Gravity effects on a solid rigid ring around earth

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Discussion Overview

The discussion revolves around the theoretical implications of a solid rigid ring constructed around the Earth, specifically focusing on its gravitational interactions, stability, and potential energy configurations. Participants explore various assumptions about the ring's weight, stability, and behavior under perturbations, engaging in both conceptual and technical reasoning.

Discussion Character

  • Exploratory
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that if the ring's center of gravity aligns with the Earth's center, its total weight relative to Earth could be considered zero due to opposing gravitational forces acting on it.
  • Others argue that the ring is fundamentally unstable, likening it to a pencil balancing on its tip, where any small perturbation would cause it to crash into the Earth.
  • A participant questions the concept of zero weight, suggesting that an object at the center of the Earth experiences zero weight but has a restoring force when displaced, contrasting this with the ring's behavior.
  • Some contributions highlight that the center of gravity of the ring is not well-defined in this scenario, which may contribute to its instability.
  • There are discussions about tethering the ring to the Earth with cables, which could theoretically prevent movement, but concerns about the structural integrity of such a massive construction are raised.
  • One participant expresses confusion about the implications of moving the ring and how it affects the center of gravity, noting that gravity is not uniform across the ring's circumference.
  • Another participant emphasizes that the lack of a restoring force means that if the ring is displaced, it will continue to move until it contacts the ground.

Areas of Agreement / Disagreement

Participants generally disagree on the stability of the ring and the implications of its weight. While some assert that the ring can be considered to have zero weight in equilibrium, others challenge this notion, arguing that the ring is inherently unstable and lacks a restoring force.

Contextual Notes

The discussion includes assumptions about the uniformity of gravitational forces and the definitions of weight and center of gravity, which may not be universally accepted or clearly defined in this context.

  • #31
D H said:
There is no restoring force for a sphere either. Google "Newton's shell theorem."
What I was not sure about was this:

Newton's Shell Theorem is normally applied to a massive shell of material and its effects on a body of negligible internal mass.

Does it also apply inside out? i.e. a negligible shell outside under the influence of a massive inside?
 
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  • #32
DaveC426913 said:
What I was not sure about was this:

Newton's Shell Theorem is normally applied to a massive shell of material and its effects on a body of negligible internal mass.

Does it also apply inside out? i.e. a negligible shell outside under the influence of a massive inside?

The shell theorem places absolutely no constraint on the mass of the object or of the shell. It doesn't matter if the object's mass is one one-millionth of the mass of the shell or a million times the mass of the shell. The gravitational acceleration due to the shell on any point mass located anywhere inside the shell is zero. For a non-point mass, use the superposition principle: Integrate the gravitational force over the volume of the object. Integrate a zero vector integrated over the volume of the object and you get zero.
 
  • #33
D H said:
The shell theorem places absolutely no constraint on the mass of the object or of the shell. It doesn't matter if the object's mass is one one-millionth of the mass of the shell or a million times the mass of the shell. The gravitational acceleration due to the shell on any point mass located anywhere inside the shell is zero.
Ah but again, that is not what is in question. What is in question is whether the gravitational acceleration due to the object is nonzero on the shell.

However, upon reflection, I realize they must be the same. You simply cannot have unequal gravitational forces between two bodies.

i.e. F(m1)=G*m2/r^2 == F(m2)=G*m1/r^2

If this were not true, you would suddenly have yourself a free propulsion system!

So, OK. The shell theorem must work both ways.
 
  • #34
D H said:
The shell theorem places absolutely no constraint on the mass of the object or of the shell. It doesn't matter if the object's mass is one one-millionth of the mass of the shell or a million times the mass of the shell. The gravitational acceleration due to the shell on any point mass located anywhere inside the shell is zero. For a non-point mass, use the superposition principle: Integrate the gravitational force over the volume of the object. Integrate a zero vector integrated over the volume of the object and you get zero.

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.
 
  • #35
Isn't the answer to this really "Zero, using the standard definition of weight but so what?"?
It's zero when in equilibrium but is that not only an unstable equilibrium. Why try to make anything more out of it? The definition of weight was not made to cope with such a strange circumstance any more than it was intended to cope with 'weight' inside an orbiting spacecraft or a falling lift. You can work out 'what will happen' and the forces involved in any similar situation without using the term 'weight'.

Edit, a bit later ::I don't think the shell theorem applies to a ring - but it may apply to a spherical shell around the earth,
 
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  • #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.
 
  • #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 (which 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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