Gauss law in gravitation Is it possible to use Gauss's law of electromagnetism, (The net electric flux through any closed surface is equal to $1⁄\epsilon$ times the net electric charge enclosed within that surface.) to calculate the gravitational field at point by making certain changes, i.e, by replacing electric flux with gravitational flux, $1⁄\epsilon$ with $1/(4\pi\,G)$, and charge with mass?
 A: Yes, you can use Gauss's law for gravity.
$$\nabla\cdot\vec{g} = 4 \pi\, G\, \rho $$
or
$$ \oint \vec{g}\cdot\mathrm{d}\vec{a} = 4 \pi\, G\, M_\mathrm{enc} $$
where $\vec{g}$ is the gravitational field (equivalently, acceleration due to gravity), $\rho$ is mass density, and $M_\mathrm{enc}$ is the total mass enclosed by the Gaussian surface.
When you make the comparison to Gauss's law for electric fields, you can see how the constants work out the way that they do:
$$E = \frac{1}{4\pi\, \epsilon_0} \frac{Q}{r^2}, \quad\quad g = G\, \frac{M}{r^2}, $$
so $1/\epsilon_0\rightarrow 4\pi\, G$.
One common use for Gauss's law for gravity is to determine the gravitational field strength at a given depth inside the Earth.  It is very similar to the calculation for the electric field inside a charged, insulating sphere.
A: Gauss Law for Gravity basically says that the total gravitational flux emanating from a sphere enclosing the Earth is $4 \pi G M$.
Now divide this by the total surface of the sphere $4 \pi R^2$ with $R$ the radius of the Earth.
The result is $\frac{GM}{R^2}$ giving the gravitational flux density. If you calculate the numerical result you get $9.81\mathrm{m/s^2}$.
A: In my opinion, should be! Unfortunately, Einstein GR does not satisfy it...
Static, (i.e. from immobility) acceleration calculated from geodesic curves formulae, does not satisfy Gauss law nor for reference space (map) - what is obvious, nor for a local observer (on the surface of a planet. For the planet, of radius equal 1.5 r_s (Schwarzschild), relativistic acceleration is (about) twice this predicted by Gauss theorem.
