Consider a thin ring of mass $M$. What is the net force on any point on the ring caused by the gravity of the other points on the ring?

My intuition for this problem was to find the linear mass density of the ring $\mu = \frac{M}{2\pi r}$. The length of a small element of the circumference of the ring made by sweeping out a small angle $d\theta$ can be given $dx = r\ d\theta$, so the mass of a small portion of the ring is $\mu d\theta = \frac{M\ d\theta}{2\pi}$ . The distance between two points separated by an angle $\theta$ is $2r\sin \frac{\theta}{2}$ (think of an isosceles triangle with two radii of the ring as two sides). The component of the force centripetal to the ring for two points separated by $\theta$ (the tangent component is cancelled by a point symmetrically across the ring) is $F_{\rm centripetal} = F\sin \frac{\theta}{2}$. The mass of a point on the ring is troublesome, but the gravitational acceleration can be found with $a = \frac{MG}{r^2}$. Inserting the terms for mass, distance, and centripetal component means that the small gravitational acceleration caused on a point on the ring by another point $\theta$ apart is $$da = \frac{(md\theta /2\pi)G\sin \frac{\theta}{2}}{4r^2\sin^2\frac{\theta}{2}}d\theta = \frac{MG\csc \frac{\theta}{2}}{8\pi r^2}d\theta$$

Factoring out constants, this can be integrated for the whole ring as $$a= \frac{MG}{8\pi r^2}\int_0^{2\pi}\csc \frac{\theta}{2}d\theta$$ However, this improper integral diverges to infinity. It does not make intuitive sense that the force on the ring by itself to be infinite.

Is there better way to approach this problem that gets a finite answer?

The problem is an offshoot of a similar problem involving gravity from discrete points arranged in a regular polygon. It gave a finite net force for finitely many objects, but slowly increased to infinity as the number of objects increased. Of course, adding objects in the case with discrete objects increased the mass of the whole system, so I was considering the result if the mass is held constant and the discrete masses were approximated by a thin ring.

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    $\begingroup$ Your idealised ring has infinite density, so it makes perfect sense for the force of the ring on itself to be infinite. $\endgroup$ – TonyK Oct 1 '15 at 22:03
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    $\begingroup$ @TonyK Even with finite density he's toast... $\endgroup$ – Kyle Oman Oct 1 '15 at 22:03
  • $\begingroup$ @KyleOman: How so? Do you mean finite linear density? That means infinite spatial density for a ring of zero thickness. $\endgroup$ – TonyK Oct 1 '15 at 22:03
  • $\begingroup$ @TonyK Pretty sure, checking and posting an answer, stay tuned. $\endgroup$ – Kyle Oman Oct 1 '15 at 22:04
  • $\begingroup$ Possibly related (although not a duplicate) question: physics.stackexchange.com/questions/41254/… $\endgroup$ – Daniel Griscom Oct 1 '15 at 22:19

It's even easier to see what's going on by simplifying further. Consider a rod of length $L$ and uniform linear density $\lambda$. If you try to calculate the acceleration caused by the rod on a point at one end of the rod, you should get the integral:

$$|a| = \left|\int_{r=0}^L\frac{G\lambda \, dr}{r^2}\right|$$

Obviously, this diverges, and the problematic part is where $r\rightarrow 0$. What you have an a line of infinite volume density, and you're integrating along it. You can get around this by using a thick ring and doing a volume integral.

$$\vec{a} = \int_V\frac{G\rho(\vec{r})\, r^2\sin^2\theta \,dr\,d\theta\,d\phi}{r^2}\hat{r}$$

That $\hat{r}$ needs to be substituted for $\sin\theta\cos\phi\;\hat{x}+\sin\theta\sin\phi\;\hat{y}+\cos\theta\;\hat{z}$. It's easy enough to see that this won't blow up for sensible choices of $\rho(\vec{r})$, for instance you can calculate the force from an octant of a sphere by taking $\rho$ to be constant and just messing with the integration bounds.


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