# Gravity effects on a solid rigid ring around earth

1. Oct 7, 2009

I was trying this mental exercise and I need help and confirmations.... so please launch your comments

Imagine (as an assumption) that the surface of the earth is completely flat with no disturbances. A perfect sphere.
Imagine that is possible to build a solid ring made of a completely rigid material all around one meridian of the earth. The ring will have a perimeter a little longer of the earth perimeter so that it hovers 1 meter over the surface of the earth.

Now assuming that it is possible to build the ring to be a completely rigid structure and because we build it to have it's center of gravity corresponding with the center of gravity of the earth, the following assumptions should be right:

1) The total weight (measured relatively to earth) of the ring is ZERO. The mass will be enormous, but of course if the weight is relative to the gravity, then for each point of the ring where a gravity vector is applied we will always have the opposite point with a completely opposite and equal vector is applied. The total weight should be therefore ZERO

2) In a stable configuration (i mean with no any other force in action) the ring will hover 1 meter over the earth surface

3) If we apply a force on one side of the ring and imagine we move the ring half meter closer on that side, then ideally we are moving the center of gravity of the ring of the same half meter on the side. This will imply that after some oscillations the ring will come back to the original position hovering 1 meter over the earth

4) The configuration of hovering over the earth is the configuration with the lowest potential energy

Is this correct or I'm losing something ?

Alex

2. Oct 7, 2009

### mgb_phys

The ring is fundementaly unstable (in the same way as a pencil balancing on it's tip) any small perturbance will cause it crash into the surface of the Earth.

The proof is a little involved - it's a result famous because of a sci-fi novel called Ringworld by Larry Niven about a solar system sized ring around a sun.

3. Oct 7, 2009

Do you know where i can find the involved proof ?

I've heard about Ringworld before. The question is, the two center of gravity of the Ring and the earth in the starting configuration are exactly at the same place, aren't they forming a system with a minimum potential energy ?
if not why ?

Thanks

4. Oct 7, 2009

### DocZaius

I find no reason why the oscillation should ever stop, if all the forces involved are conservative.

edit: Looks like the ring isn't stable so nevermind!

5. Oct 7, 2009

What if you do teeter the ring with cables to the earth surface (imagining the cables are strong enough to stop the ring from moving ?

6. Oct 7, 2009

Staff Emeritus
mgb_phys is right - the situation is unstable.

There is no restoring force if you move the ring slightly - that's what "zero weight" means. Since there's no restoring force, it won't oscillate. If you push one end, it will keep going until some part of the ring hits the ground.

7. Oct 7, 2009

### DocZaius

Is that really what "zero weight" means though? Consider a tunnel bored through the earth, passing through its center. An object at the center of the earth in that tunnel could be considered to have "zero weight" and yet if pushed in either direction will have a restoring force which will make that object oscillate indefinitely, right?

8. Oct 7, 2009

### Staff: Mentor

The problem is that the center of gravity is not quite the same as the center of mass. In most cases it doesn't make a difference, but in this case it does. The ring does not actually have a single point that can be called the center of gravity in this case. That is where the instability arises.

9. Oct 7, 2009

### Staff: Mentor

No. If you and your twin are on opposite sides of the earth, it would be nonsensical to say your total weight is 0. The total weight is the total of the two weights. Similarly, the total weight of the structure is the total of the weights of individual elements.
Yes, you could tether such a structure to earth and it wouldn't take much force to keep it positioned.

However, the structure would need to be enormously strong to support its enormous weight. Essentially, you're talking about a bridge that spans the circumference of the earth.

10. Oct 8, 2009

I do not agree on your example of the two identical twin. We are not connected together.
While the mass of the structure is certainly huge, the weight on earth is defined by the combined effect of the gravity on the total body, or in other words the force you need to apply to the body in order to hold it at rest in a gravitational field.
Assuming that the ring is in equilibrium, then the force that you need to apply to hold it at rest in the gravitational field of earth is ZERO, therefore I would conclude that the weight is ZERO.

True the reasoning on the structural integrity of the ring, that can be comparable to a bridge with 40,000 km span. But here we are not talking about the technical feasibility of it's construction.

11. Oct 8, 2009

Staff Emeritus
Yes, but once pushed in another direction, the object no longer has zero weight and now there is a restoring force. The ring as described does not have that property.

12. Oct 8, 2009

### A.T.

Is the ring unstable like a balancing pencil (local energy maximum), or indifferent like a ball on a plane (energy plateau)?

13. Oct 8, 2009

### Caesar_Rahil

I really don't get the point that if you move the ring by half meter, the center of gravity will move by half meter.
Won't it be a little complicated to calculate center of gravity as now gravity is not uniform over the circumference of the ring?

Hey, and by the way..
If i really get this thing, there is tremendous force exerted on every part of the ring (internal tension), but the ring being rigid, overcomes this as forces on opposite sides cancel each other. So now, if you move it, there will not be a balance and the resulting force will pull it on the surface.
I guess this is what is called an unstable equilibrium, force zero but potential energy maximum

14. Oct 8, 2009

### DocZaius

I always took the definition of weight as meaning that the net gravitational force acting on an object is non zero. I didn't know it had anything to do with whether that force would also be restoring or not, and would lead to oscillations.

So as soon as this unstable ring is pushed, and it is falling towards the planet, I think that it does have weight, just as my example tunnel object does, when pushed either way. Of course I agree that the ring would crash into the planet and would not oscillate, while my tunnel object would oscillate indefinitely. However my point is that the system's stability (due to a restoring force) isn't merely based on the presence or absence of weight as you seem to contend, but rather on the way the system is configured.

Of course I'm probably wrong, but would love to hear why!

Last edited: Oct 8, 2009
15. Oct 8, 2009

After some investigation I agree with the fact that the system is unstable. If it was a sphere then the balance would have restored always, but being a ring it is not true.
Now the problems is still the weight in the "equilibrium" state. So imagining we tether the ring with cables to avoid it to move (and please forget the feasibility issue), what is the weight of the entire structure.
If it is in the equilibrium state I would say ZERO.

16. Oct 8, 2009

### DaveC426913

No. A sphere is in the same boat. There is no restorative force.

What is the weight of the centre span of a bridge? The fact that it is braced so it does not fall down does not mean it has no weight. It must support its own weight (probably from compression forces, akin to a giant keystone) to prevent it from collapsing.

The ring has the same problem except that it is simply a "centre span" and is 40,000km long.

17. Oct 8, 2009

If the ring was a sphere then If the sun was at the center of the sphere it would have equal amounts of mass distributed at the same distance from it in all directions, this all cancels to a net force of zero. While if the sun was off center then the side it is closest to would have a stronger field strength due to less distance, but at the same time it the opposing side now has more mass behind the sun and though it is further away these effects perfectly cancel each other out.

Considering that the gravitational field is r^2 inverse proportional, we have the same situation of a flux, therefore if the sphere is equally dense and thick, then the gravitational attraction is proportional to the surface of the sphere again r^2 proportional to the distance from the center point mass. Easy to demonstrate that 1/r^2 and r^2 will elide each other for any given small portion of the sphere, and therefore integrating it for the entire sphere. Moving the sphere will not cause it to crash on the planet.
This is of course not true for the ring, and hence the ring instability.

Also the analogy with the Span of the bridge does not work because the bridge is hold in place by pillars that provide the force to sustain it. In this case the only force is the structural integrity of the material the ring is made of (imagining possible a material enough strong to sustain it). The case is very much different. Obviously the sphere and the ring will be subject to the gravitational pull, that will strain the structure, but we need to convene of what is the definition of weight. If it is the sum of all the force vectors that are applied to the body, then in the case of the ring and of the sphere the sum of the vectors is exactly ZERO (in a equilibrium position).

18. Oct 8, 2009

### D H

Staff Emeritus
All this talk about "center of gravity" makes me think quite a few of you have a misconception of Newton's law of gravitation. Newton's law of gravitation, F=GmM/r2, pertains to point masses only. For non-point masses blinding applying Newton's law will in general lead to the wrong answer. The gravitational force between two non-point masses is the superposition of the gravitational forces between all pairs of infinitesimal point masses that comprise the two objects.

Thanks to Newton's shell theorem, the gravitational force between a point mass and some non-point mass that has a spherical mass distribution reduces to Newton's law. The ring does not have a spherical mass distribution. Newton's law does not apply per se. You have to do an integral over all the elements of the ring.

This is a bit messy in general. The gravitational force is fairly easy to calculate when the center of mass of the Earth is on the plane of the ring and on the axis of the ring. The first case is particularly easy: The gravitational force between the ring and the (spherical) Earth is zero whenever the Earth's center of mass is on the plane of the ring. There is no restoring force in this case, nor is there a disturbing force. There is no force, period.

19. Oct 8, 2009

### Malarksb

I am thinking about atmospheric pressure and its variance being a force that would be applied to said ring. What would building a chamber that could have a vacuum drawn on to it do to equalize that pressure and other forces. My thought is that it would isolate this to the one element of gravity.

Last edited: Oct 8, 2009
20. Oct 8, 2009

### DaveC426913

What atmospheric pressure? This thing's in space, right? If not, well, we've changed the parameters.

21. Oct 9, 2009

Yes this thing originally is with no atmosphere... so only gravity matters... do not complicate the problem...

22. Oct 9, 2009

### sophiecentaur

Two points:
the 'span' is in compression, not tension
and
the term 'weight' merely describes the force on an object. It says nothing about how the force varies with position. By going to an appropriate distance from any planet you can have the same weight (or gravitational field) as on the surface of the Moon. The gradient of the field will be different in each case.

23. Oct 9, 2009

Ok I agree the Span is in compression
The point here is not to have the same weight in different gravitational fields, this is obvious that any gravitational field has it's own gradient.
The key question here is :
"Has the Ring so designed a Weight equal to ZERO or not ?"
we know that :
- The ring is unstable, but who cares because we imagine to answer the question in a case where the ring is in equilibrium
- The ring is stressed by a compression force due to gravity, but also here, who cares, because we imagine that the material the ring is composed is strong enough to maintain its structural integrity and its rigidity also in this case
- Different Planets have different gravity. Ok. Who cares we are experimenting the ring around earth, even if the same question about the weight can be valid around any other planet
- There is no atmosphere and any other force applied to the ring
- We know that the mass of the ring is huge. Who cares also in this case. The ring is constructed to withstand. and it's mass is still ridiculously small compared to the earth mass.

so the question for who wants to answer is :
IS THE WEIGHT OF THE RING ZERO ?

24. Oct 9, 2009

### A.T.

If "weight" is the net gravitational force on the entire ring, then yes it's zero. But locally the ring parts are not "weightless", as accelerometer attached to them would show. The ring is not different from the Earth's crust in that respect.

25. Oct 9, 2009