<blockquote>
<p>So where's the third one?</p>
</blockquote>
<p>As other answers indicate the Earth's equator is so circular due to hydrostatic equilibrium. But if you wanted to <em>imagine</em> that it was slightly elliptical you could think of the deviation of Earth's gravitational potential above the surface as expressed in the <span class="math-container">$J_{22}$</span> coefficient, as a roughly one part per million effect on the gravitational field that satellites in low Earth orbit experience (mentioned <a href="https://space.stackexchange.com/a/24783/12102">here</a> and elsewhere in Space SE).</p>
<p>You could also take a look at some theoretical exploration of a <em>slightly</em> triaxial Earth in <a href="https://agupubs.onlinelibrary.wiley.com/doi/full/10.1029/2009JB006564" rel="noreferrer">The rotational stability of a triaxial ice-age Earth</a> where the effect is very greatly exaggerated in the images below.</p>
<p><a href="https://i.sstatic.net/d8Otm.png" rel="noreferrer"><img src="https://i.sstatic.net/d8Otm.png" alt="enter image description here" /></a></p>
<blockquote>
<p>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).</p>
</blockquote>
<p><a href="https://i.sstatic.net/i6rvf.png" rel="noreferrer"><img src="https://i.sstatic.net/i6rvf.png" alt="enter image description here" /></a></p>
<blockquote>
<p>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 &lt; B &lt; 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.</p>
</blockquote>

CC BY-SA 4.0
2022-03-19T22:00:16.743Z
2022-03-19T22:00:16.743Z
uhoh
<p>You can consider <a href="https://en.wikipedia.org/wiki/Reference_ellipsoid" rel="nofollow noreferrer">Reference ellipsoid</a> 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:</p>
<ol>
<li>The semi-major axis of the ellipse, a, becomes the equatorial radius of the ellipsoid</li>
<li>The semi-minor axis of the ellipse, b, becomes the distance from the centre to either pole.</li>
<li>flattening f (the amount of flattening at each pole, relative to the radius at the equator.)</li>
</ol>
<p><span class="math-container">$$\mathrm{f = \frac{a-b}{a}}$$</span></p>
<p>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.</p>
<p>Of course, you can use <a href="https://en.wikipedia.org/wiki/Geodetic_coordinates" rel="nofollow noreferrer">Geodetic coordinate system</a>. Point to note that 1984 <a href="https://en.wikipedia.org/wiki/World_Geodetic_System" rel="nofollow noreferrer">World Geodetic System</a> (abbreviated as WGS84) introduced a third parameter: mean earth radius = <span class="math-container">$\mathrm{\frac{2a + b}{b}}$</span>. <a href="https://i.sstatic.net/Qznvh.jpg" rel="nofollow noreferrer">Have a look</a>.</p>
<p>Other SE posts:</p>
<ol>
<li><a href="https://astronomy.stackexchange.com/questions/38280/is-there-still-a-possibility-that-makemake-is-not-ellipsoidal-but-asteroid-shape">Is there still a possibility that Makemake is not ellipsoidal but asteroid-shaped?</a></li>
<li><a href="https://astronomy.stackexchange.com/questions/19202/what-is-meant-by-the-bifurcation-of-hydrostatic-equilibrium-shapes-in-planetary">What is meant by the bifurcation of hydrostatic equilibrium shapes in planetary formation?</a></li>
<li><a href="https://astronomy.stackexchange.com/questions/40247/what-should-be-the-poles-for-irregular-shaped-bodies">What should be the &quot;poles&quot; for irregular shaped bodies?</a></li>
<li><a href="https://astronomy.stackexchange.com/questions/39872/what-celestial-body-inside-the-solar-system-has-the-highest-flattening-ratio">What celestial body (inside the solar system) has the highest flattening ratio?</a></li>
</ol>

CC BY-SA 4.0
2022-03-20T05:45:19.737Z
2022-03-20T05:52:49.433Z
2022-03-20T05:52:49.433Z
Nilay Ghosh
<p>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 <a href="https://en.wikipedia.org/wiki/Jacobi_ellipsoid" rel="noreferrer">Jacobi ellipsoid</a>. An example in the Solar System is Haumea.</p>
<p><a href="https://upload.wikimedia.org/wikipedia/commons/9/9c/Haumea_Rotation.gif" rel="noreferrer"><img src="https://upload.wikimedia.org/wikipedia/commons/9/9c/Haumea_Rotation.gif" alt="Haumea" /></a></p>
<blockquote>
<p>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.</p>
</blockquote>
<p>However, this requires a high rotation speed. As Anders Sandberg explains <a href="https://physics.stackexchange.com/a/460817/123208">here</a>,</p>
<blockquote>
<p>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.</p>
</blockquote>
<p>For further details on this solution, please see <a href="https://farside.ph.utexas.edu/teaching/336L/Fluidhtml/node36.html" rel="noreferrer">this page</a> by
Richard Fitzpatrick, courtesy of The University of Texas at Austin.</p>

CC BY-SA 4.0
2022-03-19T18:22:36.743Z
Then why aren't neutron stars considered Jacobi ellipsoids? They have a superfast rotation speed.
Snack Exchange
CC BY-SA 4.0
2022-03-19T18:28:51.447Z
@SnackExchange They may be! Especially when they're young. As I mentioned [here](https://astronomy.stackexchange.com/questions/20823/shape-of-neutron-stars#comment35129_20823) it's highly likely that some magnetars are prolate. I've just been searching for articles on Jacobi neutron stars. I got some hits, but they're purely theoretical, just looking at the mathematics of the rotation, and not based on analysis of the emitted signals of actual pulsars. Of course, the geometry of young neutron stars is much more complicated than planets because of the powerful magnetic field.
PM 2Ring
CC BY-SA 4.0
2022-03-19T21:47:52.587Z
`+1` So far unanswered in Math SE: [How does a Maclaurin spheroid become a Jacobi ellipsoid? What happens?](https://math.stackexchange.com/q/3514142/284619) and unanswered in History of Science and Math SE: [What exactly was Lagrange's "grave mistake" with respect to rotating bodies under hydrostatic equilibrium?](https://hsm.stackexchange.com/q/11351/3869)
uhoh
CC BY-SA 4.0
2022-03-20T05:51:50.720Z
More discussion, Makemake: https://astronomy.stackexchange.com/questions/38280/is-there-still-a-possibility-that-makemake-is-not-ellipsoidal-but-asteroid-shape ... Telesto: https://astronomy.stackexchange.com/questions/39872/what-celestial-body-inside-the-solar-system-has-the-highest-flattening-ratio
Nilay Ghosh
CC BY-SA 4.0
2022-03-20T11:18:35.543Z
@SnackExchange they may could, as noted by PM2Ring, but you should also consider that the material of neutron stars is *insanely* dense. In stable form, most neutron stars are almost perfectly oblate spheroids, with at most some *meters* of hills. It is predicted that these htiny imperfections would already emit detectable levels of gravitational waves as the star rotates. I'd assume that such waves would be much more apparent from a rotating ellipsioid, and the speed of rotation would need to be insane. *(p. s., I love your username.)*
Neinstein
CC BY-SA 4.0
2022-03-21T14:03:20.197Z
Mars' moons Phobos and Deimos are also possible candidates, though they are more irregularly shaped than Haumea.
Darrel Hoffman
CC BY-SA 4.0
2022-03-21T15:47:22.557Z
@Darrel Both Phobos & Deimos are much smaller than the ["potato" radius](https://astronomy.stackexchange.com/a/24723/16685), so we don't expect them to be rounded by gravity.  However, JPL Horizons says their rotation period is synchronised to their revolution period, which is quite short. The period of Phobos is ~ 7 h 39 m. https://ssd.jpl.nasa.gov/api/horizons.api?format=text&OBJ_DATA=YES&MAKE_EPHEM=NO&COMMAND=401
PM 2Ring
CC BY-SA 4.0
2022-03-21T20:37:06.897Z
@SnackExchange I think there are several possible new questions you could ask about triaxial neutron stars! Perhaps something about observational signatures (how could one be identified) or how long they are predicted to last before going biaxial, or how the tri-to-bi-axial transition is though to take place (again theoretically)...
uhoh
CC BY-SA 4.0
2022-03-19T18:04:17.030Z
2022-03-19T18:21:55.660Z
2022-03-19T18:21:55.660Z
PM 2Ring
<p>We know that each ellipsoid has 3 diameters named <span class="math-container">$2a$</span>, <span class="math-container">$2b$</span>, and <span class="math-container">$2c$</span>.</p>
<p>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).</p>
<p>But all we have read and heard are two types of diameters: equatorial diameter, and polar diameter.</p>
<p>So where's the third one?</p>
<p>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?</p>

CC BY-SA 4.0
2022-03-19T17:13:24.023Z
Wgs84 defines the Earth as an "oblate spheroid".  Essentially an ellipsoid with two axis whic are equal.  It also goes by other names like rotational ellipsoid.  So you might just be seeing some simplified terms in your sources.
Greg Miller
CC BY-SA 4.0
2022-03-20T11:06:49.393Z
Is it a matter of words? In your taxonomy, does the word "ellipsoid" exclude an _ellipsoid of revolution_, i.e. one where two axes have equal length? This is similar to terminology questions such as: Is a circle an ellipse? Is a square a rectangle? Is an equilateral triangle also isosceles? And so on.
Jeppe Stig Nielsen
CC BY-SA 4.0
2022-03-22T02:07:03.967Z
Many small bodies, notably Phobos and Deimos, are commonly listed with three diameters.
Anton Sherwood
CC BY-SA 4.0
2022-03-22T17:18:49.150Z
*"But what about the equatorial diameter that is perpendicular to the other equatorial diameter?"*  It's pretty much the same. Absent rotation, gravity forces [planet-sized bodies](https://astronomy.stackexchange.com/q/2092/28178) into effectively perfect spheres.  With rotation and a liquid core, [centrifugal forces push the diameter out at the waist](https://en.wikipedia.org/wiki/Clairaut%27s_theorem_(gravity)), but the planet remains rotationally symmetric.
J...
<planet><rotation>
If planets are ellipsoids why don't we have 3 diameters?
Snack Exchange
# If planets are ellipsoids why don't we have 3 diameters?

(Asked by: username_0 on 2022-03-19)

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?

**Tags:** planet, rotation
> **Comments:**
> **username_1**: Wgs84 defines the Earth as an "oblate spheroid".  Essentially an ellipsoid with two axis whic are equal.  It also goes by other names like rotational ellipsoid.  So you might just be seeing some simplified terms in your sources.  
> **username_2**: Is it a matter of words? In your taxonomy, does the word "ellipsoid" exclude an _ellipsoid of revolution_, i.e. one where two axes have equal length? This is similar to terminology questions such as: Is a circle an ellipse? Is a square a rectangle? Is an equilateral triangle also isosceles? And so on.  
> **username_3**: Many small bodies, notably Phobos and Deimos, are commonly listed with three diameters.  
> **username_4**: *"But what about the equatorial diameter that is perpendicular to the other equatorial diameter?"*  It's pretty much the same. Absent rotation, gravity forces [planet-sized bodies](https://astronomy.stackexchange.com/q/2092/28178) into effectively perfect spheres.  With rotation and a liquid core, [centrifugal forces push the diameter out at the waist](https://en.wikipedia.org/wiki/Clairaut%27s_theorem_(gravity)), but the planet remains rotationally symmetric.  

---

## 3 answers

### Answer by username_7 on 2022-03-19. Score: 38 | Accepted answer

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](https://en.wikipedia.org/wiki/Jacobi_ellipsoid). An example in the Solar System is Haumea.

[![Haumea](https://upload.wikimedia.org/wikipedia/commons/9/9c/Haumea_Rotation.gif)](https://upload.wikimedia.org/wikipedia/commons/9/9c/Haumea_Rotation.gif)

> 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](https://physics.stackexchange.com/a/460817/123208),

> 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](https://farside.ph.utexas.edu/teaching/336L/Fluidhtml/node36.html) by
Richard Fitzpatrick, courtesy of The University of Texas at Austin.

> **Comments:**
> **username_0**: Then why aren't neutron stars considered Jacobi ellipsoids? They have a superfast rotation speed.  
> **username_7**: @SnackExchange They may be! Especially when they're young. As I mentioned [here](https://astronomy.stackexchange.com/questions/20823/shape-of-neutron-stars#comment35129_20823) it's highly likely that some magnetars are prolate. I've just been searching for articles on Jacobi neutron stars. I got some hits, but they're purely theoretical, just looking at the mathematics of the rotation, and not based on analysis of the emitted signals of actual pulsars. Of course, the geometry of young neutron stars is much more complicated than planets because of the powerful magnetic field.  
> **username_5**: `+1` So far unanswered in Math SE: [How does a Maclaurin spheroid become a Jacobi ellipsoid? What happens?](https://math.stackexchange.com/q/3514142/284619) and unanswered in History of Science and Math SE: [What exactly was Lagrange's "grave mistake" with respect to rotating bodies under hydrostatic equilibrium?](https://hsm.stackexchange.com/q/11351/3869)  
> **username_6**: More discussion, Makemake: https://astronomy.stackexchange.com/questions/38280/is-there-still-a-possibility-that-makemake-is-not-ellipsoidal-but-asteroid-shape ... Telesto: https://astronomy.stackexchange.com/questions/39872/what-celestial-body-inside-the-solar-system-has-the-highest-flattening-ratio  
> **username_8**: @SnackExchange they may could, as noted by PM2Ring, but you should also consider that the material of neutron stars is *insanely* dense. In stable form, most neutron stars are almost perfectly oblate spheroids, with at most some *meters* of hills. It is predicted that these htiny imperfections would already emit detectable levels of gravitational waves as the star rotates. I'd assume that such waves would be much more apparent from a rotating ellipsioid, and the speed of rotation would need to be insane. *(p. s., I love your username.)*  
> **username_9**: Mars' moons Phobos and Deimos are also possible candidates, though they are more irregularly shaped than Haumea.  
> **username_7**: @Darrel Both Phobos & Deimos are much smaller than the ["potato" radius](https://astronomy.stackexchange.com/a/24723/16685), so we don't expect them to be rounded by gravity.  However, JPL Horizons says their rotation period is synchronised to their revolution period, which is quite short. The period of Phobos is ~ 7 h 39 m. https://ssd.jpl.nasa.gov/api/horizons.api?format=text&OBJ_DATA=YES&MAKE_EPHEM=NO&COMMAND=401  
> **username_5**: @SnackExchange I think there are several possible new questions you could ask about triaxial neutron stars! Perhaps something about observational signatures (how could one be identified) or how long they are predicted to last before going biaxial, or how the tri-to-bi-axial transition is though to take place (again theoretically)...  

---

### Answer by username_5 on 2022-03-19. Score: 11

> 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](https://space.stackexchange.com/a/24783/12102) 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](https://agupubs.onlinelibrary.wiley.com/doi/full/10.1029/2009JB006564) where the effect is very greatly exaggerated in the images below.

[![enter image description here](https://i.sstatic.net/d8Otm.png)](https://i.sstatic.net/d8Otm.png)

> 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).

[![enter image description here](https://i.sstatic.net/i6rvf.png)](https://i.sstatic.net/i6rvf.png)

> 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 by username_6 on 2022-03-20. Score: 3

You can consider [Reference ellipsoid](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/Geodetic_coordinates). Point to note that 1984 [World Geodetic System](https://en.wikipedia.org/wiki/World_Geodetic_System) (abbreviated as WGS84) introduced a third parameter: mean earth radius = $\mathrm{\frac{2a + b}{b}}$. [Have a look](https://i.sstatic.net/Qznvh.jpg).

Other SE posts:

1. [Is there still a possibility that Makemake is not ellipsoidal but asteroid-shaped?](https://astronomy.stackexchange.com/questions/38280/is-there-still-a-possibility-that-makemake-is-not-ellipsoidal-but-asteroid-shape)
2. [What is meant by the bifurcation of hydrostatic equilibrium shapes in planetary formation?](https://astronomy.stackexchange.com/questions/19202/what-is-meant-by-the-bifurcation-of-hydrostatic-equilibrium-shapes-in-planetary)
3. [What should be the "poles" for irregular shaped bodies?](https://astronomy.stackexchange.com/questions/40247/what-should-be-the-poles-for-irregular-shaped-bodies)
4. [What celestial body (inside the solar system) has the highest flattening ratio?](https://astronomy.stackexchange.com/questions/39872/what-celestial-body-inside-the-solar-system-has-the-highest-flattening-ratio)


---

