How does the gravitational field behave inside a star? The interior gravitational field of a star with constant density is given by 
$ds^{2}=-\left(\frac{P_{c}+\rho_{0}}{P(r)+\rho_{0}}\right)^{2}dt^{2}+\frac{dr^{2}}{1-\frac{8\pi\rho_{0}}{3}r^{2}}+r^{2}d\theta^{2}+r^{2}\sin\theta^{2}d\phi^{2}$
Where $P_{c}$ is the pressure at the centre of the star, $\rho_0$ is the density of the star and $P(r)$ is the pressure at point $r$.
My question is does the $dr^{2}$ component $\left({1-\frac{8\pi\rho_{0}}{3}r^{2}}\right)^{-1}$ ever change sign as we go from the center of the star to the surface or does it always stay the same sign.
 A: Obviously, it does change sign when $r_{0} = \sqrt{\frac{3}{8\pi\rho_{0}}}$
Perhaps a more interesting question is "why is this?"  Well, we know that the exterior of the star has to be the Schwarzschild Metric, since this is the unique spherically symmetric vacuum solution with zero charge.  Consequently, we know, at the surface radius $R$:
$$1-\frac{2M}{R} = \left(\frac{P_{c} + \rho}{P(R) + \rho}\right)^{2}$$
and
$$1-\frac{2M}{R} = 1 - \frac{8\pi\rho}{3}R^{2}$$
Solving the second equation for $M$:
$$M = \frac{4\pi \rho R^{3}}{3}$$
putting this into the first equation:
$$1-\frac{8\pi\rho R^{2}}{3} = \left(\frac{P_{c} + \rho}{P(R) + \rho}\right)^{2}$$
But, on our first step, we worked out the value $r_{0}$ for our sign change to be equal to $r_{0} = \sqrt{\frac{3}{8\pi\rho_{0}}}$, so this really is:
$$1-\left(\frac{R}{r_{0}}\right)^{2} = \left(\frac{P_{c} + \rho}{P(R) + \rho}\right)^{2}$$
It should be clear that the right hand side of this equation is always positive and nonzero, so if $R$, the surface of our star, is ever larger than $r_{0}$, then there is no solution for our system of equations.  What does that mean?  Well, there's a theorem, that if you ever fit a mass $M$ inside of a sphere of radius $2M$, then it is impossible to have a stable body, and collapse to a black hole has to happen.  But, every time I add a sliver of radius $dR$ to the surface of a constant-volume star, I increase the radius by an amount $dR$, but I increase the volume (and therefore the mass, since it is constant density) by an amount $4\pi R^{2}dR$.  Therefore, the mass will grow faster than the radius, and once I let the radius get sufficiently large, no matter what the central density, I have eventually put a mass $M$ inside of a schwarzschild radius, the matter distribution is no longer stable, and I get a black hole.
That is what your sign change is yelling out at you about.
