Orbital Ring - Version 2

Factual Questions
1

So my 1st attempt at putting a stone ring around the earth crumbled at my choice of material. The Hamster King also linked to a discussion about why an orbital ring is inherently unstable (vs a Dyson sphere), and it’s this point that I’d like to focus on.

This time, let’s make my ring out of a continuous rod of priapusmite, an incredibly stiff material that won’t bend or sag for our purposes. Let’s say this material has the same density as aluminum. Using a 1 inch diameter rod, if we have a ring floating 3 feet above the ground, what would be its total mass?

When it’s perfectly balanced, the ring should be weightless, right? Once, this balance starts to decay, what is the rate at which one point goes from weighing nothing to a shit-ton? Like, if one side dropped by an inch, and I tried to correct it, what am I looking at lifting? And conversely, the opposite side would exhibit a lift capacity equal to that weight, right?

Now, once it does touch ground, does the point of contact weigh the entire mass of the ring, or does the weight of the opposite side provide some sort of counterbalance?

Just so we don’t get mired by little details, for this thought experiment we are assuming the earth is perfectly smooth and spherical with an equal distribution of mass.

2

Shall we also assume zero atmospheric drag?

3

Yeah, for the sake of this discussion, we’re pretty much working in the realm of spherical cows.

I plan on factoring in other variables like the moon eventually, but I don’t want to amp up the complexity yet until I can wrap my head around some of my initial questions.

4

OK, if I did the math correctly, you are talking about 121 million pounds (give or take a million pounds or so) of ring. But you are incorrect in saying that it is weightless. It weighs 121 million pounds (at least on earth. But, if it perfectly centered on a perfect sphere, it would be supporting its own weight. Being 1 inch diameter (round, I assume), we are talking about 0.785 square inches, or in other words, the ring would be under 154 million PSI of compression. Chances are, even if your priapusmite could support such a compressive load, being only 1" in diameter, it would likely buckle and come crashing to earth.

5

Wait, isn’t it rotating? I thought the idea was to counter the gravitational force by centrifugal!

In which case I still see a problem: even assuming a perfectly stiff, symmetrical ring orbiting a perfectly spherical planet in a perfectly Newtonian universe, the ring will still, by the smallest perturbation, be prone to gradually change its rotational axis to the one of maximal rotational inertia, which means it will eventually circle the Earth like a hula hoop!

6

It was circling the Earth like a hula hoop even without rotating. And it supports itself in equilibrium via its fantastic material strength. But it’s an unstable equilibrium, which means that it’d try to fall if even only an inch out of place. Which is what the OP is asking about: How much force would it take to maintain it in that just-off-equilibrium position?

7

Chronos ? I am suprised at you. Due to symmetry, the actual rule is about the centre of gravity of the mass and it matters not what the shape is.

As the centre of gravity is moved up, the net gravity must be LESSER… by that slight amount given its a meter extra distance ontop of the million meters from the surface down to the earth’s CoG.
Now if you want to justify to yourself that the CoG is all you needed to consider,
you can consider the geometry … Its not so much that the low side is heavier, as in more gravity, it is that the high side is having less gravity AND there is more of that high side… so there’s the reason for the imbalance.

What you actually have trouble with is that earth’s gravity is far from consistent, 0.7% variation, and if we guess that due to symmetry the variation is only 0.1% net, well that means that one side of this ring is pushing on the other side with a force 0.1% larger…

Now whats the mass of half of this ring ?

But the OP asked how long… well 0.1% of gravity is quite a fast acceleration… not long when its only 5,10 metres to travel…

8

No, the whole would not exhibit a force, until you tried to resist it with some external force. It doesn’t matter if you apply that force to the low side or high side … Its the same force … when you assume the ring is strong enough to remain a rigid body (thats another property of a spherical cow of course…)

The net external forces on the ring are a poor indicator of the required strength of the ring.

The internal pressure is huge… basically the eastern part of the ring is going to be pushing on the western half… so the internal pressure is if you had half a ring and you had a rocket sitting at the cut end trying to thrust the ring and keep it up… The entire weight of a quarter of the ring down onto that cross sectional area… (cause the other side is holding the other quarter… see ?) Still huge pressure…

9

For a ring, that’s not actually true. From the outside of a uniform spherical shell, you can treat it as a point mass at the center of the sphere, as far as net gravitational forces, but that’s not true of other shapes.

The whole point about the instability is that a slightly off-center ring experiences a force pulling it more off-center (even though the center of mass of the ring is being pulled away from the center of the planet).

The math for the ring is a little hairy, and I can’t be bothered, but we can get close by pretending the ring itself is massless, and it has two equal weights attached at opposite points, with the off-center movement on the line between the two points (essentially a dumbbell with a massless center rod, running through the Earth, moving only in the direction the center rod is pointing). The ‘real’ ring will have slightly less of a net force, since more of its mass is on either side of the Earth, counterbalancing, but only by a factor of something like 2 or so.

The dumbbell calc is easy, just gravity = mass /r^2 (and assume the displacement is small compared to the radius of the Earth). If I can do algebra correctly, with the dumbbell a distance D off-center, the force pulling it farther off center is the weight of the dumbbell times D divided by the radius of the earth.

Using round numbers like 6,000 km for the Earth radius, 3 cm for a displacement, and excavating for a mind’s calculation of weight, I get 120 million lb x 3 cm /6e8cm = 120 e6 lb /2e8 = 0.6 pounds of force pushing the dumbbell towards the center of the earth. That’s an upper bound for the ring, but the real force would be somewhere in that ballpark. Every three cm it moves off-center the force goes up by that amount more.

So, for the OP, if the ring drifts off-center, even if it falls the whole 3 feet to the surface of PerfectlySphericalEarth, according to my undoubtedly mistake-ridden math you’ll be able to just use your hands to push it back into place (assuming you can lift a 20 pound weight).

But, I should warn you, it will take a while (there’s only a small opposing force, but you’re still trying to accelerate all 60,000 tons of the ring). Quite a while, in fact. You might want to pack a lunch…

10

Don’t forget that once you get it moving, you have got to stop it from moving too far. 60k tons, even moving at a few centimetres an hour, would cut a pretty deep groove before it stopped.

11

I did similar calculations for the Ringworld a while back, and learned that for a thin narrow ring, a tiny amount of offcenterness will double in about 20% of the orbital period of the ring (at least at first - when the amount of offcenterness gets large, the doubling rate increases). For the kind of ring you’re talking about, the orbit period is about 90 minutes, so the doubling rate would be on the order of 20 minutes or so - so this ring isn’t going to stay in position very long.

12

Apparently doing elliptic integrals is how I like to spend my Sunday mornings. Having gone through the more detailed calculations, I can confirm that to a very good approximation, your answer via the dumbbell technique is double the “correct” answer using calculus, elliptical integrals, and Taylor series.