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Consider a large flat disc, at least two thousand miles in diameter. It is thin and has comparatively little weight or mass, but it is also opaque and does not let any light through.

This flat disc is positioned roughly 40-50 miles away from the surface of our planet, and the disc is consistently hovering above the same place on Earth (a landmass, like America or Asia). This disc blocks any sunlight hitting it from reaching the surface. This disc remains here for several weeks.

What effects would this have to our atmosphere and weather?


EDIT:

Various persons have suggested that the current question is similar to A 40km diameter alien saucer is floating 2km above the ocean for a long time. What are the effects on the sea ecosytem below?

However, I am not interested in how a much smaller disc object would effect the sea, as in my setting the much larger disc I am talking about is mostly above land - plus I want to know how this would effect the weather.

The "duplicate" question doesn't mention weather at all - which doesn't this question, which is specifically about weather (not sea ecosystems).

I have a world I am building with sunlight being blocked in a similar manner to how I have described - this disc in question is maintained and built through alien technology so advanced it is indistinguishable to magic. All I need is some input on how to describe the weather for anyone stood underneath this disc - would the weather be calmer underneath this disc? Or would there be storms? Or would the weather be the same as usual?

The core of my question is "how would the weather be affected". If anyone has any suggestions on how to make this clearer that would be greatly appreciated.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – James Aug 20 at 18:32
  • $\begingroup$ I have cleaned up the comments on this question and moved them to chat. I would suggest that you roll back your edits as they do nothing to improve the question itself. Consider that you have a greater understanding of the scenario than others users of the site. If enough users think your question is unclear or similar enough to another that it leads to closing its likely you could have written the question better. I have found that when I think I am the only one that understands what is going on and everyone else is crazy/stupid, its usually me... $\endgroup$ – James Aug 20 at 18:36
  • $\begingroup$ ...None of the users on the site are here to make people's lives difficult, everyone here, including the elected moderators are volunteers that visit to help people out. I know it can feel personal when people question your efforts, but that is kind of the point of the site...we're all here to help make people's world better and more coherent. $\endgroup$ – James Aug 20 at 18:37
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I think unless the disk isn’t synchronized with the planet and the planet with the sun it will be similar the the moon. But

1) You disk doesn’t let any light through. This means you have 2+1 option to how this thing will behave a) the disk is a black body and while not letting any light through it absorbs everything and emits it as heat radiation (look up black body and Stefan boltzman law). So in that case, while not letting light through it will act as secondary energy source in orbit and heat stuff up even if it blocks the incoming light out. b) The disk reflects all light making it into basically a mirror what will heat up your planets surface further. But if it blocks the sun it will not put any additional energy through. However due to the close proximity an increased heating of the planets surface will be still measurable. This time the reason is the infrared emitted from the planet surface and reflected back from the mirror.

c) A mix of a+b ..Moon scenario

2) the disk is quite light so it likely will not effect tides and such

3) A low orbit such as you suggested is unlikely to be stable by itself. If however it would be an orbit sustained just by speed it would fly at comparable high speed around your globe v=GM/R^2. Depending on the height and density if the atmosphere this might result in very not nice result due to friction. What in the end might bring down that disk anyway.

4) if the disc stays in one place over the surface 3) doesn’t matter. In that cases i suspect heating in the morning when the sun is low and isn’t blocked. In case 2a) Heat prior absorbed will be emitted plus what shines on the backside and gets exited as radiation. However less energy will be put through when the sun is blocked overall. In case 2b) I’d expect a steeper increase in temperature in the morning due to reflection but a steeper drop when the sun is covered than with 2a)

5) Weather will depend on the cycle and whether or not the temperature underneath is warmer or colder. See comments.

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  • $\begingroup$ In case the object stays in orbit over one place I’d expect that the area will get warmer than the surrounding due to the proximity. Hence you would get something like a low pressure region in the center making it rain. While the border areas are clear. $\endgroup$ – World Peace Aug 20 at 14:43
  • $\begingroup$ On the other hand due to stuck in one place over a longer time it might also cool down the area. In that case the weather situation should reverse. Rain at the outside where the sun still hits and clear in the center beneath the disk. $\endgroup$ – World Peace Aug 20 at 14:48
  • $\begingroup$ Thank you for this interesting answer! I will look into it as soon as I am able, but thanks for all the options and consideration you have put into this. $\endgroup$ – Jimmery Aug 20 at 15:18
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I am no geologist or weather expert but I think that if it's over an area that has water, then there might be less evaporation and thus less rain clouds at some times? Obviously not if it's over the ocean but only in that specific area maybe. Otherwise I don't think it would hav etoo much of an impact on the weather? Proabably more on the plants since they need enough sunlight... but as I said, I am no expert this is just the first thing that came to mind.

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  • $\begingroup$ Thank you for your answer! I had not considered evaporation $\endgroup$ – Jimmery Aug 20 at 13:58
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It wouldn't affect the weather, it would affect nothing

The sun still hits the spot anyway. If the disk is locked with the rotation of the Earth, it only blocks the Sun when it's between the Sun and the point on Earth it's over. For the entirety of the rest of the day, that area will still get sunlight. Spot will still get sunrise, and sunset, it'll just have a solar eclipse every day at noon. Temperature might drop a little, considering the sun is hottest as noon, but we're talking 5-10 degrees here, not 30-40.

Fermi Estimate for length of the Eclipse: The Sun has a size to distance ratio of .009, the Moon has a distance ratio of .008, your disk has a ratio of 40, meaning it cover 4444x the amount of sky the sun does. Let's say the sun covers ~ .07 square inches, giving your object the area of 311 cubic inches, with a diameter of about 20 inches. That means it blocks twenty inches worth of Sun travel - which translates horribly when I try to force it to 'length of day', but since the midpoint is when the Sun is at it's zenith (and thus, moving fastest from my perspective on Earth), I'd say no more than an hour and half on each side (3 hours total) and a minimum of 45 minutes (1 1/2 hours total.)

Second note: This would actually be effective at blocking out the sun above the Artic Circle or below the Anartic circle. But it wouldn't really affect the weather there anyway, given that it's already mostly subzero with terrifying blizzards.

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  • $\begingroup$ the disc is at least 2,000 miles in circumference - I haven't done the math, but I am assuming that a disc of that size only 50 miles away would cause a solar eclipse for longer than just noon $\endgroup$ – Jimmery Aug 20 at 14:38
  • $\begingroup$ Are you saying that, for someone underneath it, a 2,000 mile wide object (floating just 0.00005379% of the distance to the sun) would only blot out the sun for 3 hours max? $\endgroup$ – Jimmery Aug 20 at 15:12
  • $\begingroup$ Is that as in, 3 hours of zero sunlight, or are you saying 3 hours of non-direct sunlight? $\endgroup$ – Jimmery Aug 20 at 15:13
  • $\begingroup$ What I'm saying is that a five minutes Fermi estimation, part of which involved me squinting at the sun and holding coins in comparison, led me to believe that there will be 3 hours of the sun being blocked by your disk at the very center of it. The rest of the day will have normal sunlight. If you wait around, let's say 12 hours?, I can do a better size check against the moon. $\endgroup$ – Halfthawed Aug 20 at 15:15
  • $\begingroup$ my apologies, I have been saying circumference all this time when I meant diameter $\endgroup$ – Jimmery Aug 20 at 16:04

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