Assuming the atmosphere on a Niven ringworld is retained by gravity, what shape would it be?

Full details of Ringworld http://larryniven.wikia.com/wiki/Ringworld

As I understand it, Niven proposed walls on either edge that would form a sort of trough for the atmosphere to sit in to avoid it leaking into space. However keeping it in a trough assumes there is sufficient gravity to retain it. He also has oceans on the outside of the ring. These also must be retained by gravity. Why then does he need a raised edge?

Question

Assuming the cross-section of the ring is a rectangle with sufficient gravity, and there are no retaining walls, what shape would the atmosphere adopt? Would 1, 2, or 3 in my diagram be possible configurations?

Enter image description here


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    "He also has oceans on the outside of the ring" I don't remember any oceans on the outside of the ring. What do you mean by this? – Azor Ahai Dec 7 at 16:53
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    @ Azor Ahai - You're right. I turns out I misread the description in the link I gave above. It says there are two oceans opposite each other. I took this to mean on opposite surfaces. In actual fact I now see it means diametrically opposite. – chasly from UK Dec 7 at 17:05
  • One that might be more useful from what you are appearing to ask about here is the Larry Niven book Protector, where a protector from Pak brings Tree of Life root to Earth, abducts and turns a human named Brennan, who then kills the protector, and creates a very small mobius strip shaped artificial world around a core of neutronium to use as a base in his defense of Earth. This world would look a lot like your example (1). en.wikipedia.org/wiki/Protector_(novel) – AndyD273 Dec 7 at 17:26
  • @chaslyfromUK Yes, that matches my understanding. – Azor Ahai Dec 7 at 17:41
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    You might ask instead about the ring-shaped asteroid where part of Niven's Protector is set, though that was built with artificial gravity. – Anton Sherwood Dec 7 at 21:36
up vote 17 down vote accepted

Without edge walls, the spin gravity would quickly result in all the atmosphere falling off the edge of the Ringworld. Don't forget that your gravity is only outward from the center.

The "oceans" visible on the outer surface of the ring were the shapes of the under sides of basins formed into the surface for the largest, deepest oceans -- several of them many times larger than Earth.

If you built a ring that didn't spin, and it had large enough diameter relative to width and thickness that the "arch" (the opposite side of the ring) has negligible gravity effects on the local surface, you'd get a shape like your first or second illustration, an elliptical atmosphere surrounding the rectangular cross section (and all the water would tend to collect at the center of each flat). Whether the edges stick out of the atmosphere would depend on the ratio of width to thickness.

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    @chaslyfromUK That’s because the gravity caused by the mass of the Earth more than counteracts the centrifugal force. The Ringworld is the exact opposite — negligible gravity and very high centrifugal force. – Mike Scott Dec 7 at 14:56
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    Earth's spin produces about a milligee acceleration at the equator, pushing outward and opposing the one gee from the Earth's mass. The Ringworld's 770 miles per second produces one gee, "pushing" outward from the center. Its mass is negligible by comparison; if not spinning, as described (scrith base a few km thick) it would have a tiny fraction of a gee toward the surface from both sides, tapering off and tilting as you approach an edge. – Zeiss Ikon Dec 7 at 14:59
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    Ringworld doesn't orbit. It's a spin-gravity ring, held in position relative to its star by Bussard ramjets at the rim. Have you read the books? – Zeiss Ikon Dec 7 at 15:01
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    @chaslyfromUK The Ringworld spins much faster than orbital speed to create an effective 1g of gravity on the inside surface by centrifugal force. Even if that wasn’t the case, a rigid ring can’t orbit something at the centre of the ring. – Mike Scott Dec 7 at 15:05
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    A ring doesn't cancel to zero inside the way a sphere does. The ring doesn't have the coverage to do so. If the ring diameter is large relative to width/thickness, the far side effects become negligible. – Zeiss Ikon Dec 7 at 16:59

You're missing the way the ringworld simulates gravity, which is entirely centrifugal force from the rotation, the atmosphere has to be contained to remain in place.

Hence your image is more like this:

enter image description here

Where the ring has walls to contain the atmosphere and the only way onto the planet surface is through the open inside face or through airlocks in the walls or floor.

You can't have anything on the outside of the ring. While as a stationary mass the ringworld would have a centre of gravity coincident with the star it was in place around, the spin negates that gravity and would cause anything on the outer surface to fly off. This does give you a cheap space launch mechanism though as you don't need to escape the gravity well.

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    It doesn't generate gravity, it simulates it and I think the difference is important here - it's a large party of the confusion. – Tim B Dec 7 at 15:08

I think you misunderstood what a Ringworld looks like. People live on the inside of the ring with "down" pointing outwards. Gravity is mimicked by the apparent centrifugal force caused by the Ringworld spinning. This also pushes the atmosphere outwards against the "ground" (= the ring") and you only need walls on the side to keep it from leaking out there, assuming they are tall enough.

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