If all of the continental crust on Earth were eroded down, smoothed out, and distributed evenly across the whole planet, filling in all of the ocean basins and displacing the water therein, how deep would the resulting world-wide ocean be?

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    $\begingroup$ I work at the bobcat sales department. We now know not to take any orders for > 10k units from one individual, even if it means a life changing commission. $\endgroup$
    – Sidney
    Jun 27 '17 at 18:00
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    $\begingroup$ Isn't world erosion technically the opposite of worldbuilding? $\endgroup$
    – Milo P
    Jun 27 '17 at 22:55
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    $\begingroup$ @Sidney - you can take that order in good conscience. NYC has an elevation of around 33' over 303 square miles, or around 280B cubic feet. The Bobcat S130 has a bucket capacity of about 1/3 yd^3 or 3 ft^3. It would take 10,000 Bobcats around 9M bucket loads of earth to level off NYC. At 1 minute to take a load, and dump it somewhere else, it would take about 17 years just to do NYC. To do the USA, it'd take 1,000,000 bobcats around 130,000 years to level the 3M sq mi * 2500 ft average elevation of the entire USA. You'll have plenty of time to spend your commission check. $\endgroup$
    – Johnny
    Jun 28 '17 at 1:54
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    $\begingroup$ We're gonna need a bigger bobcat. $\endgroup$
    – Mazura
    Jun 28 '17 at 4:32
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    $\begingroup$ @Mazura Wouldn't that be a similodian? $\endgroup$ Mar 9 '20 at 15:47

I found this here, so it's not technically my answer, but here you go:

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.

The wiki links do have those numbers and the math seems to check out (although rounded), so there you have it :)

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    $\begingroup$ I don't think you can simply divide (volume of water)/(current surface area of earth). Changing the shape of an object while preserving volume does not preserve surface area. (If you have a cube with V = 1 u^3, s = 1 u, and SA = 6 u^2, and then flatten it into a sphere, still with V = 1 u^3, you get r = root_3(3/pi)/2^(2/3) u ~= .6 u and SA = root_3(6/pi) u^2 ~= 4.8 u^2.) I'm not sure if the approximation is good enough for Earth, but I feel that it's not, because the Earth has valleys and mountains that add lots of surface area but not much volume. $\endgroup$
    – HTNW
    Jun 27 '17 at 19:12
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    $\begingroup$ @htnw 5.1 * 10^8 square km is the surface area of a sphere with the radius of earth. Added surface area due to wrinkliness is already not accounted for or is neglible. - I suspect the latter since valleys and mountains are very small on the scale of the planet. $\endgroup$
    – Taemyr
    Jun 27 '17 at 19:32
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    $\begingroup$ @Taemyr "Very small" actually makes it worse, because the smaller something gets the greater the ratio of SA (proportional to scale factor^2) to V (proportional to scale factor^3). Indeed, as you measure more and more accurately, the surface area of a sufficiently irregular object appears to grow without bound! But some quick google-calcing starting from volumes gets very close to 5.1e8 km^2, so wrinkles were probably not accounted for. $\endgroup$
    – HTNW
    Jun 27 '17 at 19:54
  • $\begingroup$ @HTNW This is why we try to measure such things within a certain scale. Since the question is on the scale of entire oceans and continents, kilometers seem like the smallest reasonable unit to use (rounding up to several orders of magnitude is probably best). On that scale, "very small" amounts to "negligible". If you want to increase precision and use smaller units, you eventually add up the volumes of individual atoms, which is just silly in this context and ends up being nearly the same anyway. $\endgroup$
    – talrnu
    Jun 28 '17 at 3:56
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    $\begingroup$ Putting this answer another way, if you took a perfectly smooth, dry sphere with the surface area of the Earth (as you'd have if you removed all the water from Earth and leveled all of the land masses - accurate within an order of magnitude or two, I'm sure), then distributed the full volume of Earth's water over that surface (achieving the scenario in question), it would be about 2.5km deep. Since that sphere's surface area is based on the Earth with oceans, it should actually be a little smaller and you therefore might do well to round the final depth up to nearly 3km. $\endgroup$
    – talrnu
    Jun 28 '17 at 4:02

I agree with the conclusions of the current two answers, but thought a different analysis might be interesting. Since the ocean is a hollow sphere, its depth isn't given exactly by volume/surface area. However, the relationship is still simple:

$V_{ocean} = \frac43 \pi (R_{ocean}^3 - R_{earth}^3)$

Radius diagram

Taking the average radius of the Earth and volume of Earth's oceans from Google, we can solve for $R_{ocean}$:

$R_{earth} = 6371 \mbox{km}$

$V_{ocean} = 1.332 \times 10^9 \mbox{km}^3$

$R_{ocean} = 6373.61 \mbox{km}$

So the ocean's depth will be 2.61 km, or about 0.04% of the radius of the Earth - hence the similarity to the approximation using $\frac{V}{SA}$. Plug in better estimates for the average radius of the Earth to get more accuracy.

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    $\begingroup$ Nice to see someone doing the proper math I was too lazy for. Have your well-deserved upvote :) $\endgroup$
    – Syndic
    Jun 29 '17 at 6:18
  • $\begingroup$ @alex_d you did actual maths. +1. Thanks. $\endgroup$ Feb 23 at 20:40
  • $\begingroup$ Take my updoot. $\endgroup$
    – The Daleks
    Feb 23 at 20:44

If all of the dry land that lies above sea level were to be pushed into the ocean, the ocean would rise less than 300 meters.

Wikipedia summarizes the current division of land and sea as follows:

  • 510072000 km2 (196940000 sq mi)
  • 148940000 km2 land (57510000 sq mi; 29.2%)
  • 361132000 km2 water (139434000 sq mi; 70.8%)
  • The mean height of land above sea level is 0.840 km

So the volume of land above the current sea level is 148,940,000 x 0.84 = 125,109,600 km3.

Reshaping that volume so that it covers the entire earth surface, it would have a height of 125,109,600 / 510,072,000 = 0.245 km. (Of course the displaced land would sink to the bottom, and the water would be cover the entire surface.)

So the ocean would rise by 245 meters.

How deep would the current ocean be? It would be 245 meters deeper at any given point than it currently is. It's average depth (about 2.5 km) would change by less than 10%.

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    $\begingroup$ "deeper at any given point than it currently is" - except for the places you use to put current land into, of course $\endgroup$
    – Mołot
    Jun 27 '17 at 17:55
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    $\begingroup$ Ah, yes. Good point! $\endgroup$
    – Jim
    Jun 27 '17 at 18:09
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    $\begingroup$ It would be 245 meters deeper at any given point - you mean less deep or shallower, not deeper, right? $\endgroup$
    – Stephen S
    Jun 27 '17 at 19:08
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    $\begingroup$ I don't think you can just take the land above sea level, you need to put everything above median depth and use it to fill the part below median depth, making everything the same depth. $\endgroup$ Jun 27 '17 at 21:16

SO the answer is 2.5 km deep, even 3 km deep. Why? Because there are 1.3 billion cubic metres of water in the ocean and If you divide that by the total by the volume of the earth, you would get about 2.5 Km deep. As easy as that. (-:


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