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The Rocket Equation

  1. Jan 28, 2009 #1
    In deriving the rocket equation, there is one part I don't understand. The velocity of exhaust with respect to the body is assumed to be constant, where:

    v(exhaust wrt body)=v(exhaust wrt inertial)-v(body wrt inertial)

    So assuming a constant mass flow rate, the rocket propellant exerts a constant force on the rocket and hence in space, uniform acceleration. But how can an observer on the accelerating rocket observe the rocket propellant being ejected with a constant velocity?
  2. jcsd
  3. Jan 28, 2009 #2


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    This is not correct. The mass of the rocket decreases at a rate equal to the exhaust mass rate. The constant exhaust velocity should be w.r.t the rocket.
  4. Jan 28, 2009 #3
    Oops. So assuming a constant mass flow rate the propellant exerts a constant force on the rocket so the rocket's acceleration increases as follows:

    a(t)=F[1/m(t)] where F is a constant (until fuel runs out) where m(t) is the mass of the rocket at time t.

    m(t)=m(initial)-bt where b is a constant (mass flow rate)

    Hence v_rocket(t)=-Fln(m(t))/b assuming v(0)=0

    But why would an observer in the rocket observe a constant propellant velocity?
  5. Jan 28, 2009 #4


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    Because the propelling mechanism is in the rocket.
  6. Jan 28, 2009 #5
    Oh, yeah. The engine is on the back of the rocket, so it accelerates with the rocket, so if an engine ejects exhaust at a velocity v, then this is what is observed from the rocket's point of view.

    Now I feel somewhat embarrassed, but at least the rocket equation makes sense now.
  7. Jan 28, 2009 #6


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    You should not. No one knows everything. You should only feel embarrassed if you refuse to ask a question out of fear of sounding stupid. Don't let your transient embarrasment prevent your permanent understanding:
    Have fun learning.
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