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I've looked for an answer for this, but the only information I could find dealt with the ring itself. All I want to know is what would change on the Earth if all problems were solved and the ring were there to stay.

In this particular world, the geography has changed due to a total destruction of Earth's surface, with humanity evacuating to the ring. A secondary, magical humanoid race has restored the planet to nearly its original state.

What I want to know is if any biological/geographical adjustments would need to be made due to the ring's presence, and what the climate zones would be.

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  • $\begingroup$ Some Interesting Related Reading $\endgroup$ – Henry Taylor Dec 15 '18 at 18:21
  • $\begingroup$ Is the ring stationary relative to the tilt of the planet relative to its primary star? If so, any land directly under its shadow would suffer eternal night. The ring could be stationary in this way if it's axis was perpendicular to the tilt which would leave its shadow out in space, thus not affecting the planet's illumination. $\endgroup$ – Henry Taylor Dec 15 '18 at 18:24
  • $\begingroup$ I have to ask what could possibly destroy the surface of the Earth which could not be prevented by a civilization capable of building a replacement ringworld ? And what are the basic dimensions of the ringworld - radius, width, depth, mass or density ? $\endgroup$ – StephenG Dec 15 '18 at 18:32
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The effects of the ring depend on three factors:

  • The height of the ring above earth ($h_{surface}$)
  • The width of the ring ($w$)
  • The direction of the ring's rotational axis

The first two work together to give you the apparent width of the ring in the sky ($w_<$):

$$w_< = w/h_{surface}$$

They also work together with the earth's radius ($R$) to give the total surface area of the ring ($A$):

$$A = w\cdot 2\pi(R + h_{surface})$$

Thus, if we assume that the ring has the same surface area as the earth (510 million square kilometers), we get:

$$A = 510\cdot 10^6 km^2$$ $$\Rightarrow w = \frac{510\cdot 10^6 km^2}{2\pi(R + h_{surface})}$$


If we put the ring in low earth orbit, just $1000km$ above the equator, the equation above yields a width of

$$w = \frac{510\cdot10^6km^2}{2\pi(6400km + 1000km)} = 11000km$$

The "ring" would be more like a can that's almost as high as the ball that's inside. I think, we don't need to talk about the climatic effects, the ring itself won't get much sunlight on its inner surface due to its own shadow.


Ok, so we move the ring up a bit, say to $10000km$ above the equator:

$$w = \frac{510\cdot10^6km^2}{2\pi(6400km + 10000km)} = 5000km$$

Now the earth is about 2.5 times as wide as the ring, and the ring diameter is still just about 6.5 times its width. The ring basically takes all the sunlight away from the equator, and the earth takes the better part of the sunlight away from the ring. Not a workable setup.


So we move the ring even higher to a level, where it can easily be assembled using a space elevator: geostationary orbit. Now we get:

$$w = \frac{510\cdot10^6km^2}{2\pi(6400km + 36000km)} = 2000km$$ $$w_< = \frac{2000km}{36000km} = 0.0556 rad$$

For comparison, the apparent width of the sun is $0.0093 rad$, so the ring is about six times as wide in the sky as the sun.

If the ring's axis were parallel to the axis of earth's orbit, the ring would be fully in its own shadow. Thus, I will assume that the ring's axis is parallel to earth's rotational axis.

During summer and winter, the ring's shadow will entirely miss the earth by a large margin, and will wander over the globe over the course of less than two months in fall and spring.

This will severely disrupt climate, because we have a stripe of the earth that is in total darkness for about a week. Within this stripe, temperatures will drop brutally. As the air cools down, it gets heavier, so we get strong, cold winds out of the shadow region. I guess, icy storms would be a better word. This will have quite a bit of destructive effects.


You can move the ring further up, another interesting point is the hight where the apparent width of the ring matches that of the sun:

$$w_< = \frac{w}{h_{surface}} = 0.0093 rad$$ $$\Rightarrow h_{surface} = 107\cdot w$$ $$w = \frac{510\cdot 10^6 km^2}{2\pi(R + h_{surface})} = \frac{510\cdot 10^6 km^2}{2\pi(R + 107w)}$$ $$\Leftrightarrow ...$$ $$\Rightarrow w = 842km$$ $$\Rightarrow h_{surface} = 90000km$$

The radius of the ring has increase by more than a factor of two, so the speed at which the shadow will swipe over the earth in spring/fall has more than doubled. This is assuming that the ring's axis is still aligned with the earth's rotational axis. Also, we now only experience total darkness for a few hours at most, with the sun being partially occluded for a few days. The climate effects should be benign now, and they will be restricted to the time around spring/fall equinox.


If you align the axis of the ring with the axis of earth's orbit around the sun, you get a thin, unlit stripe right where the sun is in zenith.

However, the earth still rotates under this shadow: At the equinoxes, points on the equator move right across the shadow region during the day, and at the solstices the tropic circles will experience the maximum shadowing, but only around noon. I.e. shadowing follows a dayly cycle, and affects a large range of latitudes.

As such, the effects are not local enough to trigger catastrophic small scale phenomena, but rather significantly reduce the sunshine across the tropics. This will definitely reduce the strengths of a) the tropic rains, including monsoon, b) the trade winds, c) cyclones (hurricanes/taifoon/whatever they are called), and d) the aridity of the deserts.

The reduction of the strength of the tropical air cycle will have other effects on the global system of winds, but I'm not enough of a meteorologist to make a decent guess at those.

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  • $\begingroup$ You'd have a cold band. Would this in effect put a permanent cold front on either side of the shadow, creating a massive band of thunderstorms? $\endgroup$ – Sherwood Botsford Dec 16 '18 at 16:27
  • $\begingroup$ @SherwoodBotsford In the last variation of the parameters? I'm not really sure what will happen directly at the edges of this band: $842km$ width is quite small compared to the typical weather phenomens. Yes, a band of thunderstorms is likely, but we already have a band of thunderstorms in the tropics. It would basically be split into two bands of thunderstorms. The overall effect of tropical band of thunderstorms on the rest of the climate is, what would be reduced. Air would cycle quite efficiently between the shadow region and the thunderstorms, cooling the tropics in the process. $\endgroup$ – cmaster Dec 16 '18 at 16:57
  • $\begingroup$ There is a substantial chill during a total solar eclipse. Totality is short, but the partial time is long. This shades a bigger area, and does so longer, getting cycles going. Also it's not going to be stationary but will move north and south with the seasons. Except near the equinox I don't think the shadow zone is stationary with respect to the surface. $\endgroup$ – Sherwood Botsford Dec 17 '18 at 19:57
  • $\begingroup$ @SherwoodBotsford Thank you for that comment. It triggered me to reconsider the effect of the orbit-aligned ring, which is actually quite different from what I had assumed previously. The point is, that the orbit-aligned ring will never shadow the same region throughout an entire day, its effects will be smeared across the tropics in a daily cycle. So, no shadow-induced band of thunderstorms... I updated my answer to reflect this new insight. $\endgroup$ – cmaster Dec 17 '18 at 23:39
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Thanks to the shell theorem, and assuming your engineers did their jobs, the only effect is on how light gets to the surface

The Shell Theorem states that the gravity caused by a uniform mass above you (e.g., a ring or a Dyson sphere) has no impact on you, the person inside the uniform mass. Only mass "below your feet" has an impact. Think of it this way. If a straight tunnel magically existed from Earth's surface to its center, then as you descended the tunnel, only the mass still between you and the center affects you. Thus, the gravity affecting you gets less and less until it's finally zero at the center. (Which, BTW, argues quite well for something other than a liquid or solid core at the very center. Probably a really hot plasma having fun in zero-G.)

Now, this assumes your engineers did their job designing the ring. If you read Larry Niven's Ringworld Engineers you'll discover (because he was famously informed about it by a bunch of MIT students in the 70s) that rings are inherently unstable. You need engines to keep the ring in place. One assumes the engineers were bright enough to place the engines and design the size of the ring such that the engines don't fry the Earth.

That leaves sunlight, direct and reflected

Therefore, the only issue is the shadow cast by the ring during the day and the reflection of light off the ring during the night. However, for this to have a significant effect (other than on romance... I can see entire jewelry commercials being inspired by this ring) the ring would be required to have an enormous width. It would need to be, at a guess, thousands of miles wide just to create a discernible line of shadow on the Earth. This all depends on its radius compared to Earth's (for a fun exercise, see this question that deals with this from another point of view).

Conclusion

In reality, the total effect of the ring on the Earth is negligible if not outright ignorable. It might create a cool sky effect (depending on its size), but that's it. This isn't surprising as natural planetary rings (e.g., Saturn's rings) don't have an measurable impact on the planet (that I'm aware of... Fair point, I'm not perfect).

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    $\begingroup$ The shell theorem does not apply to a ring. It is only valid for spherically symmetric mass distributions. Any minor deflection from the exact center will result in the Earth being attracted to the nearest component of the ring. $\endgroup$ – Logan R. Kearsley Dec 16 '18 at 1:23
  • $\begingroup$ @LoganR.Kearsley, you're correct that the deviation can occur, which is why I mentioned the instability in my answer and the need for corrective measures. The shell theorem still applies, simply with modifications. So long as the ring is kept on axis, the Earth will feel no gravitic effects. $\endgroup$ – JBH Dec 16 '18 at 1:30
  • $\begingroup$ does the shell theorem work with rotating masses in general relativity? $\endgroup$ – Henning M. Dec 16 '18 at 3:14
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    $\begingroup$ Just because there's zero gravity at the centre doesn't mean there isn't tremendous pressure from above. $\endgroup$ – chasly from UK Dec 16 '18 at 18:15
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    $\begingroup$ @JBH - That makes no sense to me. If I descend into the ocean then the pressure increases with depth even if the gravitational pull on the water decreases with depth. It is the cumulative pressure of the column of water above you that matters. The same must be true with e.g. a liquid sphere under its own gravity. The pressure is cumulative. Every particle is attracted towards the centre - just by differing amounts. $\endgroup$ – chasly from UK Dec 17 '18 at 0:09
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It would slow down the rotation of the Earth. I'm assuming the ring is around a line connecting the sun and the Earth to prevent the ring from blocking light from getting to the Earths surface. The consequence of this is that the Earth would slowly start to drag the ring to rotate along with itself. The reason for this is complicated but we see it happening with our Moon. The Moon creates a swelling of water on the Earths surface directly beneath it (the tide) the spinning of the Earth, which is faster than the orbit of the moon, pulls that tide in front of the moon. The mass of water then imparts a small force causing the moon to accelerate and the Earth's spin to decelerate. (a quick internet search pulls up this if you want a reference https://www.bbc.com/news/science-environment-12311119). Your ring would create the same effect (this also means it would require rockets on it to prevent its spinning).

This could potentially be fixed by having the ring rotate with the Earth and eat the cost of having a solar eclipse every day.

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If the ring lies perpendicular the the planet's axis and of a certain width, it could cast a permanent shadow onto the earth, making photosynthesis and therefore plant life impossible under this shadow. The shadow could also cause other effects in non-permanent regions in the form of a short "artificial night", which may confuse fauna and flora, cause the temperature of the region to drop and mess with the weather (air temperature plays a role in the climate's behavior). Should the ring have enough mass, it may even affect the tides, further causing interesting effects together with the gravity of the moon. Again, if the ring is perpendicular to the planet's axis and has enough mass it may encourage higher trees or mountains due to the effects on gravity. Should the ring contain large, lens-like segments (I dunno, maybe for observation or heat harvesting or technobabble), it may also have an effect on the sunlight reaching earth (think giant lasers when it lines up with the sun).

In the end, the bigger the ring, the stronger the effects. Look to the moon for inspiration. Hope this helps.

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    $\begingroup$ Adding some numbers for the shadow: For full shadow, the ring would need to be as wide as the sun itself from the perspective of the earth. That's about 32 arc minutes, or about 0.009 radians. Geostationary orbit is at a height of 36000 km above the equator. As such, a ring at the height of geostationary orbit would need to about 324 km wide to be able to cast a full shadow. That would be $2\pi\cdot 42 Mm \cdot 324 km = 86\cdot 10^6 km^2$, which is about 1/6 of the earth's surface. So, full shadow would be plausible. $\endgroup$ – cmaster Dec 16 '18 at 9:18
  • $\begingroup$ Thank you, some excellent numbers there. I'm afraid I don't have the expertise to do some myself. $\endgroup$ – A Lambent Eye Dec 16 '18 at 9:19
  • $\begingroup$ However, the shadow would never affect a certain region permanently: If the ring is inclined with the earth's axis (= stationary in the sky as seen from earth), the shadow will wander with the seasons, and miss the earth entirely in summer and winter. If the ring's axis is parallel to the axis of earth's orbit, the shadow will remain on the earth at all times, but again, different regions will be affected as the seasons go by. $\endgroup$ – cmaster Dec 16 '18 at 9:22
  • $\begingroup$ Oh, the numbers are just basic geometry + looking some stuff up on wikipedia. No black magic involved ;-) $\endgroup$ – cmaster Dec 16 '18 at 9:23
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Good answers so far:

Ring vertical (polar orbit) you have a short eclipse daily.

Unlike Ringworld, where the ring is very far from the star, and having no other mass in the system, Moonworld is subject to tugs from the asymmetry of the Earth, solar tides on the ring, and nudges from Jupiter. So station keeping systems would need to be relatively more robust than the ones used by Ringworld. The same mechanism might be used, however: Collect the solar wind and use it for hydrogen fusion engines.

You have serious transport issues between the ring and the earth. The ring has to be spun really fast to give in enough centripetal force to provide gravity. This is heavy duty delta-V. If it's at syncronous orbit distance, it needs to spin at 20 km/sec to have 1 g on the inner surface.

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