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I've been told to revise the derivation that proves $\frac{\mathrm{d}P}{\mathrm{d}r} =\frac{GM(r)p(r)}{r^2}$ where brackets indicate a function of, P is pressure and p is density. Rather helpfully he hasn't given us it to revise, so if anyone knows it I'd be really grateful. Thanks.

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You should add some more information in order to let me understand your question. –  Bernhard Jan 16 '12 at 18:14
    
Yeah looking at it is pretty vague, were looking at stars and we're told to derive an expression for the rate of pressure change between the core and the surface, the pressure at the surface being 0. At a specific radius you have the mass outside pressing down, the mass inside pulling in, and the pressure from inside pushing out. The revision notes we have saying "revise this" are really badly worded so I'm afraid that's all I have to go on. –  Becky Jan 16 '12 at 19:07

1 Answer 1

I think you've mistaken $P\ $ for $\rho$.

Here's why: Since we're looking for a static solution, the gravitational force balances exactly the pressure gradient: $$\vec\nabla P +\vec f=0$$ where $\vec f$ is the gravitational force density. Since the system is spherically symmetric, you can write $\vec \nabla P=\frac{dP}{dr}\hat r$, and use $\vec f=-\rho(r) \frac{G M(r)}{r^2}\hat r$ to get $$\frac{dP}{dr}= \frac{G M(r)\rho(r)}{r^2}$$

Now in order to solve this you need to have some constitutive relation $\rho=\rho(P)$, but that is probably what your teacher is going to do in class...

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