Gravity effects on a solid rigid ring around earth

In summary, the conversation discusses the concept of building a solid ring around the Earth and its potential stability. The assumptions include the ring having a zero weight due to its center of gravity being aligned with the Earth's, the ring hovering 1 meter above the Earth's surface, and the lowest potential energy being in the hovering configuration. However, it is argued that the ring would actually be fundamentally unstable and any small disturbance would cause it to crash into the Earth's surface. The conversation also touches on the idea of tethering the ring to the Earth's surface but acknowledges the immense strength needed to support its weight. Ultimately, the concept is deemed as not feasible due to the instability of the structure.
  • #36
Dadface said:
I can take this point but only if the surroundings are ignored.In his original post Agadotti did not ignore the surroundings in that he described applying a force on one side of the ring.To apply a force, by whatever mechanism, must bring the surroundings into the analysis and the force needed to change the location of the Earth within the ring/shell depends,amongst other things, upon the geometry and structure of the whole system including that of the surroundings.If the force is applied to the Earth the same reasoning applies.If the surroundings are ignored no force can be applied and the system remains in stable equilibrium.Muddled thinking?Possibly, I am still thinking this through.
Sorry to say this, but this is indeed muddled thinking.

In the original post Agadotti misapplied Newton's theory of gravitation. He made the same mistake in post #15 with regard to a sphere as opposed to a ring. The ring is metastable if the Earth's center of mass lies along the ring axis, flat if it lines in the ring plane, plain unstable if it lies elsewhere. The sphere configuration is flat. A pencil standing upright balanced on its tip is a metastable configuration. The pencil can stay upright if it is perfectly upright. The slightest of nudges and it will fall over. A perfectly smooth air hockey table with a perfectly smooth puck is a flat configuration. The puck will not move if it is stationary, but the slightest touch sends it toward an edge of the table. Metastable and flat configurations are unstable. There is no restoring force.
 
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  • #37
D H said:
Sorry to say this, but this is indeed muddled thinking.

In the original post Agadotti misapplied Newton's theory of gravitation. He made the same mistake in post #15 with regard to a sphere as opposed to a ring. The ring is metastable if the Earth's center of mass lies along the ring axis, flat if it lines in the ring plane, plain unstable if it lies elsewhere. The sphere configuration is flat. A pencil standing upright balanced on its tip is a metastable configuration. The pencil can stay upright if it is perfectly upright. The slightest of nudges and it will fall over. A perfectly smooth air hockey table with a perfectly smooth puck is a flat configuration. The puck will not move if it is stationary, but the slightest touch sends it toward an edge of the table. Metastable and flat configurations are unstable. There is no restoring force.

Rather than describe this as "muddled thinking" I prefer now to describe it as "thinking in progress aided by the thoughts of others".:biggrin:
My post number 28 was an example of "forgetful thinking".I am familiar with the shell theorem and have indirectly referred to it on a few occasions in this forum.I just forgot it. I agree with the metastability equilibria you have described but all are sort of thought experiments and I am just taking them a bit further.
The pencil example is a good one but can apply to the real world because,in principle, a pencil can be balanced on its tip.The ice hockey puck is a real thought experiment but can easily be extended to the real world by taking resistive forces into account.All good but consider again the thought experiment as originally described by agadotti and later extended to take into account shells as well as rings.Everything has been idealised ,a perfectly spherical earth,uniform density and the rest of it and I can accept these things,its a nice thought experiment and I agree with the analysis you have presented and your conclusion about metastability.What I am thinking about now is the situation as originally described namely the system being subjected to a force.From where comes this force? I can think of many examples but all of them disrupt the idealised structure that has been analysed.Is the force applied externally?if so then other gravitational fields need to be brought into the analysis.Is it internal for example could a super fruit fly take off from the ring imparting an impulse to the system and disrupting its equilibrium as well as its idealised structure.In short I think it's a very nice thought experiment,I think you have come to a good conclusion but to do the original question full justice we have to take into account the whole thing as described and not ignore the force and its effects.Roughly speaking Agadotti described three main things,earth,ring and force.Are we allowed to ignore the earth?No there is nothing left to analyse.Are we allowed to ignore the ring? Ditto.Are we allowed to ignore the force? If we do we are not answering the question.
 
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  • #38
OK, one last thing to analyze.

When we talk about this "restoring force" (or lack of), we are talking about a negative feedback loop. In a negative feedback loop, evolution of the system in some direction results in a force opposing the evolution (speed governors on steam engines do this) Dadface called this arrangement metastable.

A positive feedback loop is when the evolution of the system results in a force encouraging that evolution (the balanced pencil does this: the farther off-centre it is, the more it is encouraged to continue falling off-centre.)

Now, the object-in-the-sphere must be neither, to satisfy Newton's Shell Theorem. ("Flat", as per Dadface's terminology.)

But what about the ring? We've established that there is no negative feedback loop. Is there a positive feedback loop? If nudged off-centre, does the ring go farther off-centre?
 
  • #39
D H said:
A pencil standing upright balanced on its tip is a metastable configuration. The pencil can stay upright if it is perfectly upright. The slightest of nudges and it will fall over.
I thought that was the definition of "unstable". I am not familiar with the term "metastable", what makes the upright pencil metastable rather than unstable.
 
  • #40
DaveC426913 said:
But what about the ring? We've established that there is no negative feedback loop. Is there a positive feedback loop? If nudged off-centre, does the ring go farther off-centre?

It continues to move farther off center, but it does not accelerate. An object in motion remains in motion and all that.
 
  • #41
Vanadium 50 said:
It continues to move farther off center, but it does not accelerate. An object in motion remains in motion and all that.

OK. So no positive feedback loop then.

Hm. This means that the ring and the sphere are identical. In both cases, there is no net force under any circumstances.

The implication is that Newton's Shell Theorem is actually Newton's Ring Theorem (which includes the Shell Theorem as the special case where ring's breadth = 100% ).

Is this what we're all agreeing on?
 
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  • #42
D H said:
Sorry to say this, but this is indeed muddled thinking.

In the original post Agadotti misapplied Newton's theory of gravitation. He made the same mistake in post #15 with regard to a sphere as opposed to a ring. The ring is metastable if the Earth's center of mass lies along the ring axis, flat if it lines in the ring plane, plain unstable if it lies elsewhere.

This isn't true. If the center of the Earth is in the plane of the ring, but not in the center, there
is a disturbing force. See here:

http://testservice-eprints.gla.ac.uk/38/1/JIBS_C_McInnes_56_308.pdf

If the disturbance is along the axis, it's easy to see that there will be a restoring force.
 
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  • #43
willem2 said:
If the disturbance is along the axis, it's easy to see that there will be a restoring force.

Y'all need to read Larry Niven's essay "Bigger Than Worlds". He goes into all sorts of ramifications. He puts the ring around the star rather than around the planet, then sets the star bobbing. Now he's got seasons.

He uses the ring to make a giant magnetic pinch, which causes the sun to lase, throwing out gouts of plasma along its axis. Now he's got a solar system-sized world with an engine. He can cruise the galaxy from the comfort of his planet-ring. (Just don't accelerate too fast or the engine will go exploring - sans planet-ring!)

And he's just getting warmed up!

Wait'll he starts talking about living on the surface of a galaxy-sized Dyson-sphere. The gravitational gradient and atmosphere is measured in light-years. You really can fly to the Moon on gossamer wings!
 
  • #44
I don't think the term 'feedback' is appropriate here. That term refers to control loops involving amplification. All that is at work here is the gradient of the potential with displacement and the shape of the potential surface. If the surface is a dip, then the force is restorative after displacement, if it is a peak, the force will be away from the initial position. There can be other shapes, such as saddles and plateaux but the same principle applies; it's nothing to do with feedback because there is not control loop..
 
  • #45
DaleSpam said:
I thought that was the definition of "unstable". I am not familiar with the term "metastable", what makes the upright pencil metastable rather than unstable.

For the pencil I am more familiar with the term unstable equilibrium,it can in principle balance but the slightest wobble and over it goes. According to my dictionary the balancing pencil can be described as being in a metastable state because it is capable of changing to a more stable state but I do feel ,however,that "metastable" is not such an appropriate word to describe the other situations described here because any changes are not necessarily towards greater stability.
 
  • #46
sophiecentaur said:
I don't think the term 'feedback' is appropriate here. That term refers to control loops involving amplification. All that is at work here is the gradient of the potential with displacement and the shape of the potential surface. If the surface is a dip, then the force is restorative after displacement, if it is a peak, the force will be away from the initial position. There can be other shapes, such as saddles and plateaux but the same principle applies; it's nothing to do with feedback because there is not control loop..

there is certainly a control loop involving amplification there

distance from the center - > acceleration away from the center -> speed away from the center ->
even larger distance from the center.

To counteract this you need to provide an artificial negative feedback mechanism.
 
  • #47
sophiecentaur said:
I don't think the term 'feedback' is appropriate here. That term refers to control loops involving amplification.

It refers to many other situations as well. Your choice is a little too specious.

Feedback occurs where the output of one state is used as part of the input for the next state.

In this case, the evolving position (of pencil or ring) is a factor in the forces applied.
 
  • #48
I agree that a control loop is needed to keep the ring in position, artificially (if you don't want to support it on pillars. But the natural forces do not constitute a feedback loop and I thought that was what the OP was discussing.
 
  • #49
Re feedback.
Would you also say that a mass hanging on a spring constitutes a negative feedback loop?
That would mean that any simple mechanical system involves 'feedback'. Is that really what feedback is?
 
  • #50
willem2 said:
This isn't true. If the center of the Earth is in the plane of the ring, but not in the center, there
is a disturbing force. See here:

http://testservice-eprints.gla.ac.uk/38/1/JIBS_C_McInnes_56_308.pdf
You're right. I did the integration wrong. The in-plane configuration is metastable.

If the disturbance is along the axis, it's easy to see that there will be a restoring force.[/QUOTE]
Thanks; I stand corrected.

Dadface said:
For the pencil I am more familiar with the term unstable equilibrium,it can in principle balance but the slightest wobble and over it goes.
That is precisely what I was taught that metastable means: An equilibrium point that is unstable.
 
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  • #51
sophiecentaur said:
Re feedback.
Would you also say that a mass hanging on a spring constitutes a negative feedback loop?
That would mean that any simple mechanical system involves 'feedback'. Is that really what feedback is?

Yes. It has a feedforward part (wich integrates i(t) to get v(t) and v(t) to get x(t), which is the output, and it has a feedback part that feeds - k ( x) back into the input.
 
  • #52
So is there any bit of kinetics that doesn't constitute feedback? What is feedback and what is an equation of motion? It's just that I have done loads of dynamics problems without needing to talk in terms of feedback and have always used the feedback concept for powered oscillators and phase locked loops. Isn't there a fundamental distinction?
 

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