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Principle of linear momentum for a rocket

  1. Jan 10, 2015 #1
    1. The problem statement, all variables and given/known data
    vertical rocket start:
    F_thrust (simple F)
    Δm (fuel consumed per Δt)
    m_tot

    the relative velocity of the gas w is demanded.

    2. Relevant equations
    vgas=v+Δv+w, w<0
    m(t)=mtot-Δm*t

    3. The attempt at a solution
    rocket acceleration.
    F-G=m(t)*a(t)
    a(t)=F/m(t)-g

    first:
    v*m=Δm*vgas+(v+Δv(t))(m-Δm)
    ⇒w=-Δv(t)*m(t)/Δm
    Δv(t)=a(t)*Δt
    or second direct way:
    w=F/Δm*Δt

    so the first one should lead to the direct solution if I'm not mistaken, but I struggle to get the correct solution.
    To be more specific, it seems in order to get the same result Δv=(a+g)Δt but this wouldn't make much sense for me.

    thanks in advance
     
    Last edited: Jan 10, 2015
  2. jcsd
  3. Jan 10, 2015 #2

    ehild

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    Welcome to PF!

    What is given in the problem?

    Without the external force (gravity) :
    The thrust can be considered as force F=ma=mΔv(t)/Δt=-w Δm(t)/Δt.

    If you know the thrust force F and the fuel consumed in unit time, Δm/Δt, .you can calculate the velocity of the gas with respect to the rocket.

    ##w=-\frac{F}{\Delta m/\Delta t} ## .
     
  4. Jan 10, 2015 #3
    Maybe I wasn't clear enough, given are thrust, Δm and the total mass.

    So to calculate the relative velocity w $$ w=-\frac{F}{\Delta m / \Delta t} $$ could be used, I know that, but I would prefer to calculate it with the conservation of linear momentum, since I try to explain the concept. The last time I had to solve an exercise like this one is a little bit in the past.. Now my problem is that I get two different solutions and I can't find my mistake.
    Did I forget to include the gravity in the conservation somewhere?
     
    Last edited: Jan 10, 2015
  5. Jan 10, 2015 #4

    ehild

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    It came from conservation of momentum and it is the same (but a sign ) what you wrote w=F/Δm*Δt, if you meant (F/Δm)*Δt. I do not see two different solutions.
    The thrust does not depend on gravity. The motion of the rocket depends on it.
     
  6. Jan 10, 2015 #5
    So then my assumption $$\Delta v=a*\Delta t \text{ where } a=\frac{F-G}{m}$$ is wrong, because I am only interested in the acceleration and Δv which came from the thrust only?
     
    Last edited: Jan 10, 2015
  7. Jan 10, 2015 #6

    ehild

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    You said the thrust F was given and w was the question. You determined Δv from conservation of momentum, and made m (Δv/Δt) equal to the thrust force. It was correct.
    The motion of the rocket is an other question. If gravity is involved, the acceleration depends on it. You can consider the whole motion as repeated "collisions" ( acceleration because of the ejected gas) followed by motion under gravity.
     
  8. Jan 10, 2015 #7
    I finally got my mistake.
    Thank you, very much.
     
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