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Troublesome Equation

  1. May 11, 2004 #1

    Kurdt

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    Just looking for some advice on where my maths is going wrong with this. I have the following equation.

    [tex] L_{orb}=(\frac{GD}{M})^\frac{1}{2}M_sM_p [/tex]

    and information that the time derivatives of L and M are zero. Also M_s varies with time along with D. I am supposed to arrive at the following equation.

    [tex] \frac{\dot{D}}{D}=-2(1-\frac{M_s}{M_p})\frac{\dot{M_s}}{M_s}[/tex]

    I first brought the M over to be on the same side as the L as when I take the time derivative they will be 0 and then after taking the time derivative of what is left on the left hand side and rearranging a little I can only get

    [tex]\frac{\dot{D}}{D}=-2\frac{\dot{M_s}}{M_s^2}[/tex]

    Any pointers as to where my maths fails. I realise it could have something to do with a substitution of variables but I'm assuming not as it seems unlikely at the minute and I wouldn't like to type out all the possibilities :wink: . Any help is much appreciated.
     
    Last edited: May 11, 2004
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  3. May 11, 2004 #2

    arildno

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    What's r and b?
    (This looks like some sort of spin equation in a gravitational field, but..)
    And M? Do you mean [tex]M_{p}[/tex]?
     
  4. May 11, 2004 #3

    Kurdt

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    sorry r and b was supposed to be part of the subscript I will change it and M is the sum of M_p and M _p.
     
  5. May 11, 2004 #4

    Kurdt

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    Its just basically the orbital angular momentum in the Roche model of binary stars which I am studying at the minute. Trying to extract useful information on the rate of orbital decay.
     
  6. May 11, 2004 #5

    arildno

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    Remove the [tex]G^{\frac{1}{2}}[/tex] over as well.
    Then we have:
    [tex]0=\frac{\dot{D}}{2\sqrt{D}}M_{s}M_{p}+\sqrt{D}\frac{d}{dt}(M_{s}M_{p})\rightarrow\frac{\dot{D}}{D}=-2\frac{\frac{d}{dt}(M_{s}M_{p})}{M_{s}M_{p}}[/tex]

    The desired expression is now readily obtained
     
  7. May 12, 2004 #6

    Kurdt

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    Thanks for your help. I knew it was something simple that I'd missed.
     
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