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Rockets burning in terms of momentum conservation

  1. Feb 26, 2007 #1
    Can some one help me with this question plz. dont realy understand

    mathematically analyse, using diagrams, rockets burning in terms of momentum conservation and the rate of fuel consumption R and derive the acceleration for a rocket during its launch stage.
  2. jcsd
  3. Feb 26, 2007 #2


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    Well, what do you know about the topic? What equations do you know? Show your work, and people will help you; however, we will not simply do your homework for you!
  4. Feb 26, 2007 #3
    lemme give u a tip....you need to remember that the mass is not constant and that the fuel used is continuously reducing the mass of the body.....try to a get a start, use calculus and post how far you get....
  5. Feb 26, 2007 #4
    Be careful when you're looking for an explanation -- many high school/introductory undergraduate textbooks derive the rocket expressions with scalars, although they use vector tactics along the way. As I recall, Serway was one of the few that had a decent, correct explanation.

    Also, the Internet is your friend. We're not going to just tell you how to do the problem, but nothing's stopping you from hunting down explanations online. That way, you'll learn by working through the derivations on your own.
  6. Feb 27, 2007 #5
    sry about that

    we know that rockets momentum final= R

    then the F= R/t

    were R is the rate of fuel consumption and F is the total thrust

    am i gettin there ?
  7. Feb 27, 2007 #6
    Well.. you might want to start by considering that momentum is always conserved.

    If we imagine the rocket sitting in outer space it can make the thought process simpler.

    If we assume initially the rocket is stationary its initial momentum will be zero, after it starts accelerating the total momentum of the system will still be zero (1st clue).

    Secondly... Momentum = Mass * Velocity, however, in this case the mass of the rocket is not constant! Try working out the Thrust via differentiation.
  8. Feb 27, 2007 #7
    Why do you describe "R" as both the rate of fuel usage and the change of momentum?
  9. Feb 27, 2007 #8
    ok i think R=M/T

    and therefore F=(R)(Vex)

    Vex is the velocity when the rocket loses the fuel because it goes faster with less weight

    total momentum intial= (M+m)V=0
    total momentum final=M(V+v)=0
    therefore total momentum: (M+m)V=M(V+v)

    m : this is delta m is the fuel (before and after)
    but wat do i do with the fuel that is expelled
  10. Feb 28, 2007 #9
    When you write F=R*Vex, if as you also say R=M/T

    This is the thrust. However you then describe Vex as something other than the exhaust velocity (then it is not the thrust). Also, its very unclear what you mean by M and m.

    Before you try solving this for the general case in which R might vary take R to be constant. Lets also assume that the exhaust velocity is constant.

    I say this because your momentum equations look rather confused and we should sort that out first.

    Vex = exhaust vel.
    Mf = mass of Fuel.
    M = Initial mass of rocket.
    V = final mass of rocket

    Once the rocket has used all its fuel...

    Vex*Mf = (M-Mf)*V

    Which is the bog standard cons. of mom. eqn.

    If your ok with this then we can set up the "rocket equation".
    Last edited: Feb 28, 2007
  11. Mar 1, 2007 #10
    yea i get wat ur doin

    ur sayin momentum before=momentum after
  12. Mar 1, 2007 #11


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    I notice that you still haven't said exactly what the question is. That is important in deciding how to answer it!
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