Water planet

Due to spherical symmetry, the fields $\vec g(\vec r) = g(r) \vec e_r$ and $p(\vec r) = p(r)$ depend only on radial distance $r = |\vec r|$ and the unit vector $\vec e_r = \vec r/r$ . If we integrate over a sphere with radius $r \leq R$ , Gauss' law results $\begin{aligned} \oint \vec g(\vec r) \cdot d\vec A &= g(r) \cdot 4 \pi r^2 \\ - 4 \pi G \int \vec \rho(\vec r) \cdot dV &= -4\pi G \rho_0 \cdot \frac{4}{3} \pi r^3 \\ \Rightarrow \quad g(r) &= - \frac{4}{3} \pi G \rho_0 r \end{aligned}$ Integration of the differential equation determines the gravitational pressure: $\begin{aligned} \vec \nabla p(\vec r) &= \frac{\partial p}{\partial \vec r} = \frac{\partial p}{\partial r}\frac{\partial r}{\partial \vec r} = p'(r) \left(\begin{array}{c} x/r \\ y/r \\ z/r \end{array} \right) = p'(r) \vec e_r = \rho_0 g(r) \vec e_r \\ \Rightarrow \quad p(r) &= \int_{0}^r \rho_0 g(r) dr + p_0 = - \frac{2}{3} \pi G \rho_0^2 r^2 + p_0 \end{aligned}$ Since $p(R) = 0$ , the pressure in the center of the planet results $\begin{aligned} p_0 &= \frac{2}{3} \pi G \rho_0^2 R^2 \\ \Rightarrow \quad R &= \sqrt{ \frac{3 p_0}{2 \pi G \rho_0^2} } \approx 131 \,\text{km} \end{aligned}$