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Rocket Equation Derivation

  1. Nov 6, 2005 #1
    I'll write the equation first, and then state my problem with it.

    mv=(m+dm)(v+dv)-dm(v-v')

    The book says that in the term (m+dm), that dm is a negative quantity. If I believe this, then what about the -dm(v-v') term? Why is dm negative?

    If I were to derive it I would have (m-dm) and +dm(v-v'), then I wouldn't have to say "dm" is a negative quantity. But if I use this method, then my result is different by a minus sign.

    Will someone please explain this in a more intuitive sense. Thanks
     
  2. jcsd
  3. Nov 6, 2005 #2

    Tide

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    dm is negative because mass is being lost by the rocket. Of course, you're free to set up the momentum conservation either way. It might be helpful if in setting up your equations if you used absolute value signs around dm and write the explicit sign in front of it. You'll just have to be careful when you finally attempt to integrate the differential equation.
     
  4. Nov 6, 2005 #3
    Using the textbook method their result: mdv=-v'dm

    My method I get: mdv=v'dm

    And if I were to integrate to find the velocity as a function of mass, I would integrate from m(t) to m. This is exactly opposite the textbook method, so I end up with the same result.

    So does the differential form really matter then, if I am off by a negative? i.e. Would a professor be just in docking off points?
     
  5. Nov 6, 2005 #4

    Tide

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    A good prof wouldn't dock you if you make it clear what you are doing and it is valid.
     
  6. Nov 7, 2005 #5

    arildno

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    Hi, brent:
    You might want to have a look at
    https://www.physicsforums.com/showthread.php?t=72176
    where the issues of geometric systems (including the rocket equation) are dealt with in a general manner.
     
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