# Gauss law in gravitation

by iitjee10
Tags: gauss, gravitation
 P: 534 Yes. Notice how gravity corresponds with electrostatics; for point masses/charges you have $$\vec E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat r \leftrightarrow \vec g = -G \frac{m}{r^2} \hat r.$$ Thus you have the correspondences $$q \leftrightarrow m, 1/4\pi\epsilon_0 \leftrightarrow -G$$. From Gauss's law (in integral form) for electrostatics, you can get the corresponding equation for gravity: $$\oint \vec E \cdot d \vec a = \frac{q_{encl}}{\epsilon_0} \leftrightarrow \oint \vec g \cdot d \vec a = -4\pi Gm_{encl}.$$ In differential form you get $$\nabla \cdot \vec E = \frac{\rho_e}{\epsilon_0} \leftrightarrow \nabla \cdot \vec g = -4\pi G \rho_m$$ where $$\rho_e$$ is the charge density, and $$\rho_m$$ is the mass density.