Gauss’s flux theorem for gravity, Poisson’s equation and Newton’s law

Gauss’s flux theorem for gravity (also known as Gauss’s law for gravity) in differential form says that

$\nabla \cdot \mathbf{g} = -4 \pi G \rho$

where $\mathbf{g}$ is the gravitational force, $G$ is the Newtonian gravitational constant and $\rho$ is the (differential) mass density. The link with the scalar potential through $\mathbf{g} = -\nabla \phi$ gives us

$\nabla^2 \phi = 4 \pi G \rho$

(a well known type of partial differential equation known as Poisson’s equation).

In this short note I want to record the observation that one can get quite far towards deriving Gauss’s law for gravity without knowing Newton’s law of universal gravitation, 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