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Rocket Thrust Balancing Gravity

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data
    A rocket (initial mass m0) needs to use its engines ot hover stationary, just above the ground. If it can afford to burn no more than a mass (lamda)m0 of its fuel, for how long can it hover? If the exhaust velocity is 3000 m/s and lambda is 10% how long can the rocket hover?

    2. Relevant equations
    Thrust = [tex] v_{ex} \frac{dm}{dt} [/tex]

    [tex] v_{ex} [/tex] is the exhaust velocity

    3. The attempt at a solution

    So starting off with the basic sum of forces equals zero:

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

    [tex] \frac{dm}{dt} = \frac{mg}{v_{ex}} [/tex]

    [tex] \frac{dm}{m} = \frac{g dt}{v_{ex}} [/tex]

    [tex] \frac{g}{v_{ex}} \int_{0}^{t} dt = \int_{m_{0}}^{m_{0} - \lambda m_{0}} \frac{dm}{m}[/tex]

    [tex] \frac{g}{v_{ex}} t = ln(m) ]_{m_{0}}^{m_{0} - \lambda m_{0}} [/tex]

    [tex] \frac{g}{v_{ex}} t = ln [m_{0} - \lambda m_{0}] - ln [m_{0}] [/tex]

    [tex] \frac{g}{v_{ex}} t = ln [\frac{m_{0} - \lambda m_{0}}{m_{0}}] [/tex]

    [tex] \frac{g}{v_{ex}} t = ln [ 1 - \lambda] [/tex]

    [tex] t = \frac{v_{ex}}{g} ln [ 1 - \lambda] [/tex]

    And here is where I encounter an issue. Since lambda will always be positive, regardless of how much fuel the rocket can expend this equation will give a negative time. Does anyone see where I went wrong with the calculations, or are my original assumptions wrong?

    Thanks for any and all help
     
  2. jcsd
  3. Sep 21, 2009 #2

    jambaugh

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    Science Advisor
    Gold Member

    The thrust equation just gives magnitude. Note that dm/dt should be a negative number since mass is decreasing.
     
  4. Sep 21, 2009 #3
    So if I carry along the negative associated with the decreasing mass, then the last few steps would be:

    [tex] (-) t = \frac{v_{ex}}{g} ln [ 1 - \lambda] [/tex]

    [tex] t = - \frac{v_{ex}}{g} ln [ 1 - \lambda] [/tex]

    [tex] t = \frac{v_{ex}}{g} ln [ (1 - \lambda)^(-1)] [/tex]

    [tex] t = \frac{v_{ex}}{g} ln [ \frac{1}{1 - \lambda}] [/tex]

    And that would give a positive value for time. This equation makes sense, but is it a proper solution?
     
  5. Sep 21, 2009 #4

    D H

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    Staff Emeritus
    Science Advisor

    That's correct.
     
  6. Sep 21, 2009 #5
    Thanks!
     
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