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B Separation of variables - rocket equation

  1. Nov 27, 2016 #1
    hello there
    Im trying to do a derivation of tsiolkovsky's rocket equation, but i got stuck at the step when i have to use separation of variables (marked with red in the pic), i used maple to solve it, so i could get on with it, but i want to understand what is happening to solve this, so can anyone explain how to solve this step with separation of variables?
    Thanks :)
    upload_2016-11-27_15-19-15.png
     
  2. jcsd
  3. Nov 27, 2016 #2

    Orodruin

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    It is exactly what they have done. What step in particular do you have problems with?
     
  4. Nov 27, 2016 #3
    ok thanks so i have done some right ;)

    im having trouble explaining what is happening, or i think i do.
    I can explain the first steps, just isolate the variables on each side of the equation.
    But what rules are used/how is this integrated (the bordered step)
    Thanks :D
    upload_2016-11-27_15-59-54.png
     
  5. Nov 27, 2016 #4

    Orodruin

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    You integrate both sides between the same points. It is essentially making an integration and then making a change of variables.
     
  6. Nov 27, 2016 #5
    Thanks i see now :D
     
  7. Nov 27, 2016 #6

    Orodruin

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    Oh, and the integration boundaries on the LHS should be ##v_f## to ##v_i##. In this particular example it does not matter for the result because the integrand is constant. You then make a change of variables to ##m(v)## and use ##m_f = m(v_f)## and ##m_i = m(v_i)##.
     
  8. Nov 27, 2016 #7
    Hello
    Do you mean like this?
    upload_2016-11-27_16-20-29.png
    im not 100% sure what you mean with the change of variables
    Again thank you for helping
     

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  9. Nov 27, 2016 #8

    Orodruin

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    Yes. Consider the integral
    $$
    \int_{v_f}^{v_i} dv.
    $$
    Now, you know that ##m## is a function of ##v## so change variables to ##m##. The integral changes to
    $$
    \int_{m(v_f)}^{m(v_i)} \frac{dv}{dm} dm.
    $$
    Insert the known differential equation and perform the new integral.
     
  10. Nov 27, 2016 #9
    so i have it like so:
    upload_2016-11-27_16-46-29.png
    or do i get -1/u*m(vi)-m(vf)?
    thank you for your patience and help
     
  11. Nov 27, 2016 #10

    Orodruin

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    No, you have the wrong integration boundaries in the first integral. They are what I said in my post.
     
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