If planets are ellipsoids why don't we have 3 diameters?

Question

We know that each ellipsoid has 3 diameters named $2a$, $2b$, and $2c$.

The Earth and all planets, in general, are ellipsoids (Saturn is the best example because it's the most oblate planet in the solar system).

But all we have read and heard are two types of diameters: equatorial diameter, and polar diameter.

So where's the third one?

I mean, the equatorial diameter itself should differ. One is 12756 km in the case of Earth. But what about the equatorial diameter that is perpendicular to the other equatorial diameter?

Answer 1

It is possible for a rotating body in hydrostatic equilibrium to be a triaxial ellipsoid. This solution was found by Jacobi in the mid 1800s, thus it's known as the Jacobi ellipsoid. An example in the Solar System is Haumea.

The dwarf planet Haumea is believed to rotate in just under 4 hours. This rapid rotation causes the dwarf planet to be elongated in appearance.

However, this requires a high rotation speed. As Anders Sandberg explains here,

If a deformable self-gravitating initially spherical body rotates it will become an ellipsoid. For low rates of rotation this is an oblate spheroid with a circular cross-section, the Maclaurin case. As the rate of rotation becomes higher this state becomes unstable, and it turns into an elongated Jacobi ellipsoid. For even higher angular momentum these become unstable, and the object does break in two.

For further details on this solution, please see this page by Richard Fitzpatrick, courtesy of The University of Texas at Austin.

Answer 2

So where's the third one?

As other answers indicate the Earth's equator is so circular due to hydrostatic equilibrium. But if you wanted to imagine that it was slightly elliptical you could think of the deviation of Earth's gravitational potential above the surface as expressed in the $J_{22}$ coefficient, as a roughly one part per million effect on the gravitational field that satellites in low Earth orbit experience (mentioned here and elsewhere in Space SE).

You could also take a look at some theoretical exploration of a slightly triaxial Earth in The rotational stability of a triaxial ice-age Earth where the effect is very greatly exaggerated in the images below.

Figure 5. Predictions of the present-day ice-age-induced TPW generated using the “global” loading geometry and the sawtooth loading history. The predictions are based on a rotational stability theory valid for a triaxial Earth [equations (13) and (16)] with nonhydrostatic inertia differences given by equations (3) and (4). Moreover, the calculations adopt an Earth model with an elastic lithospheric thickness of 100 km, upper mantle viscosity of 1021 Pa s, and a lower mantle viscosity of Y × 1021 Pa s, where Y is specified by the label adjacent to the arrowhead of each line. Also shown, for the sake of comparison, is the direction of TPW associated with all predictions based on the biaxial rotation theory of Mitrovica et al. [2005] (dotted line).

Figure 3. Schematics illustrating the shape and principal axis orientation of the triaxial Earth in both a (a) 3-D and (b) top view. The figures reflect, using an exaggerated scale, the relative sizes of the principal moments of inertia (A < B < C). The x3 axis is coincident with the rotation vector, while x1 and x2 equatorial axes point toward −14.93°E and 75.07°E, respectively.

Answer 3

You can consider Reference ellipsoid which gives us a close approximation of the geoid (the imperfect figure of the Earth, or other planetary body). The shape of an ellipsoid is determined by three shape parameters:

1. The semi-major axis of the ellipse, a, becomes the equatorial radius of the ellipsoid
2. The semi-minor axis of the ellipse, b, becomes the distance from the centre to either pole.
3. flattening f (the amount of flattening at each pole, relative to the radius at the equator.)

$$\mathrm{f = \frac{a-b}{a}}$$

So, shape of almost all of the planetary bodies can be determine by just equatorial radius and polar radius and you necessarily don't need a third one.

Of course, you can use Geodetic coordinate system. Point to note that 1984 World Geodetic System (abbreviated as WGS84) introduced a third parameter: mean earth radius = $\mathrm{\frac{2a + b}{b}}$. Have a look.

Other SE posts:

1. Is there still a possibility that Makemake is not ellipsoidal but asteroid-shaped?
2. What is meant by the bifurcation of hydrostatic equilibrium shapes in planetary formation?
3. What should be the "poles" for irregular shaped bodies?
4. What celestial body (inside the solar system) has the highest flattening ratio?