I am trying to derive Newton’s gravitational potential $\phi_N = -\frac{GM}{r}$ from Poisson’s equation $\Delta \phi_N = 4\pi G\rho$, where G is the gravitational constant, M is the mass of which the potential is being measured, r is point of measurement, $\rho$ is the density of the mass.
Here is my working so far: For a spherically symmetrical source with constant density $\rho = \rho_0 = const.$, where $\phi_N = \phi_N(r)$ (only r-dependence), we have:
$\Delta \phi_N = \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d\phi_N}{dr}\right) = 4\pi G \rho_0$
Where the Laplacian in spherical coordinates is given by $\Delta = \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\right) + \frac{1}{r^2\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d}{d\theta}\right) + \frac{1}{r^2\sin^2\theta}\frac{d^2}{d\varphi^2}$
Rearranging the expanded Poisson’s equation, we have: $$d\left(r^2\frac{d\phi_N}{dr}\right) = 4\pi G \rho_0 r^2 dr $$ $$r^2\frac{d\phi_N}{dr} = 4\pi G \rho_0 \int r^2 dr $$ $$r^2\frac{d\phi_N}{dr} = 4\pi G \rho_0 \frac{r^3}{3} $$ $$d\phi_N = 4\pi G \rho_0 \frac{r}{3} dr $$ $$\phi_N = \frac{4}{3}\pi G \rho_0 \int r dr $$ $$\phi_N = \frac{4}{3}\pi G \rho_0 \frac{r^2}{2} $$
Rearranging, we have: $$\phi_N = \frac{4}{3}\pi r^3 G \rho_0\frac{1}{2r} $$
We know from basic geometry that the volume of the spherically symmetric source is equal to the volume of a sphere: $\frac{4}{3}\pi r^3$. Therefore, the equation can be written as:
$$\phi_N = V G \rho_0\frac{1}{2r} $$
A mass’ volume times its (constant) density is equal to the mass therefore:
$$\phi_N = M G\frac{1}{2r} $$
This result does not coincide with Newton’s gravitational potential $\phi_N = -\frac{GM}{r}$. Where did I go wrong in my working?