Circular wire center of gravity

Click For Summary
SUMMARY

The center of gravity of a circular wire is not stable at its geometric center. When a marble is placed near a circular wire in space, gravitational forces act according to the inverse square law, causing the marble to be attracted to each segment of the wire. This results in a net gravitational force of zero at the center of the wire, but any deviation from this point leads to instability, causing the marble to drift towards the wire. The discussion highlights the importance of understanding gravitational interactions in non-spherical geometries, as demonstrated by the principles of Stoke’s theorem and Gauss’ law.

PREREQUISITES
  • Understanding of gravitational forces and the inverse square law
  • Familiarity with Stoke’s theorem and Gauss’ law
  • Basic knowledge of circular and spherical geometries
  • Concept of stability in physical systems
NEXT STEPS
  • Research the implications of Stoke’s theorem in gravitational contexts
  • Explore the stability of various geometric configurations in gravitational fields
  • Study the applications of Gauss’ law in electrostatics and gravity
  • Investigate the dynamics of oscillatory systems in gravitational fields
USEFUL FOR

Physicists, engineers, and students studying gravitational mechanics, particularly those interested in the stability of systems involving circular and spherical geometries.

Cassis
Messages
2
Reaction score
0
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?
 
Physics news on Phys.org
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?
 
  • Like
  • Haha
  • Love
Likes   Reactions: davenn, Vanadium 50, russ_watters and 1 other person
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.
 
  • Like
Likes   Reactions: vanhees71
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.
 
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.
 
Last edited:
  • Like
  • Wow
Likes   Reactions: Nugatory, russ_watters, vanhees71 and 3 others
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.
 
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.
 
  • Like
Likes   Reactions: nasu and A.T.
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.
 
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.
 
Last edited:
  • #10
Thank you all for your answer, it made very clear where I was wrong, thank you again!
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 51 ·
2
Replies
51
Views
5K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 49 ·
2
Replies
49
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K