Core pressure of sphere of fluid in a vacuum Suppose that we have some known fluid floating in space with no outside gravitational forces. How would we calculate the pressure of the outside fluid on acting on the "core" of the fluid (in other words, the gravitational force the sphere of fluid is exerting on itself)? (This would also assume that the fluid would not evaporate, boil, or change state in general.)
 A: Assuming the fluid forms a sphere of radius $R$ and is incompressible with constant density $\rho$, we could calculate the pressure it exerts on a smaller sphere of radius $r_1$ by first getting the weight of the fluid above this smaller sphere as follows: 
$$W=\int_0^{\pi}\int_0^{2\pi}\int_{r_1}^R\dfrac{G(\rho*\dfrac{4}{3}\pi r^3)}{r^2}\rho r^2\sin(\zeta)dr d\theta d\zeta$$
Where $\dfrac{G(\rho*\dfrac{4}{3}\pi r^3)}{r^2}$ is the strength of the gravitational field at the radius of interest (taking into account the matter inside the sphere of radius of interest). Also, $\rho dV = \rho r^2\sin(\zeta)dr d\theta d\zeta$ in spherical coordinates and constitutes the mass of the portion of mass immediately above the radius of interest.
Solving this integral yields $W=\dfrac{4}{3}\pi ^2 G\rho ^2 (R^4-r_1^4)$ and then the pressure would be the weight divided by the surface area of the sphere of radius $r_1$ $$P=\dfrac{\pi}{3r_1^2} G\rho ^2 (R^4-r_1^4)$$
As you can see, taking the limit when $r_1 \rightarrow 0$, we have $P \rightarrow \infty$ and that is because the area of this sphere becomes infinitely small.
