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View Full Version : How deep would the water covering be on a perfectly spherical Earth?

dogfood
04-13-2009, 12:33 AM
So I basically daydream all day and come up with ridiculous hypotheticals to think about. The other day i started to wonder, if the Earth was perfectly spherical so all the available water on the planet would distribute evenly, how deep would the water level be? I thought about this for quite some time before I realized I suck at math and don't understand physics, and there are about a billion crazy variables such as the moon, ice, and water in the atmosphere. So I figured I would ask people who are smarter than I.

If anyone wants to try to work this out feel free to ignore said variables.

Chronos
04-13-2009, 02:23 AM
The total volume of the oceans is 1.3 billion cubic kilometers (http://en.wikipedia.org/wiki/Ocean#Physical_properties). The surface area of the Earth is 510,072,000 square kilometers (http://en.wikipedia.org/wiki/Earth). Dividing the volume by the surface area, we get a depth of 2.5 kilometers. Physics doesn't enter into it, nor do a billion crazy variables.

Omniscient
04-13-2009, 02:49 AM
Well, that answer is overly simplistic and the subtleties and physics do indeed matter. First off, we might want to assume that the OP means "smooth" as opposed to spherical. The earth as most realize is not spherical but an oblate spheroid and if the ocean floor were perfectly smooth and level the depth would be greater at the equator than at the poles. I'm not sure what the math would be on that, but certainly you'd have to include centripetal forces and angular momentum on the water as well as the amount of distortion rotation causes to the crust of the earth. The changed distribution of mass with a smooth surface would mean the planet would have a slightly different shape than it does now.

More simply, we can assume that your given surface area of the planet is calculated at sea level, with all the land that's above sea level moved to the bottom would increase what we currently understand to be sea level and increasing the diameter and circumference of the globe.

Next, we have to realize that that "surface area" is probably at sea level. The math you did would need to be adjusted for the bottom of the "smooth" sea instead of the existing surface of the oceans.

Also, simply dividing the volume by the area works for a flat plane, but not for a sphere. There's more water in the top 1000 feet of the ocean than there is at the bottom 1000 feet because the layers of water get bigger. I'm not sure how to do that math off hand.

The OP probably needs to clarify what he means. Are we taking "smooth" and in the same shape as now? Are we talking about a true sphere that isn't rotating? Are we shaving the existing land masses off and throwing them away or are we grading everything out to be flat?

tim314
04-13-2009, 04:08 AM
Also, simply dividing the volume by the area works for a flat plane, but not for a sphere. There's more water in the top 1000 feet of the ocean than there is at the bottom 1000 feet because the layers of water get bigger. I'm not sure how to do that math off hand.If the surface of the water gives a sphere of radius RW, its volume is (4/3)pi RW3.

Of course, that volume includes the Earth (meaning the rock/dirt part) at the middle, so we have to subtract that out. If the radius of the Earth is RE, then its volume is (4/3)pi RE3.

Thus, the volume of the water alone is:
VW = (4/3)pi (RW3 - RE3)

We solve for RW:

RW = ((3/(4 pi)) VW + RE3)1/3

The OP wants the depth of the water, which is D = RW - RE. This gives:

D = ((3/(4 pi)) VW + RE3)1/3 - RE

Prettying this up a bit, we have:

D = (((3 VW) / (4 pi RE3) + 1)1/3 - 1) RE

The value of VW is given by Chronos above. In order to determine RE, I need to know if the OP means a sphere with the same surface area as the Earth, or a sphere with the same volume as the Earth, or something else.

tim314
04-13-2009, 04:22 AM
However, for the case where the volume of Earth (given by (4/3)pi RE3) is much larger than the volume of water (VW), the above expression is approximately:
D = VW / (4 pi RE2)

We can see that this is just the volume of the water divided by the surface area of the Earth, just as Chronos said. This approximation is indeed valid for the Earth. To put it another way: the Earth is sufficiently large that if you aren't too far above the surface it looks like a plane.

Hirka T'Bawa
04-13-2009, 05:17 AM
Physics doesn't enter into it, nor do a billion crazy variables.

The OP was right about some of the variables, mainly that of the moon. If the entire surface was water, I would think the moons effect on the tides would make a significant difference in the water depth during high and low tides. I also don't know exactly how much effect the sun would have on the tides of this "water world".

Now of course, I could be totally wrong, and it won't have any effect due to some weird physics law... Dammit Jim! I'm a pharmacist (in training) not an engineer!

MikeS
04-13-2009, 08:52 AM
Well, that answer is overly simplistic and the subtleties and physics do indeed matter. Knowing Chronos, I'm sure that if the physics did matter he'd know about it. First off, we might want to assume that the OP means "smooth" as opposed to spherical. The earth as most realize is not spherical but an oblate spheroid and if the ocean floor were perfectly smooth and level the depth would be greater at the equator than at the poles. I'm not sure what the math would be on that, but certainly you'd have to include centripetal forces and angular momentum on the water as well as the amount of distortion rotation causes to the crust of the earth. The changed distribution of mass with a smooth surface would mean the planet would have a slightly different shape than it does now.The oblateness of the Earth is (more or less) caused by the centrifugal forces due to its rotation; to a very good approximation the Earth is "flat", in the sense that the surface is perpendicular to the total force due to gravity and centrifugal force. So the oblateness of the Earth would not significantly change if we flattened it out.
More simply, we can assume that your given surface area of the planet is calculated at sea level, with all the land that's above sea level moved to the bottom would increase what we currently understand to be sea level and increasing the diameter and circumference of the globe.

Next, we have to realize that that "surface area" is probably at sea level. The math you did would need to be adjusted for the bottom of the "smooth" sea instead of the existing surface of the oceans.The height of Mt. Everest is about 8800 metres. The depth of Challenger Deep is about 10,900 metres. The mean radius of the Earth is about 6,370,000 metres  five or six hundred times larger than either of these numbers. To a very good approximation, the Earth is already smooth.

Also, simply dividing the volume by the area works for a flat plane, but not for a sphere. There's more water in the top 1000 feet of the ocean than there is at the bottom 1000 feet because the layers of water get bigger. I'm not sure how to do that math off hand.This was already addressed by tim314.

If you took all the above factors into account, I would estimate that the correction that you would find would be no more than 1% different from Chronos' estimate of 2.5 km. In this case, back-of-the-envelope is probably good enough for most purposes.

If the entire surface was water, I would think the moons effect on the tides would make a significant difference in the water depth during high and low tides.Not really. The tides you see at the seashore are typically much larger than they would be if the earth was perfectly flat and covered with water; if you figure out the gravitational forces of the Moon and the Sun when they're aligned, and calculate what the tides would actually be in this case, you get "spring tides" of about 86 cm (or about 2' 10".) Again, a very small correction to the 2.5 km estimate Chronos obtained.

Xema
04-13-2009, 08:54 AM
If the entire surface was water, I would think the moons effect on the tides would make a significant difference in the water depth during high and low tides.
Probably not. On the Earth as it's currently arranged, tides in the open ocean are small - they become significant only when there's land to confine the water (a bay, for example).

ETA: Beaten by MikeS.

Xema
04-13-2009, 08:58 AM
A factor that hasn't been mentioned is the volume of the Earth's fresh water, which is apparently about 3% of the volume of salt water.

Chief Pedant
04-13-2009, 09:19 AM
Well, that answer (by Chronos) is overly simplistic...<snip>
Well, I liked the answer, and I suspect it's the type of answer the OP was looking for. Straightforward and with the assumption that the assorted variables don't add up to a very significant change in the final depth.

If they do, is someone willing to take into account the variables you mention and actually come up with more precise alternate number so we can see how far off the approximation was?

If the variables you mention do not significantly change the answer, then what we have here is the too-frequent SDMB practice of proposing to add significant figure decimal places when only a rough estimate was called for, but not actually delivering. That's not omniscience; it's kibitzing. ;)

Polycarp
04-13-2009, 10:48 AM
Does the (rather minimal) oblateness of Earth from precise sphericity have any significant bearing on the answer? Would the water be deeper at the equator than at the poles to any noticeable degree? I suspect not, but it would be nice to have that hunch confirmed with numbers.

Hey, brigning up the poles -- on this water world, do we take into account water's physical characteristics? If so, I presume that the poles would be covered by floating ice with the displacement value equivalent to water (i.e., that the fact some of the water is frozen would not influence the depth to any significant degree) -- but what would happen as regards insolation and evaporation on a water world? Would there be greater or lesser cloud cover, more or less formation of cyclonic storms, etc.?

(I don't have the answers, but I do have good questions! :D)

Chronos
04-13-2009, 12:27 PM
First of all, note that the figure I was using for the volume of the ocean only has two significant figures, so I rounded my answer to the same degree. So unless someone finds a better figure for ocean volume, we might as well completely ignore anything that makes a difference of less than 50 meters.

Second of all, the OP asked for a single number for depth, so even to the extent that some effects would cause a variation in depth in different locations, what we really want is the average depth. Tides, whether solar, lunar, or Jovian, would have no effect on the average depth, and the oblateness of the Earth would have only a negligible effect (it'd be second or third order in the oblateness, which is already down around 1%).

Third, I also didn't take into account any quantum mechanical, general relativistic, or quantum gravitational effects. Everything is always an approximation: It's just a matter of realizing which approximations are justified in any given situation, and which aren't.

Hari Seldon
04-13-2009, 01:07 PM
Given the approximations and uncertainties involved, the first answer is as good as you can do. I would have expected somewhat deeper, but I cannot fault his math. All those other things are secondary or tertiary effects. In fact, only the oblateness matter to any significant degree and given the approximations, even that probably doesn't matter.

Don't carry your analysies further than the data allow.

DesertDog
04-13-2009, 01:47 PM
The height of Mt. Everest is about 8800 metres. The depth of Challenger Deep is about 10,900 metres. The mean radius of the Earth is about 6,370,000 metres  five or six hundred times larger than either of these numbers. To a very good approximation, the Earth is already smooth.Isaac Asimov (I think it was) commented that if the Earth were shrunk to the size of a billiard ball, it would be rounder and smoother than a billiard ball.

tdn
04-13-2009, 01:59 PM
I also don't know exactly how much effect the sun would have on the tides of this "water world".

The Sun accounts for about a third of tidal activity.

tim314
04-13-2009, 07:20 PM
The depth of Challenger Deep is about 10,900 metres.But how many times have people been there, and in what year, and for how long? ;)

04-13-2009, 09:13 PM
Is a significant amount of the existing water distribution deeper than 2.5 klicks? The compression of water below that depth will make some difference, since--when it's averaged out--it won't be as compressed any more.

I don't think is will change the answer much, but at least it's a question that hasn't been asked yet. :)

dogfood
04-13-2009, 10:47 PM
Hey, brigning up the poles -- on this water world, do we take into account water's physical characteristics? If so, I presume that the poles would be covered by floating ice with the displacement value equivalent to water (i.e., that the fact some of the water is frozen would not influence the depth to any significant degree) -- but what would happen as regards insolation and evaporation on a water world? Would there be greater or lesser cloud cover, more or less formation of cyclonic storms, etc.?

This is more what I was thinking of about the ice and atmospheric water. I have no idea what the temperature of Earth would be in the surface was entirely water, and I have no idea how much water would then wind up locked up in polar ice caps. Then the issue of how much water would be (more or less) constantly in the atmosphere, and how much fresh water not included in the ocean volume figure, ect.

These kind of variables just make the answer(s) overly complicated. I am content with Chronos's answer.

Colophon
04-14-2009, 08:46 AM
Isaac Asimov (I think it was) commented that if the Earth were shrunk to the size of a billiard ball, it would be rounder and smoother than a billiard ball.
Smoother, yes, but not rounder, surely?

DesertDog
04-16-2009, 01:21 AM
Smoother, yes, but not rounder, surely?Well, according to Wiki (http://en.wikipedia.org/wiki/Earth), the difference between the polar radius and equatorial radius of the Earth is 22km with a mean radius of 6371km That's a difference of .00345 (22/6371)

An American pool ball (http://en.wikipedia.org/wiki/Billiard_ball#American-style_pool) is 2.25in +/- .005in in diameter. Dividing the diameter by two to get the radius and dividing that into the tolerance, you get a difference of .00444 Doubling the tolerance. because it is plus or minus after all, doubles the difference to .00888 (.005/1.125 or .010/1.125).

If you want to see an oblate planet, go look at Saturn.

FRDE
04-16-2009, 05:24 AM
I vaguely remember something similar turning up here some time ago.

The answer given was that if earth were the size of a base/basket/billiard ball then the vertical distance between the deepest trench and the highest mountain would be about the thickness of a fingernail.

I can't remember which ball, but with that sort of ratio it doesn't seem to matter that much.

wolfstu
04-16-2009, 01:41 PM
Now that the basic mathematical answer has been given, it may be of interest to consider that the Earth is not perfectly spherical, or of uniform density. For the real earth, a global ocean wouldn't have a spherical surface.The irregular gravity field (http://www.esa.int/esaLP/ESA1XK1VMOC_LPgoce_0.html) shapes a virtual surface at mean sea-level called the 'geoid'. This is the surface of equal gravitational potential of a hypothetical ocean surface at rest and so defines 'the horizontal' - meaning that if a ball were placed on this hypothetical surface it would not roll, despite the presence of virtual slopes. The geoid is important for understanding more about ocean currents and is also employed as a reference for traditional height systems and monitoring sea-level change. In addition, the geoid is used for levelling and construction.

Determining the shape of the geoid to very high precision is the goal of the recently launched GOCE mission (http://www.esa.int/esaLP/ESAYEK1VMOC_LPgoce_0.html) (Gravity field and steady-state Ocean Circulation Explorer).

Snarky_Kong
04-16-2009, 01:55 PM
Well, according to Wiki (http://en.wikipedia.org/wiki/Earth), the difference between the polar radius and equatorial radius of the Earth is 22km with a mean radius of 6371km That's a difference of .00345 (22/6371)

An American pool ball (http://en.wikipedia.org/wiki/Billiard_ball#American-style_pool) is 2.25in +/- .005in in diameter. Dividing the diameter by two to get the radius and dividing that into the tolerance, you get a difference of .00444 Doubling the tolerance. because it is plus or minus after all, doubles the difference to .00888 (.005/1.125 or .010/1.125).

If you want to see an oblate planet, go look at Saturn.

2.25 +/- 0.005 gives radii of 1.1225 and 1.1275 respectively for a variation of 0.005 or +/- 0.0025. The difference is then 0.00444, not 0.0088.

Chronos
04-16-2009, 03:57 PM
Of course, for the real Earth, if you had a global ocean the shape of the geoid would be different.

DesertDog
04-18-2009, 08:56 AM
2.25 +/- 0.005 gives radii of 1.1225 and 1.1275 respectively for a variation of 0.005 or +/- 0.0025. The difference is then 0.00444, not 0.0088.Oops. I forgot to halve the tolerance when I halved the diameter to get the radius. Still, .00444 is less round than .00345

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