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?

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    $\begingroup$ earthscience.stackexchange.com/questions/2214/… $\endgroup$ – arkaia Aug 16 '17 at 10:25
  • $\begingroup$ @aretxabaleta There are many interesting links there, thanks! The question here is short, but needs an answer with some insight, but the comparison is interesting and I'll read further. $\endgroup$ – uhoh Aug 16 '17 at 10:41

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