# Gravitational force due to a ring but not at it's center!

While I was reading a proof on Newton's Shell Theorem, an idea struck my mind. It was that if there is a 2-D ring of mass $$M$$ and radius $$R$$ with constant linear mass density and we keep a point mass $$m$$ anywhere inside the ring other than it's center at a distance $$r$$ away from it, then what would be the net gravitational force on the mass $$m$$ due to the ring? I feel like it should be zero but I couldn't prove it. I tried taking an element of the ring and the gravitational force due to it but the function becomes complicated.

I also searched the same on physics stackexchange but couldn't find any convincing proofs.

• If you made an attempt you should show what you've done. Commented Mar 26, 2023 at 7:30
• The best 2-d analog of a sphere (in a 3-d space) is an infinite cylinder. A ring does not work the same. Commented Mar 26, 2023 at 10:24
• @Triatticus I wanted to, but I did it in rough and formalising it would require a lot of time. So, instead I just wrote a few words as a hint. Commented Mar 26, 2023 at 12:35
• Yes well that doesn't help people help you if you don't tell us what you did, it helps to see if it was just a minor math mistake or a total conceptual error. Commented Mar 26, 2023 at 17:41
• Commented Oct 14, 2023 at 16:19

This problem can be solved by reasoning, as in the other answers, or by too much math, as in this answer.

The full 3D potential due to a ring of mass or charge cannot be expressed in elementary functions. (Ciftja Babineaux & Hafeez 2006) gives the solution for the potential $$V(\rho,z)=\frac{4G\lambda R}{\sqrt{(\rho+R)^2+z^2}}K\left(\frac{4\rho R}{(\rho+R)^2+z^2}\right)$$ where $$\lambda$$ is the mass per unit length (I translated from charge to mass, the result is identical), $$R$$ is the ring radius and $$\rho,z$$ cylindrical coordinates. $$K$$ is the complete elliptic integral of the first kind $$K(m)=\int_0^{\pi/2}\frac{d\theta}{\sqrt{1-m\sin^2(\theta)}}.$$

To get the force we "just" need to calculate the gradient of the potential. The derivative of the integral is: $$K'(m)=\frac{E(m)}{m(1-m^2)}-\frac{K(m)}{m}.$$ (E is the complete elliptic integral of the second kind.) Good news, we can in principle calculate the $$\rho$$ and $$z$$ derivatives. The bad news is that calculating this turns into reams of algebra. The good news is that this is what computer algebra systems are good at. I am not going to try to reproduce it here (life is too short to keep track of nested parentheses and square roots in LaTeX), but the key thing is that it is clearly nonzero in general.

Addendum: (Escalante 2021) actually works out the field explicitly. They find the slightly neater (but still messy): $$F_\rho = \frac{G\lambda}{R}\frac{1}{\rho \sqrt{(1+\rho)^2+z^2}((\rho-1)^2+z^2)}\left\{ (\rho^2-1-z^2)E\left[\frac{4\rho}{(1+\rho)^2+z^2}\right] +((1-\rho)^2+z^2)K\left[\frac{4\rho}{(1+\rho)^2+z^2}\right] \right\}$$ and $$F_z = \frac{G\lambda}{R} \frac{2z}{\sqrt{(1+\rho)^2+z^2}((\rho-1)^2+z^2)}E\left[\frac{4\rho}{(1+\rho)^2+z^2}\right].$$ (Here $$\rho$$ has been scaled so the ring has radius 1).

Since $$|F_z|>0$$ everywhere except for $$z=0$$ we can conclude that there will be a net force everywhere above and below the ring. For $$z=0$$ $$F_\rho$$ is nonzero, with a sign change at the ring $$\rho=1$$.

A numerical calculation also bears this out, as shown in the figure above showing the potential and gradient in a cross section of the ring. The field strength inside the ring is small, but nonzero.

Indeed, calculating the force along the z-axis (a much easier exercise) is enough to show that it is nonzero except for at the center of the ring.

• Thanks for the wonderful answer! Just one more help: The first link requires me to login through an institution, so could you please insert the screenshots of that journal. Commented Mar 26, 2023 at 12:48
• @TanmayGupta - I found a new reference, the Escalante paper, that has a free preprint and roughly contains the same calculation. Commented Mar 26, 2023 at 15:19
• Thank you so much for helping me out! Commented Mar 26, 2023 at 15:38
• The vector plot has a difficulty I've had many times - you've let the vector lengths be normalized by the field strength right in the middle of the mass, so they're tiny everywhere else. I'd recommend in the future using color for magnitude or limiting the normalization to some distance away from the maximum. Commented Mar 26, 2023 at 17:37

You must be careful because what happens in a $$3$$-D "world" ids not necessarily the same in a $$2$$-D "world".

A "hand-waving" answer to you question is as follows.

The left hand diagram is my attempt to illustrate a spherical shell with two spherical caps with areas $$A$$ and $$a$$.

The ares of those caps are approximately proportional to the distance from a point $$X$$ within the shell, squared, ie $$A \propto R^2$$ and $$a \propto r^2$$.

So the mass of a cap divided by the distance from point $$x$$ is proportional to $$\dfrac {A}{R^2} = \dfrac {R^2}{R^2} = 1$$, ie a constant.
This means that the gravitational attraction $$(\propto \dfrac {\text{mass}}{\rm separation ^2}$$ - Newton gravitation), at point $$X$$ due cap $$A$$ has the same magnitude as the gravitational attraction at point $$X$$ due cap $$B$$ and the two forces are opposite in direction.
Thus the net gravitational field at point $$X$$ is zero.

In two dimensions you have two arcs of length $$L$$ and $$\ell$$ at distances $$R$$ and $$r$$ from point $$Y$$.
This time the length, and hence mass, of an arc is approximately proportional to the distance from point $$Y$$.
The gravitational attraction due to an arc is proportional to $$\dfrac {\text{mass}}{\rm separation ^2}\propto \dfrac {L}{R^2} \propto \dfrac 1R$$ and so is not independent of the position of $$Y$$.
Thus there is a gravitational field within a ring.

• Actually the problem with this approach is that $L$ is too large to be approximated as a point mass while writing forces. So, I wasn't completely satisfied when I previously saw this method too. Commented Mar 26, 2023 at 12:31

I think we can use E(4πr^2)= 4πG( mass inside gaussian surface) For gaussian surface imagine a circle passing the object and do not include mass of assumed element

• Can you please show some math from the scratch which carries your idea. Actually, I am in high school so I know only a little about Gauss's Theorem. Commented Mar 26, 2023 at 12:59