by

    [tex]m\dot{v} = -\dot{m}v_{ex}-mg[/tex]

    Assume the rocket ejects mass at a constant rate, [itex]\dot{m}=-k[/itex] (where k is a positive constant), so that [itex]m=m_{0}-kt[/itex]. Solve equation for v as a function of t, using separation of variables.

    2. Relevant equations

    [tex]m\dot{v} = -\dot{m}v_{ex}-mg[/tex]

    3. The attempt at a solution

    [tex]m\dot{v} = -\dot{m}v_{ex}-mg[/tex]

    [tex](m_{0}-kt)\frac{dv}{dt}=-kv_{ex}-(m_{0}-kt)g[/tex]

    [tex]\frac{dv}{dt}=\frac{-kv_{ex}}{m_{0}-kt}-g[/tex]

    [tex]dv=(\frac{-kv_{ex}}{m_{0}-kt}-g)dt[/tex]

    [tex]v(t)=v_{ex}ln(m_{0}-kt)-gt-c[/tex]

    Setting t = 0, it appears that [itex]c=v_{ex}ln(m_{0})[/itex].

    However, the next step of the problem is to plug in values, which revealed that this clearly isn't the right solution. I attempted to solve it again and ended up with the same solution, so I'm obviously having some difficulty recognizing my mistake.
     
  2. jcsd
  3. Sep 18, 2010 #2
    You mistook [tex]v_{ex}[/tex] (I don't know what ex stands for). That is the relative velocity of the rocket relative to the mass dm it has just ejected, NOT some constant.
    Derive yourself the differential equation for the motion of the rocket and you will see why :wink:
     
  4. Sep 18, 2010 #3
    Thanks! That has gotta be what I was missing.
     
  5. Sep 18, 2010 #4
    Wait, no, it is most definitely a constant. That is the velocity of the exhaust relative to the rocket, which is constant. They take this into account during the derivation of the differential equation by noting that the velocity of the mass ejected with respect to the ground is velocity of the rocket minus the exhaust velocity.

    I think my mistake may lie in something much simpler. [itex]\dot{m} = -k[/itex], not just k. This flips the sign of the two natural logs, and the equation makes much more sense. This way v grows as a function of t, though it eventually becomes asymptotic on a large enough domain of t. This is because the rocket isn't entirely made out of fuel, so we have to limit t to a domain such that 0 < k*t <= mass of fuel < initial mass of rocket. I think the solution is as follows.

    [tex]v(t) = -v_{ex}*ln(m_{0}-kt)-gt+v_{ex}ln(m_{0})[/tex]
     
  6. Sep 18, 2010 #5
    This is deadly wrong. Not so obvious. Don't state without proof.
    Most textbooks when analyzing the rocket motion assume that the exhaust velocity is constant, but don't assume that dm/dt = const. Here the problem assumes that dm/dt = const, but not exhaust velocity. Therefore, the results in 2 cases (textbooks vs this problem) are different.

    Look back again. You DIDN'T make that mistake :biggrin: You got [tex]m=m_o-kt[/tex]. Isn't it the same as [tex]dm/dt = -k[/tex]?
     
  7. Sep 18, 2010 #6
    Actually I did. Though what you say is correct, when I substituted [itex]\dot{m}=-k[/itex] into [itex]-v_{ex}\dot{m}[/itex] I got [itex]-v_{ex}k[/itex] instead of [itex]v_{ex}k[/itex]. This is the mistake I was referring to.

    This may be so, but this one seems to assume both. Let me give you the relevant portions of the problem to explain why I am assuming so.

    "Assume that the rocket ejects mass at a constant rate, [itex]\dot{m}=-k[/itex]..."

    "Using rough data from Problem 3.7, find the space shuttles speed two minutes into flight...

    (From problem 3.7) "...the average exhaust speed [itex]v_{ex}[/itex] is about 3000 m/s..."

    It provides no other indication of what v_ex is or should be. It is also fairly obvious that without an explicit formula for v_ex (or even if we had one that was non-constant), that the differential equation would not be solvable by separation of variables because of the addition of -g. Therefore, given the weight of these factors, I can only assume that both dm/dt and v_ex are constant.
     
  8. Sep 18, 2010 #7
    I will, however, grant that the use of the word "average" implies that v_ex isn't constant, but again, without any other kind of knowledge of what v_ex may be, I don't think I have another choice.
     
  9. Sep 18, 2010 #8
    I'm fairly certain my corrected solution is the right one.

    In a later problem it gives the following as the solution for y(t):

    [tex]y(t)=v_{ex}t-\frac{1}{2}gt^{2}-\frac{mv_{ex}}{k}ln(\frac{m_{0}}{m})[/tex]

    If we make the substitution for [itex]m=m_{0}-kt[/itex] and then differentiate, you get the solution I posted above.
     
  10. Sep 18, 2010 #9
    If that's so, then I can only assume that the problem misses a point about v_ex :smile:
     
  11. Sep 18, 2010 #10
    Could be! But if it's the textbook's fault and not mine, then I can just submit this and remember the point you raised.
     
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