Theoretically, from Newton's equation for the force between two bodies if we take two ordinary rings in a distance there will be a force according to the square inverse law of their distance (from their mass center) such as case 1 from the below picture. If i place their mass center so as to coincide (case 2) the Force will be infinite, although they can separate from each other with a small force. What's wrong with that supposition?
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  • $\begingroup$ Simple symmetry arguments can be used to prove that the forces between the two rings are zero. But to understand more the answer by @sammy gerbil is very complete. $\endgroup$ – Bob Bee Oct 16 '17 at 0:23

First, in principle Newton's law of gravitation holds for two POINT PARTICLES only.

Second I think you'd better consider spherical shells instead of rings.

For rings, there is not a theorem as you stated: "....there will be a force according to the square inverse law of their distance (from their mass center)..."

For spherically symmetric objects, there is a similar theorem though, which is a result of the shell theorem, that the force between two spherical symmetric objects COMPLETELY OUTSIDE ONE ANOTHER is the same as if the two objects's masses were all concentrated at the center. But even for spherically symmetric objects, the above is not true if one is inside the other.

  • $\begingroup$ Indeed, that's why we consider their mass concentrated to a single point which is the center mass point. Also, there is no difference between 2D or 3D shapes. $\endgroup$ – Vaggelis Kyrilas Oct 15 '17 at 17:00
  • $\begingroup$ There is crucial difference between 2D and 3D!!! $\endgroup$ – velut luna Oct 15 '17 at 17:02
  • $\begingroup$ Either they are rings or spheres their mass center points remain the same. $\endgroup$ – Vaggelis Kyrilas Oct 15 '17 at 17:04
  • $\begingroup$ You are using something you called theorem without proof. You just take it for granted. Please google shell theorem! $\endgroup$ – velut luna Oct 15 '17 at 17:06
  • $\begingroup$ Newton's law of gravitation is for two point particles separated by a distance. That's it. Any other statements you claimed require proof! $\endgroup$ – velut luna Oct 15 '17 at 17:08

If instead of two circular rings you used two spherical shells, one inside the other, the resultant gravitational force between them would be zero. This is a consequence of Newton's Shell Theorem. This result is the same whether or not the centres of the two spherical shells coincide. The inner mass does not even have to be spherically symmetric - any shape will do. The only requirements are that the inner mass is completely inside the outer mass, and the outer mass is spherically symmetric.

This works for a sphere because the geometry of 3D space matches the $1/r^2$ nature of gravity. It does not work for a ring because the ring is 2D so it cannot be spherically symmetric. It would work for a 2D ring if gravity were a $1/r$ force instead of $1/r^2$. Then any mass of any 2D shape would experience no resultant gravitational attraction from the outer ring (provided that it remained in the same plane aas the ring). See Gravitational field of thin 2D ring - numerical simulation.

If you do use 2 rings in (real) 3D space, the total force on the inner ring will be zero if the rings lie in the same plane and their centres coincide exactly. Every particle in the inner ring is pulled towards the nearest point on the outer ring, but the inter-molecular forces in the inner ring prevent it from being torn apart. This means that there is a tension in the inner ring. If the inner ring were inside a spherical shell, there would be no resultant force on every particle, and no tension in the ring.

If the centres of the two rings do not coincide, then the inner ring would be attracted to the nearest point of the outer ring. The resultant force between the 2 rings is zero when their centres coincide and reaches a maximum when the rings touch. The force between every 2 point masses in the two rings obeys the $1/r^2$ law, but the resultant force between the 2 entire rings does not necessarily obey the $1/r^2$ law ($r$ being measured between centres).


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