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Fuel burn rate

  1. Oct 29, 2006 #1
    A rocket, mass m=4000kg, lifts off the ground vertically with acceleration of a=0.2g. The velocity of gass emmiting from it is u=1200m/s (with respect to the rocket). Find engine's reactive force and then fuel burn rate (mu).

    This is what i managed to do:
    rocket's speed at any moment t (with respect to the ground) is a*t
    gass's speed at any moment t (with respect to the ground) is u-a*t


    Mass of rocket at any time t is m-mu*t
    Mass of fuel that is burned is mu*t
    rocket's mass multiplied by it's speed and gass's mass multiplied by its speed are equal, so

    (m-mu*t)*a*t=mu*t*(u-a*t)

    and i get that m*a*t=mu*t*u

    mu=(m*a)/u=6.53 kg/s

    but... I was given 4 answers (5.9kg/s 9.8kg/s 39.2kg/s 19.8kg/s) and mine is wrong...

    Any ideas anyone?
     
  2. jcsd
  3. Oct 30, 2006 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    You are forgetting about gravity. Use Newton's third law: The rate of change of momentum of the rocket + force of gravity = rate of change of momentum of the gas being expelled:

    [tex]F_{rocket} = mg + ma = - F_{gas} = - vdm/dt[/tex]

    AM
     
    Last edited: Oct 30, 2006
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