The Ring Of Steel

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The Other Other Rope Around the Earth

So we return to The Other Other Rope Around The Earth.

See the Rope Around The Earth Refined
If you recall, our friend has encircled the Earth with Rope, discovered that it's 6 metres longer than needed, so expanded the circle by 0.95 metres to take up the slack. Thus we have a ring of rope around the Earth at height just about 1 metre.

Don't forget, next time we'll be moving away from a rope around the Earth to a sheet around the Earth. See the previous post - Rounding Up The Ropes - for a description.
This is pretty clearly preposterous, so I guess the next step, suggested by Bill Mullins, isn't really too much of a stretch. Our friend, having done all that work, goes to bed and sleeps the sound sleep of the satisfied. But overnight, somehow, magically, the rope gets converted to steel! And then, over breakfast, a slight Earth tremor gently rattles the props, and they all fall down, leaving the ring of steel completely unsupported.

OK, so at this point it's worth saying that we're in "Puzzle World". In truth, the "steel rope" would just collapse ignominiously, losing its shape in a random and unpredictable manner. But for interest's sake, let's assume that it remains a perfect circle.

The magic word in physics is "rigid" - that's a code word to say that "it doesn't change shape". Our personal experience is that, on our scale, steel is "rigid", although as mentioned above, at large scales that's not true. So for the purpose of this exercise, to use the "magic code phrase," we're assuming the ring is rigid.

If you want to live in "Crypto World" and look at all the real life details, I'd be interested to hear of any conclusions you find. Let me know.

What happens next?

Picture the situation - we now have a rigid ring of steel that encircles the Earth at a height of about 1 metre, completely unsupported, just floating in space. If you're there standing next to it, you might expect it to fall.

But wait! If it falls here, it would have to rise on the other side of the Earth, and it seems unreasonable for someone there to see this bar of steel rise into the air.

So really, what happens?

It's in equilibrium

We're in puzzle world, so we're assuming the Earth is a perfect, homogeneous sphere of constant density, and we'll ignore the rotation, etc. Obviously you can come back to these things if you like, but for now, let's just go with this very simple situation.

Clearly it's currently in equilibrium. But is it stable? A pencil, if perfectly balanced on its point, is in equilibrium, but just sneeze and it will move from that position, never to return. A pendulum, on the other hand, if disturbed from hanging straight down, will eventually return to that position, because that position is stable.

Does the ring, if disturbed, return to being the same height all round? Or once it has moved "off centre", does it then get further off centre?

A brief diversion - The Shell Theorem.

There's a wonderful theorem in mathematical physics that says that the interior of homogeneous shell has no gravitational gradient.

Let me put that into other words.

Imagine a shell, like the Earth's crust, and imagine that, unlike the Earth, it's hollow. Imagine further that, also unlike the Earth, every part of it has the same thickness and density. No matter where you look, the shell is the same.

Because of symmetry, a small particle placed at the very centre of the interior will have no inclination to move in one direction compared with another. The slightly amazing thing is that this is true for every location in the interior. A particle placed anywhere in the interior will have no inclination to move.

There is no gravitational gradient.

To some that seems a little unexpected. The centre of mass of the shell is at the ... well ... centre, and it might be reasonable to assume that a particle would be attracted to the centre of mass, and hence want to migrate to the centre. But no, that turns out not to be the case.

The shell theorem can be proved fairly easily with a little bit of calculus and a little bit of geometry. You can probably find a proof online with a few minutes searching, but what I want to do is to give a plausibility argument. Then we'll see how it applies to the Other Other Rope Around The Earth problem.

How the Shell Theorem works.

Now imagine that you're at the centre of the shell. Hold a picture frame out at arm's length and look at the piece of shell framed by it. You are experiencing some gravitational attraction from that small section of shell. Now turn to look in the exact opposite direction and again look through the picture frame. Again, you are experiencing gravitational attraction from that small piece of shell. Those two forces are equal and opposite, and so balance each other, leaving you with no net force (from those bits of shell).

Well, symmetry would tell you that anyway, so there's no surprise there. But now imagine moving closer to one part of the shell. As you look through the picture frame you can see less shell in that direction, so there is less mass there exerting its influence. But it's closer, and gravity increases according to the inverse square law.

And as luck would have it, these two effects exactly cancel each other out. As you get closer the area falls, the influence rises, each according to a square law.

It's worth mentioning as an aside that a solid body is the sum of an infinite stack of nested shells, like a Russian doll. In doing so we can start to break down a complex system into a number of separate pieces that can be analysed separately.
So as you move off centre, the balance of forces remains, and there is no gravitational gradient.

Obviously it's more complex than that, but in essence, that's how it works, and that's how the calculus ends up working.

Very satisfying.

Back to the ring of steel

Given the shell theorem, can we solve the $(Other)^2$ Rope Around the Earth problem?

Yes. (Obviously, otherwise I wouldn't be asking)

Imagine now instead of being in the interior of a shell, you are at the centre of a ring. Again, being exactly in the middle means that you are in equilibrium, but now consider what happens as you move off-centre, even by the tiniest amount.

You move further from one part of the ring, and as before, its gravitational influence decreases according to the inverse square law. And as before, you can see more of it, but now the amount of extra that you see only grows linearly, not quadratically. Similarly, the section of ring that you get closer to attracts you more strongly, again as an inverse square, but now while it's true that there is less of the ring attracting you, the decrease is only linear, but the increase in gravitational attraction of that piece is quadratic.

So as you get off centre the gravitational forces drag you further in the same direction.

And so it is with the ring of steel around the Earth. Even the smallest displacement from being exactly centered means that it will carry on in the same direction, accelerating gently, until some part of the ring impacts the Earth. The section of ring on the opposite side will rise accordingly, defying all reason for the people who might be standing nearby.

Cultural references

Some of my readers will be familiar with "Ringworld" by Larry Niven. After it was published, apparently lots of people wrote to Niven, some with indecent glee, pointing out that the Ringworld system was unstable. I don't know if this was intentional, or if Niven made a virtue of a necessity, but the stability of the system was a major plot point in the sequel, explaining how it all worked.

But at the time I never did understand why people said the system was unstable, the explanations via calculus were ... unsatisfying. I find this qualitative explanation really useful to see how and why it works, and I'm grateful to Bill for proposing this variant of the classic problem.

There's a further reference in "Dragon's Egg" by Robert L Forward. Tidal forces are stretching in the direction of the centre of mass, and compressive perpendicular to that. This is exactly opposite to the gravitational forces at the centre of a ring, so tidal forces in orbit can be (in theory!) counteracted by having a suitably massive ring of material in the right orientation at the right distance. This is important when you're in orbit around a neutron star.



My thanks to Bill Mullins (Twitter) for suggesting the problem.

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