A quarter-section of a cross-section of Earth. Not to scale.

In the rotating frame of Earth, the centrifugal pseudoforce generates its own pseudopotential:

where *r* is the distance to Earth’s axis, *not* the distance to the center. The gravitational potential energy of a particle at a distance *r* from the axis and height *h* above the plane of the equator is given by

The exact formula for an oblate spheroid like Earth is very complicated and differs, in the case of Earth, only in decimal places beyond what we are concerned with.

A ball placed on the surface of a spinning planet in hydrostatic equilibrium does not roll; thus, the potential energy–including the centrifugal pseudopotential–in the rotating frame must be constant. In particular, the potential energy at the pole must equal that at the equator:

We note that we can rewrite the mass of Earth in terms of its density and volume. The volume of an oblate spheroid is , so

Note that we have taken the opportunity to clear the signs and divide out the mass of the ball. To simplify life somewhat, we define a quantity such that :

Rearranging this, we find that

Earth’s density is about 5515 and rotates once per day, giving it an angular frequency of . Thus,

The equatorial radius is 6378.1 km; the calculated polar radius is therefore 6367.2 km. This gives us an equatorial bulge of r_{e}-r_{p} = 10.9 km. The polar radius is actually 6358.9 km, making the real equatorial bulge 19.2 km.

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