The question If the Earth was a smooth spheroid how deep would be the ocean? has an excellent answer which explains the shape and surface area of an oblate spheroid before calculating the average depth of the Earth's surface water (about 2.7 km). The difference in radius, and surface area compared to a sphere look to be of the order of parts-per-thousand compared to a sphere.

But even within the approximation of an ellipsoid, the Earth's gravity is not uniform in direction nor magnitude, and so the assumption of a constant water depth from pole to equator may also require some correction of the order of parts-per-thousand.

How would one calculate the variation of depth of the water from pole to equator taking into account some kind of first order variation of gravitational force and rotational effects (centrifugal force) to get the shape of the "water ellipsoid"? Does it turn out that the thickness would end up being uniform because the Earth ellipsoid is already a response to rotational effects, or would there be a pole-to-equator variation in depth of order parts-per-thousand?


A complicating factor will be the change in density with temperature. So for this I will assume a constant density of water.

As the thickness of the water layer is quite small compared to the size of Earth (given as 2.6 km by the answer to the other question) I think an approximation with a constant g at any particular location should give a fairly reasonable answer.

Water naturally finds its level based upon pressure. So if we have a constant pressure at the top of the surface, the pressure at the bottom must also be constant. And this should work as a reasonable approximation for any number of layers.

Noting that for any column of water, the pressure increase will be given by:
$P=\rho \cdot g \cdot h$

Taking the values of g from Wikipedia:

At the equator, g, due to Earth being an oblate spheroid and rotating is ~$9.78 \frac{\mathrm{m}}{\mathrm{s}^2}$

At the poles, g is ~$9.83 \frac{\mathrm{m}}{\mathrm{s}^2}$

Equating the 2 we end up with: $P=\rho \cdot 9.78 \frac{\mathrm{m}}{\mathrm{s}^2} \cdot h_e=\rho \cdot 9.83 \frac{\mathrm{m}}{\mathrm{s}^2} \cdot h_p$

Factoring out common terms leaves us with: $h_e = \frac{9.83}{9.78} h_p$ So the equator will have water that is roughly 0.5% deeper than the poles.

For 2.6 km, that works out to be roughly 13 m.

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    $\begingroup$ I have updated the answer to note that I took the values from Wikipedia. $\endgroup$ – Jack Black Jan 22 at 3:30
  • $\begingroup$ Great, thanks! :-) $\endgroup$ – uhoh Jan 22 at 3:42
  • $\begingroup$ I guess surface tension is not a factor, right? Isn't that usually important only for small droplets where the surface/volume ratio is larger? $\endgroup$ – uhoh Jan 22 at 3:43
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    $\begingroup$ Yes, surface tension is insignificant at this level. The pressure of a curved droplet is inversely proportional to the radius, so for a 6371 km radius Earth, even with water's high surface tension of ~70 mN/m, you end up with a pressure of roughly 11 nPa, or roughly a 10 trillionth of an atmosphere. Conversely, the pressure for our 2.6 km high water would be 25 MPa, or 250 atm. You're welcome. And thanks for the bounty. $\endgroup$ – Jack Black Jan 22 at 7:05

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