Proof of Gauss' Law of gravitation without reference to Newton? Gauss' Law of gravity is:
$$\bigtriangledown \cdot \mathbf{g}=  4\pi G\rho$$
This can be shown to be equivalent to Newton's Law of gravity via the divergence theorem.  However, this does not really constitute a proof.  Where does the $4\pi$ come from?  I would like to derive Gauss' Law from the notion of solid angle and/or the definition of the scalar potential.
Specifically, just using the following facts:
$$\mathbf{g}=\bigtriangledown\cdot\phi$$
Where $\phi$ is the scalar potential, and the definition of the scalar potential (from the wikipedia article):
$$4\pi\phi=\int_V \frac{\bigtriangledown\cdot\mathbf{g}}{r}\:\:dV$$
(Where the RHS is a volume integral, and the necessary assumptions about asymptotic vanishing towards infinity are made).
 A: One can get quite far towards Gauss' Law of gravity without knowing Newton's Law, but not all the way. To explore this, suppose that all we know is that the gravitational force depends on mass and radial distance:
$\mathbf{g} = k(M, r) \cdot \mathbf{e_r}$
Here, $k$ is an unspecified function, $M$ is a mass which can be taken as being located at the origin, $r$ is the radial distance from the origin, and $\mathbf{e_r}$ is a radial unit vector. 
Now we imagine a closed spherical surface $\delta V$ of radius $r$ centered at the origin. The total flux of the gravitational field $\mathbf{g}$ over the closed surface $\delta V$ is 
$\oint_{\delta V} \mathbf{g} \cdot d\mathbf{A} = \oint_{\delta V} k(M, r) \cdot \mathbf{e_r} \cdot d\mathbf{A}$
= $k(M, r) \oint_{\delta V} \mathbf{e_r} \cdot \mathbf{e_r} \cdot dA$ 
= $k(M, r) \oint_{\delta V} dA$
=  $k(M, r) \cdot 4 \pi r^2$  (this explains where the $4 \pi$ comes from).
The total flux is independent of $r$ so to eliminate $r^2$ we must have $k(M, r) = \frac{k^*(M)}{r^2}$ where $k^*(M)$ is some unspecified function of $M$, and therefore 
$\oint_{\delta V} \mathbf{g} \cdot d\mathbf{A} = k^*(M) \cdot 4 \pi$
By the divergence theorem we can write this as 
$\oint_{V} \nabla \cdot \mathbf{g} dV = k^*(M) \cdot 4 \pi$
and therefore differentiating both sides with respect to $V$ we get 
$\nabla \cdot \mathbf{g} = k^{* \prime}(M)\frac{dM}{dV} 4 \pi$
If we set $\frac{dM}{dV} = \rho$ we see that this is nearly Gauss' Law: 
$\nabla \cdot \mathbf{g} = k^{* \prime}(M) 4 \pi \rho$
We only need Newton's Law to tell us that $k^{* \prime}(M) = -G$ at this final stage. 
The link with the scalar potential comes through $\mathbf{g} = -\nabla \phi$ which gives us 
$\nabla^2 \phi = 4 \pi G \rho$ 
(a well known type of partial differential equation known as Poisson's equation). 
