Understanding the Troublesome Equation: Tips for Solving Math Problems

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In summary, the conversation is about a mathematical equation involving orbital angular momentum and variables that change over time, such as M_s and D. The goal is to arrive at a certain equation involving the time derivatives of these variables. After some discussion and corrections, the desired expression is obtained.
  • #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 realize 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.
 
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  • #2
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]?
 
  • #3
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.
 
  • #4
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.
 
  • #5
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
 
  • #6
Thanks for your help. I knew it was something simple that I'd missed.
 

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