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If the water on earth were displaced to surround the earth in the atmosphere to create a water canopy, how thick would that canopy be? I apologize I should have been more specific in my original question, how thick would the shell of water be if it was in the upper thermosphere around the 80 mile mark (125ish km)?

Another way to look at this without asking whether it is plausible or not, if the Earth's oceans were to evaporate into cloud and then be pushed by force into the upper atmosphere (thermosphere) and then condense back to water around the Earth how thick would that shell of water surrounding the earth be?

So I’ve been thinking about this water canopy still, lol. I’m stuck for some reason wanting to use this in my book, and I keep thinking about the improbabilities of it and I’m still struggling with the explanations I have been given. The average depth of the oceans is around 2.7km and the height of the lower thermosphere at the 100 degree temperature mark is around 110 km from sea level. So to distribute the sea water around the globe at that height (110 km or higher) I can’t seem to grasp that the water would be as thick as every one is telling me. At that height and dispersal around the globe I would think it would be much shallower than what has been explained to me. I’m sorry to pester you, when you already explained the answer, I’m hoping you will be able to explain in simpler terms for me to grasp. It just seems to me that expanding a two kilometre depth outwards to 110 km would greatly reduce the depth. Like taking a full cup of water and then spreading it over a larger surface makes the liquid appear much shallower. I’m sure I’m just missing something or just not understanding, please help me to understand better lol.

So now that I have my answer clearly defined, thank you for your patience, I have another question to put forth on the same subject. If said water barrier or shell or canopy was in place, at the thickness it was calculated to be(I understand this to be around the 2km mark) would the gravitational pull of the moon be able to affect the thickness, like the tides currently, allowing for a thin enough barrier for sunlight to penetrate giving a day/night scenario?

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    $\begingroup$ What is a "water canopy"? By "displaced" you mean "evaporated"? $\endgroup$ – AlexP Feb 9 '18 at 5:43
  • $\begingroup$ Don't forget that the atmosphere would also be displaced down into the now empty oceans. I surmise there would be places where the water canopy would touch the mountains, a form of inverted shoreline.. $\endgroup$ – Joe Bloggs Feb 9 '18 at 6:53
  • $\begingroup$ It is suddenly moved there and then, you want it to stay? $\endgroup$ – Vincent Feb 9 '18 at 17:50
  • $\begingroup$ Note, intuitivelty, you should not expect it to change by much. When visualizing how much water there is, don't forget that our oceans are basically a 2.7km thick shell, 6300km in radius, and about 71% full (the rest is full of land). You are now making it a X km thick shell, 6410km in radius, and 100% full. That increase in radius is not all that much, when you get down to it, but the fact that you're not having to account for land-masses once the ocean is up in the air will make the shell thinner $\endgroup$ – Cort Ammon Feb 21 '18 at 19:11
  • $\begingroup$ The earth is 6300 km in radius, adding another 100 km onto that is not a lot really, this is why it's not as thin as you were expecting. $\endgroup$ – Tim B Feb 21 '18 at 20:38
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The average depth of the oceans is around 2.7km

This seems to be untrue. According to the United States National Ocean Service, the average depth of the oceans is approximately 12,100 feet or 2.3 miles or 3.7 kilometers. Oceans cover about 71% of the Earth, so that's enough water (around sea level) for a depth of around 2.6 km.

Where you may be having trouble is that you think of 110 km as a great distance. In some ways it is. I wouldn't want to walk it, as it would take me days. But in terms of the size of the Earth, it's not so much. The Earth's radius is about 6371 km. That's about fifty-eight times your 110 km. So what you're basically saying is that you're moving from 6371 to 6481 km. That's less than 2% growth. Yes, that will reduce the shell's depth, but it won't do it by that much. Perhaps from 2.6 km to 2.5 km, a 4% decrease.

If we make modest changes in the estimates of the average depth of the ocean or the amount of ocean or the radius of the Earth or the height of the thermosphere, we'll get different answers. Since those estimates are necessarily a bit fuzzy, the variation from that is at least as great as the expected change when moving from sea level to the thermosphere.

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  • $\begingroup$ You are totally correct lol, I didn’t check the figures and got miles and kilometres all Mixed up. Thank you for the correction $\endgroup$ – P.reid Feb 21 '18 at 21:42
  • $\begingroup$ 2.7 km is more or less correct if you consider the water as spread over all the Earth's surface. Since it is not, and 30% of the Earth's surface is land, the average depth goes up by 1/(70%), and that's the about 4 km depth that NOAA reports. Perversely, the ratio of km to miles is similar (62.1%), so the waterworld average depth in kilometers is almost equal to the actual Earth's average depth in miles :-D $\endgroup$ – LSerni Feb 25 '18 at 8:08
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A magic liquid shell above Earth.

Taking a4android's figure of 1.3 billion km³ for the water volume, the shell is the difference (s kilometers thick) between two spheres at height of h kilometers, where

$V = \frac{4\pi}{3}((R+h+s)^3-(R+h)^3)$

with R being the radius of Earth at 6378 km.

Explanation: a spherical shell is simply a sphere that's has been hollowed out by a smaller sphere. So the volume of the shell is the volume of the outer minus the volume of the inner sphere. The volume of a sphere is $\frac{4\pi}{3}$ times the cube of its radius; since these spheres extend from height h to height h+s above the Earth's surface that has radius R, after some math we arrive at the equation above, that gives the volume for any height and thickness.

The equation has infinite solutions depending on the altitude h (for example, if you're satisfied with a very thin shell, you can put it at very large distances).

We can expect the figure to be not too distant from seven tenths of the average water depth for the oceans (the water surface of Earth being seven tenths of the planet's surface). The average depth of the seas is given at around 4 km (Source: NOAA); so, if that were spread all around Earth at a zero height, ignoring mountains and valleys, it would be 2.8 km deep. By plugging the known V and the value zero for h in the equation above, we verify that the value for s is indeed around 2.8 km.

For a height of 3 km above ground, I then calculate a thickness of about 2.707 kilometers (so, the shell extends from an altitude of 3000 to 5700 m), thick enough to keep all the Earth beneath 3000m in total, dry darkness. The magic would need to withstand the pressure of 2707 m of saltwater, somewhere around 270 atmospheres (there is not enough air to be compressed at 270 atm in a shell 3 km thick; and if there was, living with an air pressure of 270 atm isn't recommended anyway).

By lowering the required thickness from 2.7 to only 2 km we are able to have a larger shell, that can rise well above the atmosphere and low Earth orbit, at around 1047 km altitude. A thickness of two kilometers is still more than enough to ensure perpetual darkness.

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  • $\begingroup$ I appreciate your answer where I could follow it, I understand your answer in a basic way but I am not experienced in the maths department lol. so your equation was lost on me, but appreciated nonetheless. I think I can gain an understanding from you answer, enough to figure out what I’m looking for. Thank you $\endgroup$ – P.reid Feb 9 '18 at 17:11
  • $\begingroup$ Sorry, I sometimes get carried away :-). I added some explanations about how those numbers come about. Feel free to add to your question if you need something specific (for example, for the water shell to be transparent enough to allow sunlight to pass through). $\endgroup$ – LSerni Feb 9 '18 at 19:30
  • $\begingroup$ No need to apologize:) it’s the best answer yet! I’m trying to find a somewhat plausible idea to form an earth that no longer has any oceans, but has a somewhat temperate climate, but you brought a very valid point about the sunlight not getting through, I’m sure most readers won’t think to hard about it as it’s sci fiction writing, but I want it somewhat believable. Thanks for your input so far and feel free to throw out any ideas you might have:) $\endgroup$ – P.reid Feb 13 '18 at 4:47
  • $\begingroup$ You might posit some kind of organism - somewhere between a mould and a grass - that sequesters water in some hard-to-extract form, not unlike Dune's sandtrout, and yet has a wet surface (for oxygenation maybe) from which water can evaporate. Functionally a large prairie of such an organism would be similar to a shallow sea. Most water should be sequestered underground though. $\endgroup$ – LSerni Feb 13 '18 at 10:15
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Simply very big.

The total volume of water on Earth is estimated at 1.386 billion km³ (333 million cubic miles), with 97.5% being salt water and 2.5% being fresh water.

Source: Water distribution of Earth

To estimate the size of the water canopy the volume of water should be converted into the equivalent volume of a cloud.

Clouds that have low densities, such as cirrus clouds, contain very little water, thus resulting in relatively low liquid water content values of around .03 g/m3. Clouds that have high densities, like cumulonimbus clouds, have much higher liquid water content values that are around 1-3 g/m3, as more liquid is present in the same amount of space.

For simplicity the hypothetical can have the density of 1 g/m^3, at the low end of the cumlonimbus range. The rest is arithmetic.

A volume of 1.33 billion km^3 (V) = 1.33 x 10^9 km^3 = 1.33 x 10^18 m^3

The density of water (D) = 10^3 g/m^3

Therefore, the volume of water canopy (V2) = Volume (V) x Density (D)

V2 = 1.33 x 10^18 x 10^3 = 1.33 x 10^21 m^3

Without further calculation, this suggests that the hypothetical water canopy would be ridiculously big. Probably, reaching well above the atmosphere of the planet Earth.

Please note: this answer only considers the volumes and densities involved. It has assumed that other physical factors can be ignored. Of course, for any physically realistic model this would be a nonsense.

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  • $\begingroup$ One thing - density of water is 10^6 g/m^3 (one ton per cubic meter), not 10^3. Also, if you increase the density the volume V2 has to go down; you need to divide, not to multiply. $\endgroup$ – LSerni Feb 9 '18 at 7:27
  • $\begingroup$ @LSerni Your correction is much appreciated. Congratulations on making better use of the volume of Earth's water than I did. The fact that it was day three of a heatwave with 30 degrees Centigrade temperatures here didn't help. Even so, I probably would have mucked up such straightforward calculations. It usually takes several drafts at calculations get things right. $\endgroup$ – a4android Feb 10 '18 at 3:57
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I may be a little too late to be helpful to the OP, but I'll go ahead and expand upon LSerni's answer with a quick estimate.

If we assume that the whole Earth is a sphere with radius $R_{e}$ with the thickness of our oceans being $t_{o}$ then the formula for the volume of water would be: $$V_{w}=\frac{4\pi}{3}(R_{e})^3-\frac{4\pi}{3}(R_{e}-t_{o})^3$$ If you then expand the cubic term then the $R_{e}^3$ bits cancel out. Then, since $R_e>>t_o$ we can ignore the last two terms and make an estimate of $$V_{w}\approx4\pi R_{e}^2t_{o}$$ which is much easier to work with. If we do a similar estimate for the volume of water in the water canopy, then we get $V_{w}\approx4\pi R_{s}^2t_{s}$. Then lastly, since we're using the same volume of water both times we can set the two estimates equal to each other and rearrange things by solving for $t_s$. $$t_{s}\approx t_{o}\Big(\frac{R_e}{R_s}\Big)^2$$ This final equation really tells us what we need to know, which is how increasing $R_s$ changes the thickness. What it tells us is that if the outside of the "water canopy" shell is twice as big as the radius of the Earth, then the canopy thickness will be a quarter as thick as the oceans currently are.

Unfortunately that means that the canopy is going to be very very thick. The ISS orbits at about 400km above the Earth's surface, which is a very small addition to the ~6400km radius of the Earth. Its orbital radius is only about 106% of Earth's radius, so from the above estimates your shell would still be about 88% as thick as the oceans currently are. In order to get a thin shell, you'd have to zoom way way out. To get a thin shell, something like 30m deep, you'd have to have the canopy approximately halfway to the moon!

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