The Rocket Equation

  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. turin

    turin 2,326
    Homework Helper

    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. 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. turin

    turin 2,326
    Homework Helper

    Because the propelling mechanism is in the rocket.
  6. 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. turin

    turin 2,326
    Homework Helper

    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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