<p>An approximation can be obtained quite simply by dividing the volume of water in the oceans by the surface area of an <a href="https://en.wikipedia.org/wiki/Reference_ellipsoid" rel="noreferrer">ellipsoid</a> with a smooth surface representing the idealized Earth in your question.</p>

<p>The volume of <a href="http://www.wolframalpha.com/input/?i=volume+of+oceans" rel="noreferrer">Earth's oceans, seas and bays</a> is $1.332 \times 10^9 \text{ km}^3$.</p>

<p>The <a href="http://www.wolframalpha.com/input/?i=equatorial+radius+of+earth" rel="noreferrer">equatorial radius of Earth</a> (semi-major axis of the spheroid) is $a = 6378.1 \text{ km}$. The <a href="http://www.wolframalpha.com/input/?i=polar+radius+of+earth" rel="noreferrer">polar radius of Earth</a> (semi-minor axis) is $c = 6356.8 \text{ km}$.</p>

<p>The <a href="https://en.wikipedia.org/wiki/Ellipsoid#Surface_area" rel="noreferrer">surface area</a> of the oblate ($c &lt; a$) spheroid is:</p>

<p>$$S = 2 \pi a^2 \left( 1 + \frac{1 - e^2}{e}\tanh^{-1} e \right)$$</p>

<p>where $e^2 = 1 - \frac{c^2}{a^2}$.</p>

<p>Which gives us $\approx 0.51 \times 10^9 \text{ km}^2$.</p>

<p>Dividing the volume of the oceans by this results gives us $\approx 2.6 \text{ km}$.</p>

<p><strong>Note</strong>: <a href="https://earthscience.stackexchange.com/q/108/5293">Earth is not a sphere</a>. An ellipsoid is a better <a href="https://en.wikipedia.org/wiki/Figure_of_the_Earth" rel="noreferrer">representation of our Earth</a>. Nevertheless, the answer to your question would have been approximately the same had I used a sphere instead, as suggested in the title of your question.</p>

CC BY-SA 3.0
2016-02-03T20:27:26.267Z
Do you think that because of all the mountains and seadephts the sizes you used (6378 and 6356) are wright?
Marijn 
CC BY-SA 3.0
2016-02-03T20:34:35.613Z
These are average numbers, and the "true figure of Earth" is actually more complicated. On the other hand, consider that the highest mountains and the deepest chasms under the sea are actually small in comparison to the size of our planet.
carnendil
CC BY-SA 3.0
2016-02-03T20:44:00.887Z
Ok, it should be al little deeper than? about 2.7km? When I compare it with snowball earth, where is told that the ice was about 3km thick, we are close to it?!
Marijn 
CC BY-SA 3.0
2016-02-03T20:54:01.190Z
You can try and add the water of all lakes and lagoons, as well as the water in rivers and ice in glaciers and mountaintops, and see how much a variation you get. I am not much familiarized with the Snowball Earth hyphotesis.
carnendil
CC BY-SA 3.0
2016-02-04T04:00:46.707Z
The 3KM estimate for snowball earth is that, an estimate, also, it's important to remember, at least during the more recent snowball earths, there were already continents and oceans.  3 KM thick ice was likely over the oceans with some ocean remaining liquid on the bottom, the growing salt concentration would prevent freezing all the way down.  On land, there were glaciers, likely less thick.   It probably wasn't a smooth 3K all the way around.
userLTK
CC BY-SA 3.0
2016-02-07T14:15:38.053Z
@userLTK -- It's the other way around. If this hypothesis is correct, ice sheets up to 3 km thick ice covered the continents, but not the oceans. The oceans would have been covered by a much thinner layer of ice, possibly with partially open water near the equator.
David Hammen
CC BY-SA 3.0
2016-02-08T04:40:50.593Z
@DavidHammen   Thanks, and that makes sense.   Ice would build up on land but perhaps stay liquid in the salty oceans, with, as you said, a thinner ice layer on top.    Sea level must have dropped perhaps as much as a mile in that case.
userLTK
CC BY-SA 3.0
2016-04-11T03:43:14.147Z
Even with a smooth Earth, the oceans wouldn't be a uniform depth. For the same reason Earth is an oblate spheroid, the Earth + water would be a slightly larger oblate spheroid. The ocean would be shallower at the poles and deeper at the equator by the same ratio as the Earth's axes, so 1/290.
kwknowles
CC BY-SA 3.0
2017-08-15T05:33:23.163Z
Without digging into the math, would've thought it might be easier and slightly more correct to set the integral that makes a spherical shell equal to the volume.  Doubt the answers are much different at all, I guess.
JeopardyTempest
CC BY-SA 3.0
2017-08-16T03:39:19.417Z
`+1` I think it's great you've included a better representation of the shape of the Earth than a sphere. But the assumption of *uniform depth* may not then be compatible with this level of detailed treatment. Would it not be better to use a second ellipsoid to represent the shape of the potential due to variations of gravity and centrifugal force as well? (sorry, I don't know the precise term for this). Wouldn't there be a relative difference in depth between the poles and the equator *at least* of order the parts-per-thousand variation in radius and area the ellipsoid treatment evokes?
uhoh
CC BY-SA 3.0
2017-08-16T03:47:17.337Z
I've just asked [What would be the depth variation of a “water ellipsoid” above a reference ellipsoid Earth?](https://earthscience.stackexchange.com/q/12096/6031) I'm beginning to think that the uniform depth assumption is based in reality and not just a shortcut, but not sure yet, so I've added a new question to make more space to address this.
uhoh
CC BY-SA 3.0
2016-02-03T20:19:24.657Z
2017-08-16T03:17:27.770Z
2017-08-16T03:17:27.770Z
carnendil
<p>510,100,000 square kilometers of <a href="https://www.google.co.nz/search?q=earth+surface+area&amp;oq=earth+surface+area&amp;aqs=chrome..69i57.4295j0j1&amp;sourceid=chrome&amp;ie=UTF-8" rel="nofollow noreferrer">surface area</a>, and a total of 1,386,000,000 cubic kilometers of <a href="https://water.usgs.gov/edu/earthhowmuch.html" rel="nofollow noreferrer">water</a> gives you a 2.717 kilometer column of water across the whole planet if it was billiard ball smooth, but the same basic shape.</p>

CC BY-SA 3.0
2017-08-16T11:16:55.690Z
I should note that there will be some variance, in the range of metres due to tidal and rotational effects, and very _very_ minor variances due to local thermal conditions.
Ash
CC BY-SA 3.0
2018-02-11T01:16:50.217Z
Hopefully, you mean _ideally_ snooth rather than the much rougher surface of a billiard ball. [Earth is already smoother than a billiard ball](http://www.curiouser.co.uk/facts/smooth_earth.htm).
Spencer
CC BY-SA 3.0
2017-08-15T12:56:07.797Z
2017-08-15T13:59:16.330Z
2017-08-15T13:59:16.330Z
Ash
<p>At the moment there are deep seas and high mountains. But imagine that the land elevation of the Earth is equal everywhere. How deep would the ocean be in that case?</p>

CC BY-SA 3.0
2017-08-15T03:29:04.880Z
Related: https://skeptics.stackexchange.com/questions/10763/is-earth-as-smooth-as-a-billiard-ball and http://www.improbable.com/airchives/paperair/volume9/v9i3/kansas.html
Spencer
CC BY-SA 4.0
2018-10-16T17:05:15.877Z
The newer question [Will the oceans swallow all of the land?](https://earthscience.stackexchange.com/questions/15340/will-the-oceans-swallow-all-of-the-land/15359#15359) is likely worth related viewing, asking a about the feasibility rather than the values.
JeopardyTempest
CC BY-SA 4.0
2018-10-16T17:05:42.930Z
Also the consideration in my answer about some of the dirt becoming additional mud may be a realistic consideration that may need to be made to the answer here as well.
JeopardyTempest
<geology><ocean><world-building>
If the Earth were a smooth spheroid, how deep would the ocean be?
Marijn 
# If the Earth were a smooth spheroid, how deep would the ocean be?

(Asked by: username_0 on 2016-02-03)

At the moment there are deep seas and high mountains. But imagine that the land elevation of the Earth is equal everywhere. How deep would the ocean be in that case?

**Tags:** geology, ocean, world-building

> **Comments:**
> **username_1**: Related: https://skeptics.stackexchange.com/questions/10763/is-earth-as-smooth-as-a-billiard-ball and http://www.improbable.com/airchives/paperair/volume9/v9i3/kansas.html  
> **username_2**: The newer question [Will the oceans swallow all of the land?](https://earthscience.stackexchange.com/questions/15340/will-the-oceans-swallow-all-of-the-land/15359#15359) is likely worth related viewing, asking a about the feasibility rather than the values.  
> **username_2**: Also the consideration in my answer about some of the dirt becoming additional mud may be a realistic consideration that may need to be made to the answer here as well.  

---

## 2 answers

### Answer by username_3 on 2016-02-03. Score: 19

An approximation can be obtained quite simply by dividing the volume of water in the oceans by the surface area of an [ellipsoid](https://en.wikipedia.org/wiki/Reference_ellipsoid) with a smooth surface representing the idealized Earth in your question.

The volume of [Earth's oceans, seas and bays](http://www.wolframalpha.com/input/?i=volume+of+oceans) is $1.332 \times 10^9 \text{ km}^3$.

The [equatorial radius of Earth](http://www.wolframalpha.com/input/?i=equatorial+radius+of+earth) (semi-major axis of the spheroid) is $a = 6378.1 \text{ km}$. The [polar radius of Earth](http://www.wolframalpha.com/input/?i=polar+radius+of+earth) (semi-minor axis) is $c = 6356.8 \text{ km}$.

The [surface area](https://en.wikipedia.org/wiki/Ellipsoid#Surface_area) of the oblate ($c < a$) spheroid is:

$$S = 2 \pi a^2 \left( 1 + \frac{1 - e^2}{e}\tanh^{-1} e \right)$$

where $e^2 = 1 - \frac{c^2}{a^2}$.

Which gives us $\approx 0.51 \times 10^9 \text{ km}^2$.

Dividing the volume of the oceans by this results gives us $\approx 2.6 \text{ km}$.

**Note**: [Earth is not a sphere](https://earthscience.stackexchange.com/q/108/5293). An ellipsoid is a better [representation of our Earth](https://en.wikipedia.org/wiki/Figure_of_the_Earth). Nevertheless, the answer to your question would have been approximately the same had I used a sphere instead, as suggested in the title of your question.

> **Comments:**
> **username_0**: Do you think that because of all the mountains and seadephts the sizes you used (6378 and 6356) are wright?  
> **username_3**: These are average numbers, and the "true figure of Earth" is actually more complicated. On the other hand, consider that the highest mountains and the deepest chasms under the sea are actually small in comparison to the size of our planet.  
> **username_0**: Ok, it should be al little deeper than? about 2.7km? When I compare it with snowball earth, where is told that the ice was about 3km thick, we are close to it?!  
> **username_3**: You can try and add the water of all lakes and lagoons, as well as the water in rivers and ice in glaciers and mountaintops, and see how much a variation you get. I am not much familiarized with the Snowball Earth hyphotesis.  
> **username_4**: The 3KM estimate for snowball earth is that, an estimate, also, it's important to remember, at least during the more recent snowball earths, there were already continents and oceans.  3 KM thick ice was likely over the oceans with some ocean remaining liquid on the bottom, the growing salt concentration would prevent freezing all the way down.  On land, there were glaciers, likely less thick.   It probably wasn't a smooth 3K all the way around.  
> **username_5**: @username_4 -- It's the other way around. If this hypothesis is correct, ice sheets up to 3 km thick ice covered the continents, but not the oceans. The oceans would have been covered by a much thinner layer of ice, possibly with partially open water near the equator.  
> **username_4**: @username_5   Thanks, and that makes sense.   Ice would build up on land but perhaps stay liquid in the salty oceans, with, as you said, a thinner ice layer on top.    Sea level must have dropped perhaps as much as a mile in that case.  
> **username_6**: Even with a smooth Earth, the oceans wouldn't be a uniform depth. For the same reason Earth is an oblate spheroid, the Earth + water would be a slightly larger oblate spheroid. The ocean would be shallower at the poles and deeper at the equator by the same ratio as the Earth's axes, so 1/290.  
> **username_2**: Without digging into the math, would've thought it might be easier and slightly more correct to set the integral that makes a spherical shell equal to the volume.  Doubt the answers are much different at all, I guess.  
> **username_7**: `+1` I think it's great you've included a better representation of the shape of the Earth than a sphere. But the assumption of *uniform depth* may not then be compatible with this level of detailed treatment. Would it not be better to use a second ellipsoid to represent the shape of the potential due to variations of gravity and centrifugal force as well? (sorry, I don't know the precise term for this). Wouldn't there be a relative difference in depth between the poles and the equator *at least* of order the parts-per-thousand variation in radius and area the ellipsoid treatment evokes?  
> **username_7**: I've just asked [What would be the depth variation of a “water ellipsoid” above a reference ellipsoid Earth?](https://earthscience.stackexchange.com/q/12096/6031) I'm beginning to think that the uniform depth assumption is based in reality and not just a shortcut, but not sure yet, so I've added a new question to make more space to address this.  

---

### Answer by username_8 on 2017-08-15. Score: 5

510,100,000 square kilometers of [surface area](https://www.google.co.nz/search?q=earth+surface+area&oq=earth+surface+area&aqs=chrome..69i57.4295j0j1&sourceid=chrome&ie=UTF-8), and a total of 1,386,000,000 cubic kilometers of [water](https://water.usgs.gov/edu/earthhowmuch.html) gives you a 2.717 kilometer column of water across the whole planet if it was billiard ball smooth, but the same basic shape.

> **Comments:**
> **username_8**: I should note that there will be some variance, in the range of metres due to tidal and rotational effects, and very _very_ minor variances due to local thermal conditions.  
> **username_1**: Hopefully, you mean _ideally_ snooth rather than the much rougher surface of a billiard ball. [Earth is already smoother than a billiard ball](http://www.curiouser.co.uk/facts/smooth_earth.htm).  

---

