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Separation of variables problem

  1. Jan 8, 2006 #1
    Hey,
    I need some guidance in this problem. Consider a rocket taking off vertically from rest in a gravitational field g, the equation of motion (which I had to derive in the previous part of this problem) is:

    [tex]

    m \dot{v} = -\dot{m}v_{ex} - mg

    [/tex]

    where

    m is the mass of the rocket
    [tex] v_{ex} [/tex] is the the speed at which the exhaust fuel is being ejected relative to the rocket


    Also, assume that the rocket is ejecting mass at a constant rate, so [tex] \dot{m}=-k [/tex] (where k is a positive constant) so that [tex]m = m_{0} - kt [/tex]

    Solve the equation for v as a function of t, using separation of variables (rewriting the equation so that all the terms involving v are on the left and all the terms involving t on the right)

    Now what is confusing me is at what point to I have to substitute for
    [tex] \dot{m} [/tex] and m ?

    Can I start by saying, [tex] m \frac{dv}{dt} = k v_{ex} - mg [/tex]

    and then plug in [tex] m = m_{0} - kt [/tex] and take it from here?
     
    Last edited: Jan 8, 2006
  2. jcsd
  3. Jan 8, 2006 #2

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    Yes. Do what you suggest. Then it's just a case of separation of variables.

    Edit: You can reduce a little bit of the work by dividng by [tex]m[/tex] first.
     
  4. Jan 8, 2006 #3
    Thanks for the reply.
    so, dividing by m,
    [tex] \frac{dv}{dt} = \frac{kv_{ex}}{m} - g[/tex]

    then, [tex] \frac{dv}{dt} = \frac{kv_{ex}}{m_{0}-kt} - g[/tex]

    [tex] \int dv = \int \frac{kv_{ex}}{m_{0}-kt} dt - \int g dt[/tex]

    is this right so far?
     
    Last edited: Jan 8, 2006
  5. Jan 8, 2006 #4

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    Perfectly correct, yes.
     
  6. Jan 8, 2006 #5
    Thanks a lot for the help. Excuse my rusty math skills but I haven't done calculus in a long time. So would it be correct to integrate the first term on the right like this:

    [tex] \int \frac{kv_{ex}}{m_{0}-kt} dt = k v_{ex} \int \frac{1}{m_{0}-kt} dt [/tex]

    then let [tex] u = m_{0} - kt [/tex]

    [tex] du = -k dt [/tex]

    [tex] \frac {-1}{k} du = dt [/tex]

    so we get [tex] k v_{ex} \int \frac {1}{u}. \frac{-1}{k}du [/tex]

    taking the constant k outside the integral:

    [tex] -v_{ex} \int \frac{1}{u}du [/tex]

    which is equal to [tex] -v_{ex} \ln u + C [/tex]

    which is [tex] -v_{ex} \ln (m_{0} - kt) + C [/tex]

    I needed the absolute value signs but didn't know how to write them using latex

    does this look correct for that first integral on the right hand side?
     
    Last edited: Jan 8, 2006
  7. Jan 9, 2006 #6

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    Your maths skills may be rusty, but you ceratinly haven't forgotten procedure. Your integration method is spot on, and once again everything is correct.
    Also, you can check your work by differentiating your result to ensure that you get back to what you started with. This will help to confirm it.
    Just one little comment. Since this is just a partial integral (you have yet to integrate the other terms), then you don't actually need to bother with a constant of integration at this point.
     
  8. Jan 9, 2006 #7
    thanks

    yeah i was aware that I need to integrate the other parts and then I won't need the constant.

    I should use absolute value though when I take the natural log of u, right? Or can I get away without using absolute values? I am not sure how that works and when/why do we get to omit them. If you can clarify it that would be great :)

    cheers mate
     
  9. Jan 9, 2006 #8

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    You should take absolute values if you know that the argument of the log function could be negative. E.g. if you have a trig function,
    [tex]\ln|\sin x|[/tex]
    In your case however, u is the current mass of the rocket minus propellant and will never be negative, so you don't need to use the modulus symbols.
     
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