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#1




How deep would the water covering be on a perfectly spherical Earth?
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. 
#2




The total volume of the oceans is 1.3 billion cubic kilometers. The surface area of the Earth is 510,072,000 square kilometers. 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.

#3




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? 
#4




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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 R_{E}, then its volume is (4/3)pi R_{E}^{3}. Thus, the volume of the water alone is: V_{W} = (4/3)pi (R_{W}^{3}  R_{E}^{3}) We solve for R_{W}: R_{W} = ((3/(4 pi)) V_{W} + R_{E}^{3})^{1/3} The OP wants the depth of the water, which is D = R_{W}  R_{E}. This gives: D = ((3/(4 pi)) V_{W} + R_{E}^{3})^{1/3}  R_{E} Prettying this up a bit, we have: D = (((3 V_{W}) / (4 pi R_{E}^{3}) + 1)^{1/3}  1) R_{E} The value of V_{W} is given by Chronos above. In order to determine R_{E}, 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. 


#5




However, for the case where the volume of Earth (given by (4/3)pi R_{E}^{3}) is much larger than the volume of water (V_{W}), the above expression is approximately:
D = V_{W} / (4 pi R_{E}^{2}) 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. 
#6




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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! 
#7




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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, backoftheenvelope is probably good enough for most purposes. Quote:

#8




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ETA: Beaten by MikeS. Last edited by Xema; 04132009 at 08:55 AM. 
#9




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.



#10




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 toofrequent 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. The answer then becomes a contest over who can add the most fine points instead of...an answer. 
#11




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! ) Last edited by Polycarp; 04132009 at 10:49 AM. Reason: fixed ifrst paragraph 
#12




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. 
#13




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. 
#14




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#15




The Sun accounts for about a third of tidal activity.

#16




But how many times have people been there, and in what year, and for how long?
Last edited by tim314; 04132009 at 07:20 PM. 
#17




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, sincewhen it's averaged outit 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. 
#18




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These kind of variables just make the answer(s) overly complicated. I am content with Chronos's answer. Last edited by dogfood; 04132009 at 10:50 PM. 
#19




Smoother, yes, but not rounder, surely?



#20




Well, according to Wiki, 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 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. 
#21




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. 
#22




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.
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#24




Of course, for the real Earth, if you had a global ocean the shape of the geoid would be different.



#25




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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