<h3>If hydrostatic equilibrium IS required</h3>
<p>As far as I know, the only effect that can make solar system objects <em>that are in hydrostatic equilibrium</em> deviate from a sphere is rotation. So <a href="https://astronomy.stackexchange.com/questions/39872/what-celestial-body-inside-the-solar-system-has-the-highest-flattening-ratio#comment124176_39872">as JamesK points out</a>, Telesto would probably not count if hydrostatic equilibrium were required. I say &quot;probably&quot; because right now I can't proove that it's not in hydrostatic equillibrium - I can't find a rotational period for Telesto. But if you melted Telesto so that hydrodynamics could happen, I think it would become much more spherical.</p>
<p>In this case, Saturn wins.</p>
<h3>If hydrostatic equilibrium IS NOT required</h3>
<p>On the other hand, if hydrostatic equilibrium is not required then probably <a href="https://en.wikipedia.org/wiki/%CA%BBOumuamua" rel="nofollow noreferrer">ʻOumuamua</a> is the winner with a longest to shortest ratio somewhere in the ballpark of 6 to 8 according to that article. There are no direct measurements of its dimensions, but careful analyses of its reflected sunlight curve as it tumbles has been used to estimate its 3D shape.</p>
<p>That huge ratio and narrowest dimension of tens of meters to of order 100 meters evoked the &quot;ancient &amp; dead(?) rocket ship&quot; meme.</p>
<ul>
<li><a href="https://astronomy.stackexchange.com/q/33932/7982">If we&#39;ve only observed Oumuamua as a single pixel, how can we know its shape?</a></li>
<li><a href="https://astronomy.stackexchange.com/q/23620/7982">Could asteroid &#39;Oumuamua actually be round?</a></li>
<li><a href="https://astronomy.stackexchange.com/q/23644/7982">Could we detect if Oumuamua asteroid wasn&#39;t an asteroid?</a></li>
</ul>
<p>One caveat:</p>
<blockquote>
<p>What celestial body <em>(inside the solar system)</em></p>
</blockquote>
<p>ʻOumuamua is currently inside the solar system, but it is slightly gravitationally unbound, so it will not be coming around the inner solar system again. Instead it will leave the solar system in the far future and continue on to generate sensational headlines for other species to get excited about.</p>

CC BY-SA 4.0
2023-10-09T22:31:38.363Z
2023-10-10T01:55:16.840Z
2023-10-10T01:55:16.840Z
uhoh
<p>I believe besides Telesto the dwarf planet Haumea in the Kuiper belt could also be in the mix for the highest flattening of objects in our solar system. Haumea is a triaxial ellipsoid where the 'f' value would around 1/2 between the largest and smallest major axis.</p>

CC BY-SA 4.0
2020-11-16T12:46:43.517Z
2020-11-16T13:47:13.457Z
2020-11-16T13:47:13.457Z
Benny
<p>As a planet, Saturn has the highest <a href="https://en.wikipedia.org/wiki/Flattening" rel="nofollow noreferrer">flattening(ellipticity)</a> <a href="https://www.smartconversion.com/otherInfo/Ellipticity_Flattening_of_planets_and_the_sun.aspx" rel="nofollow noreferrer">which is 0.09796</a>. So, it makes Saturn with the largest equatorial bulge as a planet and as such Saturn is the <em>flattest planet</em>.</p>
<p>However, I have encountered this statement <a href="https://en.wikipedia.org/wiki/Reference_ellipsoid" rel="nofollow noreferrer">from here</a>:</p>
<blockquote>
<p>Earth's Moon is even less elliptical, with a flattening of less than
1/825, while Jupiter is visibly oblate at about 1/15 and <strong>one of
Saturn's triaxial moons, Telesto, is highly flattened, with f between
1/3 to 1/2 (meaning that the polar diameter is between 50% and 67% of
the equatorial).</strong></p>
</blockquote>
<p>So, Telesto has a flattening value even higher than Saturn itself. I couldn't find any other body that has higher flattening value than Telesto. Does that mean Telesto has the highest recorded 'f' value?</p>

CC BY-SA 4.0
2020-11-17T03:35:58.197Z
Does stating a "flattening ratio" require that the flattening be due to rotation? Can crazy-shaped objects like [ʻOumuamua](https://en.wikipedia.org/wiki/%CA%BBOumuamua) or ravioli-like moons be considered as well? (see [this answer](https://astronomy.stackexchange.com/a/34890/7982) and [this answer](https://astronomy.stackexchange.com/a/36444/7982))
uhoh
CC BY-SA 4.0
2020-11-17T05:20:24.410Z
@uhoh It is hard to measure f value for 'crazy shaped objects' as they can have irregular dimensions as such it is suitable for objects having the ellipsoid/spheroid shape(similar to rugby ball) or any other shape as long as it maintains _hydrostatic equilibrium_. The flattening can be due to any reasons. It can be due to rotation, gravitational perturbation, orbital eccentricity, tidal locking etc. I want to know the highest known value which has been recorded by astronomers and not just a value from 'back-of-the-envelope' calculation.
Nilay Ghosh
CC BY-SA 4.0
2023-10-07T08:13:46.763Z
Telesto is not in hydrostatic equilibrium, so it should be excluded by that criterion.
James K
<planet><natural-satellites>
What celestial body (inside the solar system) has the highest flattening ratio?
Nilay Ghosh
# What celestial body (inside the solar system) has the highest flattening ratio?

(Asked by: username_0 on 2020-11-16)

As a planet, Saturn has the highest [flattening(ellipticity)](https://en.wikipedia.org/wiki/Flattening) [which is 0.09796](https://www.smartconversion.com/otherInfo/Ellipticity_Flattening_of_planets_and_the_sun.aspx). So, it makes Saturn with the largest equatorial bulge as a planet and as such Saturn is the *flattest planet*.

However, I have encountered this statement [from here](https://en.wikipedia.org/wiki/Reference_ellipsoid):

> Earth's Moon is even less elliptical, with a flattening of less than
> 1/825, while Jupiter is visibly oblate at about 1/15 and **one of
> Saturn's triaxial moons, Telesto, is highly flattened, with f between
> 1/3 to 1/2 (meaning that the polar diameter is between 50% and 67% of
> the equatorial).**

So, Telesto has a flattening value even higher than Saturn itself. I couldn't find any other body that has higher flattening value than Telesto. Does that mean Telesto has the highest recorded 'f' value?

**Tags:** planet, natural-satellites

> **Comments:**
> **username_1**: Does stating a "flattening ratio" require that the flattening be due to rotation? Can crazy-shaped objects like [ʻOumuamua](https://en.wikipedia.org/wiki/%CA%BBOumuamua) or ravioli-like moons be considered as well? (see [this answer](https://astronomy.stackexchange.com/a/34890/7982) and [this answer](https://astronomy.stackexchange.com/a/36444/7982))  
> **username_0**: @username_1 It is hard to measure f value for 'crazy shaped objects' as they can have irregular dimensions as such it is suitable for objects having the ellipsoid/spheroid shape(similar to rugby ball) or any other shape as long as it maintains _hydrostatic equilibrium_. The flattening can be due to any reasons. It can be due to rotation, gravitational perturbation, orbital eccentricity, tidal locking etc. I want to know the highest known value which has been recorded by astronomers and not just a value from 'back-of-the-envelope' calculation.  
> **username_2**: Telesto is not in hydrostatic equilibrium, so it should be excluded by that criterion.  

---

## 2 answers

### Answer by username_1 on 2023-10-09. Score: 3 | Accepted answer

### If hydrostatic equilibrium IS required

As far as I know, the only effect that can make solar system objects *that are in hydrostatic equilibrium* deviate from a sphere is rotation. So [as username_2 points out](https://astronomy.stackexchange.com/questions/39872/what-celestial-body-inside-the-solar-system-has-the-highest-flattening-ratio#comment124176_39872), Telesto would probably not count if hydrostatic equilibrium were required. I say "probably" because right now I can't proove that it's not in hydrostatic equillibrium - I can't find a rotational period for Telesto. But if you melted Telesto so that hydrodynamics could happen, I think it would become much more spherical.

In this case, Saturn wins.

### If hydrostatic equilibrium IS NOT required

On the other hand, if hydrostatic equilibrium is not required then probably [ʻOumuamua](https://en.wikipedia.org/wiki/%CA%BBOumuamua) is the winner with a longest to shortest ratio somewhere in the ballpark of 6 to 8 according to that article. There are no direct measurements of its dimensions, but careful analyses of its reflected sunlight curve as it tumbles has been used to estimate its 3D shape.

That huge ratio and narrowest dimension of tens of meters to of order 100 meters evoked the "ancient & dead(?) rocket ship" meme.

* [If we've only observed Oumuamua as a single pixel, how can we know its shape?](https://astronomy.stackexchange.com/q/33932/7982)
* [Could asteroid 'Oumuamua actually be round?](https://astronomy.stackexchange.com/q/23620/7982)
* [Could we detect if Oumuamua asteroid wasn't an asteroid?](https://astronomy.stackexchange.com/q/23644/7982)

One caveat:

> What celestial body *(inside the solar system)*

ʻOumuamua is currently inside the solar system, but it is slightly gravitationally unbound, so it will not be coming around the inner solar system again. Instead it will leave the solar system in the far future and continue on to generate sensational headlines for other species to get excited about.


---

### Answer by username_3 on 2020-11-16. Score: 1

I believe besides Telesto the dwarf planet Haumea in the Kuiper belt could also be in the mix for the highest flattening of objects in our solar system. Haumea is a triaxial ellipsoid where the 'f' value would around 1/2 between the largest and smallest major axis.


---

