# Newton's original proof of gravitation for non-point-mass objects

Suppose we have two bodies, one very large (Earth), and one very small (a cannon ball). If the cannon ball is some distance away from the Earth, to find out the force produced on the cannot ball, we need to compute an integral in three dimensions of how each infinitesimal piece of Earth pulls on the ball, i.e.

$$\int_{\Omega}G \frac{m\delta(\mathbf{r})}{||\mathbf{r}||^2}dV,$$

where $\Omega$ is region of the large mass, $\delta$ is the density of the large mass, and $\mathbf{r}$ is the distance from the small mass (which we consider a point object) to each point of the large mass.

Newton showed that, provided that the large mass is a perfect sphere with uniform density, we can treat it as a point mass as well. This is not hard to do with an iterated triple integral.

Newton did not know about triple iterated integrals, and according to a physics book I have, devised a clever proof of this using a single integral only. Does someone know how Newton did this?

The third integral is over all the annuli in the sphere, over $0\leq\phi\leq\tau/2$ or over $R-r\leq s\leq R+r$. This one is a little bit hairy, even with the advantage of modern notation.
Newton's clever trick is to consider the relationship between the force due to the smaller, nearer annulus HI and the larger, farther annulus KL defined by the same viewing angle (in modern notation, $d\theta$). If I understand correctly he argues again, based on lots of similar triangles with infinitesimal angles, that the smaller-but-nearer annulus and the larger-but-farther annulus exert the same force at P. Furthermore, he shows that the force doesn't depend on the distance PF, and thus doesn't depend on the radius of the sphere; the only parameter left is the distance PS (squared) between the particle and the sphere's center. Since the argument doesn't depend on the angle HPS, it's true for all the annuli, and the theorem is proved.