Circular wire center of gravity

  • #1
Cassis
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The center of gravity of a circular wire should be in its center? What would happens if we put a marble and a wire into an empty space. Would the gravity force skyrocket as the marble approach the wire center of gravity?
 
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  • #2
Cassis said:
The center of gravity of a circular wire should be in its center? What would happens if we put a marble and a wire into an empty space. Would the gravity force skyrocket as the marble approach the wire center of gravity?
This isn't difficult to do. A key ring is a circle of wire - have you ever noticed anything like infinite forces trying to take your keys put of your pocket?
 
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  • #3
Cassis said:
The center of gravity of a circular wire should be in its center? What would happens if we put a marble and a wire into an empty space. Would the gravity force skyrocket as the marble approach the wire center of gravity?
The inverse square law applies to the exterior of spherically symmetrical sources.
 
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  • #4
Cassis said:
What would happens if we put a marble and a wire into an empty space.
Depending on the relative size and the orientation of the ring, the system could oscillate about the barycenter of the pair, or they would collide, to finally settle, looking like a miniature Saturn with one wire ring.
 
  • #5
Baluncore said:
Depending on the relative size and the orientation of the ring, the system could oscillate about the barycenter of the pair, or they would collide, to finally settle, looking like a miniature Saturn with one wire ring.
Hm. Larry Niven thought so too. He wrote an essay on it, then followed up with a novel. That is, until his fans (many of whom of whom are mathematicians) disagreed. Niven had to write an entire sequel to Ringworld just to ret-con the engineering.

A mass at the centre of a ring of mass is not stable. Given any nudge it will drift away from the centre until the centre and the ring touch.
 
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  • #6
DaveC426913 said:
A mass at the centre of a ring of mass is not stable. Given any nudge it will drift away from the centre until the centre and the ring touch.
OK, but the OP did not specify a relative diameter, nor any nudge away from perfect stability.

If the ring is small, it will seat on the spherical surface of the marble.

If the ring is big, it may oscillate for some time, but in the end the marble will rest against the inside of the ring.
 
  • #7
I’m not sure anyone but @A.T. got at what the OP was asking, and I’m not sure even that was completely clear. I think the point is that the inverse square law applies ONLY in the case of the exterior of a spherically symmetric mass.

All the mass of the ring is NOT at the center of the ring. The test marble will be gravitationally attracted to each little piece of the ring with the inverse square law. The total force is the integral of all those contributions. In the center of the ring the gravitational attraction is outward towards each piece of the ring and by symmetry adds up to zero. There is no force on the marble at the center of the ring.

Instead of a ring consider a spherical shell. When the marble is outside the shell all those contributions from all the little bits of the shell happen to add up to act exactly as if all the mass was at the center of the shell. However, that happenstance is a happy quirk of symmetry, nature, and mathematics described by Stoke’s theorem. Once the marble reaches the shell, that trick that makes all the contributions add up to act as if they are at the center of the ring no longer applies. In fact, the same Stokes theorem can be used to show that the net force on the marble is zero anywhere inside the sphere (surprisingly, not just the center!). This is equivalent to Gauss’ law in electrostatics.

So, bottom line, you can’t assume the gravity of a distributed mass will act the same as if all the mass were located at the center of gravity. That happens to be true only in certain specific circumstances. Those circumstances are very common (astronomical bodies are often nearly spherically symmetric) but not universal.
 
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  • #8
Yeah, that was kind of what i was trying to point out. Unless I'm wrong, a ringed sphere is not only unstable, it is a positive feedback loop. Like a ball balancing on another ball. Even a miniscule deviation from exact centre is magnified and accelerated.
 
  • #9
DaveC426913 said:
Yeah, that was kind of what i was trying to point out. Unless I'm wrong, a ringed sphere is not only unstable, it is a positive feedback loop. Like a ball balancing on another ball. Even a miniscule deviation from exact centre is magnified and accelerated.
Yes this is what unstable usually means. If you place the mass at rest inside the ring, off-center, in the plane of the ring, there should be no oscillation, just a crash into the inside of the ring.

A more complex case is placing it out-of-plane or giving it out-of-plane initial velocity.
 
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  • #10
Thank you all for your answer, it made very clear where I was wrong, thank you again!
 

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