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A Rocket + air resistence

  1. May 4, 2012 #1
    Because this is not a 'formal' question I won't use the template.

    I was wondering, what if I have a rocket (varying mass) with air resistance acting upon it?
    Let's say the [itex]F=-kv[/itex]
    u is the speed of the rocket relative to the gas, and the rate of mass/second extracted is
    b
    without g it would look like this:
    [itex]\frac{dv}{v} =-u \frac{dm}{mv}-\frac{k}{m}dt [/itex]


    My problem is that I don't know how I can integrate the expressions in the right side.
    I'm sure I can't use the variable v as a constant in dm/mv but maybe I'm wrong.

    The only book I have about mechanics is Berkeley's book, if this type of problems are in another, please point me to it. Thanks!
     
  2. jcsd
  3. May 4, 2012 #2

    tiny-tim

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    Hi BitterX! :wink:
    dm/dt = -b …

    does that help? :smile:
     
  4. May 9, 2012 #3
    Ok, so:
    [itex]F= ub - kv[/itex]

    [itex]m\frac{dv}{dt}= ub - kv[/itex]

    [itex]m\frac{dv}{dt}=ub-k\frac{dx}{dt}[/itex]

    now I'm still stuck

    [itex]m\frac{dv}{v}=(\frac{ub}{v}-k)dt[/itex]

    [itex]vdt=dx \Rightarrow \frac{dt}{v}=\frac{dx}{v^2} = \frac{dv}{v^2 dt} [/itex]

    how can I isolate v to be only with dv?
    should I use [itex]m=M_0 - bt[/itex]?
     
  5. May 9, 2012 #4

    tiny-tim

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    Hi BitterX! :smile:
    Why did you introduce x in the next line?? :rolleyes:

    Just separate the variables, and solve! :biggrin:
     
  6. May 9, 2012 #5
    I'm really sorry, but that's exactly my problem.
    I can't see how can I separate v and m,
    If I divide by v and m I'm still stuck with [itex] \frac{ ub}{mv}dt [/itex]
    and [itex] \frac{ k}{m}dt [/itex]

    how can I integrate [itex]\frac{dt}{m}[/itex] or [itex]\frac{dt}{mv}[/itex]?

    and more generally, is there a text about how to do these things?

    Thanks for the help :)
     
    Last edited: May 9, 2012
  7. May 9, 2012 #6

    tiny-tim

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  8. May 9, 2012 #7
    Thanks! I guess I'm blind :)
     
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