I'm working on a problem where I have to prove that the gravitational field outside a sphere of varying density looks from outside of the sphere the same as a point mass at origin:

$$\int_{}^{} \int_{}^{} \int_{B}^{} \frac{\rho(\sqrt{x^2+y^2+z^2})}{|(x,y,z) - (0,0,z_{0})|}dV=\frac{m}{z_{0}}$$

where B is a sphere of radius $R$, $\rho$ is density (a function of only radial position, and is unspecified), $m$ the mass of the sphere and $z_{0}$ the distance of a point from origin that lies outside the sphere ($z_{0} > R$). The denominator in the integrand is the distance between point $(x,y,z)$ and the point $z_{0}$. A hint is given to use the cosine law to express the distance.

Converting to spherical coordinates and using the cosine law:

$$\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{R}\frac{\rho(r)}{\sqrt{r^2+z^2-2rz\cos(\phi)}}r^2\sin(\phi)dr d\phi d\theta$$

Changing the order of integration, pulling out density and one of the $r$s out of the $\phi$ integral (since radius is not a function of angle) we get the derivative of the inner function (inside the root):

$$\frac{1}{2z}\int_{0}^{2\pi} \int_{0}^{R} \rho(r)r \int_{0}^{\pi}\frac{1}{\sqrt{r^2+z^2-2rz\cos(\phi)}}2rz\sin(\phi)d\phi dr d\theta$$


$$\frac{1}{2z}\int_{0}^{2\pi} \int_{0}^{R} \rho(r)r 2\sqrt{r^2+z^2-2rz\cos(\phi)} \vert_{0}^{\pi} dr d\theta$$


$$\frac{1}{2z}\int_{0}^{2\pi} \int_{0}^{R} 2\rho(r)r (\sqrt{r^2+z^2+2rz} - \sqrt{r^2+z^2-2rz}) dr d\theta$$

But now I'm stuck. I need to integrate w.r.t $r$ next, but the function $\rho$ is not given, and I can't get anywhere trying to integrate by parts $2\rho(r)r (\sqrt{r^2+z^2+2rz} - \sqrt{r^2+z^2-2rz})$.

How should I proceed? Thank you!

  • $\begingroup$ I don't think you completely transformed to spherical coordinates -- your integrand still contains the variable $z.$ $\endgroup$
    – Allawonder
    Commented Sep 12, 2019 at 17:48
  • $\begingroup$ Assume $z > r$, one has $\sqrt{r^2 + z^2 + 2rz} - \sqrt{r^2+z^2-2zr} = (z+r)-(z-r) = 2r$ $\endgroup$ Commented Sep 12, 2019 at 18:06
  • $\begingroup$ @achille hui Thanks! With that step I was able to solve it. I would accept your answer if I could :) $\endgroup$
    – S. Rotos
    Commented Sep 12, 2019 at 19:13

1 Answer 1


There is a much simpler way of looking at this. Gauss' Law of Gravitation can be read as

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

where $\bf g$ is the gravitational acceleration caused by the field, $G$ is the gravitational constant, and $\rho$ is the mass density. By integrating over some volume $\Omega$, we have

$$\int_{\Omega}\nabla\cdot{\bf g}dV=-4\pi G\int_{\Omega}\rho dV=-4\pi GM_\Omega$$

where $M_\Omega$ is the total mass contained in the volume $\Omega$. The left side of the equation can use the Divergence Theorem, so

$$\int_{\partial\Omega}{\bf g}\cdot d{\bf S}=-4\pi GM_\Omega$$

From here, we can't do much without specifying a solid $\Omega$. As per usual, it is useful to consider $\Omega$ as a sphere of radius $r$, and we will assume that $r>R$ such that $M_\Omega=m$, the mass of the object.

If, as you say, the density is angularly symmetric, it is fair to say that the direction of gravity is always radial, as all other forces would balance out. The force of gravity is directed inward, while the surface normal is outward, so the dot product is negative and

$$\int_{\partial\Omega}gdS=4\pi Gm$$

Since we assume that $g$ is constant everywhere on the surface of $\Omega$, then

$$4\pi r^2g=4\pi Gm\implies g=\frac{Gm}{r^2}$$

which is precisely the definition of the gravitational acceleration of a point mass of mass $m$ at the origin. Thus, the acceleration has no dependence on the radius $R$ (so long as $r>R$) nor the density (so long as we can write $\rho$ as a function of the radius $r$ alone).


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