4
$\begingroup$

As a planet, Saturn has the highest flattening(ellipticity) which is 0.09796. 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:

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?

Improve this question
$\endgroup$
3
  • $\begingroup$ Does stating a "flattening ratio" require that the flattening be due to rotation? Can crazy-shaped objects like ʻOumuamua or ravioli-like moons be considered as well? (see this answer and this answer) $\endgroup$
    – uhoh
    Nov 17, 2020 at 3:35
  • 1
    $\begingroup$ @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. $\endgroup$
    – Nilay Ghosh
    Nov 17, 2020 at 5:20
  • 1
    $\begingroup$ Telesto is not in hydrostatic equilibrium, so it should be excluded by that criterion. $\endgroup$
    – James K
    Oct 7 at 8:13

2 Answers 2

Reset to default
2
$\begingroup$

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 JamesK points out, 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 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.

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.

Improve this answer
$\endgroup$
1
$\begingroup$

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.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .