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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?

Sunny
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  • If you want to get the $-GM/r$ then you need a Dirac delta as the source $\rho$ i.e when integrating, $\int \rho(r) 4\pi r^2, dr=\int \rho, dV= M$, the total mass. Of course this way of doing things is hand-wavy but it’s a start. Now solve the ODE, and be mindful of the integration constant. – peek-a-boo Nov 30 '23 at 10:22
  • @peek-a-boo What do you mean by the source density $\rho$ is a Dirac delta function? Could you elaborate more on that? How does defining the Dirac delta function allow you to equate $\int \rho(r) 4\pi r^2 dr = \int \rho dV$? Also, in the derivation method I use, $\rho$ is not a function of r, $\rho$ is a constant $\rho_0$. – Sunny Dec 01 '23 at 13:25
  • A Dirac delta source very roughly means density is zero everywhere except for the exact point where the mass is located where it is “infinite”, in such a manner that when you integrate over all space, you get the mass $M$. So really, you should first solve Laplace’s equation, i.e 0 on the RHS. Solving this ODE will give you two “constants of integration”. Work this out first. Then one of the constants you’ll set to 0, and the other will give you the $M$ (after you take into account that the integral over space must give $M$). – peek-a-boo Dec 01 '23 at 13:30

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