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?
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ Related: skeptics.stackexchange.com/questions/10763/… and improbable.com/airchives/paperair/volume9/v9i3/kansas.html $\endgroup$– SpencerAug 15, 2017 at 3:29
-
$\begingroup$ The newer question Will the oceans swallow all of the land? is likely worth related viewing, asking a about the feasibility rather than the values. $\endgroup$– JeopardyTempestOct 16, 2018 at 17:05
-
$\begingroup$ 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. $\endgroup$– JeopardyTempestOct 16, 2018 at 17:05
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
11
An approximation can be obtained quite simply by dividing the volume of water in the oceans by the surface area of an ellipsoid with a smooth surface representing the idealized Earth in your question.
The volume of Earth's oceans, seas and bays is $1.332 \times 10^9 \text{ km}^3$.
The equatorial radius of Earth (semi-major axis of the spheroid) is $a = 6378.1 \text{ km}$. The polar radius of Earth (semi-minor axis) is $c = 6356.8 \text{ km}$.
The 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. An ellipsoid is a better representation of our 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.
-
1$\begingroup$ Do you think that because of all the mountains and seadephts the sizes you used (6378 and 6356) are wright? $\endgroup$– MarijnFeb 3, 2016 at 20:27
-
1$\begingroup$ 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. $\endgroup$ Feb 3, 2016 at 20:54
-
5$\begingroup$ @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. $\endgroup$ Feb 7, 2016 at 14:15
-
1$\begingroup$ @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. $\endgroup$– userLTKFeb 8, 2016 at 4:40
-
1$\begingroup$ 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. $\endgroup$ Apr 11, 2016 at 3:43
$\begingroup$
$\endgroup$
2
510,100,000 square kilometers of surface area, and a total of 1,386,000,000 cubic kilometers of water gives you a 2.717 kilometer column of water across the whole planet if it was billiard ball smooth, but the same basic shape.
-
1$\begingroup$ 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. $\endgroup$– AshAug 16, 2017 at 11:16
-
3$\begingroup$ Hopefully, you mean ideally snooth rather than the much rougher surface of a billiard ball. Earth is already smoother than a billiard ball. $\endgroup$– SpencerFeb 11, 2018 at 1:16