Density of star from hydrostatic equilibrium and pressure

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Dazed&Confused
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Homework Statement


Assume that the pressure [itex]p[/itex] in a star with spherical symmetry is related to the density [itex]\rho[/itex] by the (distinctly unrealistic) equation of state [itex]p= \tfrac12 k\rho^2[/itex], where [itex]k[/itex] is a constant. Use the fluid equilibrium equation obtained in Problem 23 to find a relation between [itex]\rho[/itex] and [itex]\Phi[/itex]. Hence show that Poisson's equation yields
[tex] \frac{d^2[r\rho(r)]}{dr^2} = -\frac{4 \pi G}{k}r \rho(r)[/tex]
Solve this equation with the boundary conditions that [itex]\rho[/itex] is finite at [itex]r=0[/itex] and vanishes at the surface of the star. Hence show that the radius [itex]a[/itex] of the star is determinde solely by [itex]k[/itex] and is independent of its mass [itex]M[/itex]. Show also that [itex]M =(4/ \pi )a^4 \rho(0)[/itex].

Homework Equations


[itex]\nabla p = \rho \textbf{g}[/itex] and [itex]p + \rho \Phi = \textrm{constant}[/itex]. And [itex]\nabla^2 \Phi = 4\pi G \rho[/itex].

The Attempt at a Solution


Assuming the differential equation, I am able to do the rest of the question. The relation they are looking for is [itex]\Phi = -k\rho[/itex] and this can be easily shown by
[tex]\nabla p = k\rho'(r) \rho(r) \hat{\textrm{r}} = \rho \textrm{g}[/tex] and so
[tex]-\nabla \Phi = \textrm{g} = k\rho'(r) \hat{\textrm{r}}[/tex]

from which you can get [itex]\Phi[/itex]. However by the second relevant equation this must also mean that [itex]-\tfrac12 k \rho^2 = \textrm{constant}[/itex], or I'm missing something. I've also not been able to obtain the differential equation.
 
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Not that it matters too much but it should be [itex]M = (4/\pi)a^3\rho(0)[/itex]. Any ideas anyone?

Edit 2: the equation [itex]p + \rho \Phi = \textrm{constant}[/itex] is only for an incompressible fluid, so there is no contradiction.
 
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Dazed&Confused said:
[tex]-\nabla \Phi = \textrm{g} = k\rho'(r) \hat{\textrm{r}}[/tex]

Note that this is just ##\nabla \Phi = -k \nabla \rho##.

To relate this to Poisson's equation, take the divergence of both sides.
 
Hi thanks for responding. I did realize this and found the relation they wanted. I took the Laplacian of this which did not work, so unsurprisingly taking the divergence of what you have said does not work either.
 
Yes I did use them. Could you try to see if you get the correct differential equation?
 
Certainly. I used Poisson's equation on the relation, so [tex] \nabla^2 \Phi = 4 \pi G \rho = -k \nabla^2 \rho = -k\frac{1}{r^2} \frac{d}{dr} \left ( r^2 \frac{d\rho}{dr} \right )[/tex]

from which I can't get the correct equation.
 
Thanks I was able to get it.