# What is the most oblate astronomical object known?

There are some very vast spinning objects out there. How high can the eccentricity get, or, which I think is easier to understand, how small can the ratio between the polar and equatorial diameter be?

For clarification, I'm asking about objects composed composed of particles (not some boundary or field etc) and that are devoid of non-microscopic empty spaces. I.e. a planet or star etc; not a galaxy or rings or a heliosphere etc. For the purpose of this question, I just don't find galaxies or nebulae interesting because they can have almost any shape.

I'm talking about structures that are oblate spheroids essentially.

• What's an object? Would Saturn's rings qualify? -- they are pretty flat. Otherwise, what about the Galactic disc? or other astrophysical discs (accretion discs have aspect ratios 1:1000)? None of these are fully self-gravitating: there is always a round component contributing or even dominating gravity... – Walter Dec 9 '14 at 17:35
• @Walter: I understand the point, but I'm not looking for trick answers. To me an object is one composed of particles (not some boundary or field etc) that is devoid of non-microscopic empty spaces. I.e. a planet or star etc; not a galaxy or rings or a heliosphere etc. Perhaps someone has an appropriate term or expression. – ThePopMachine Dec 9 '14 at 20:50
• Well, that's not what you've asked (originally). Perhaps you should qualify your answer then. IHMO, my suggestions are not trick answers, but honest and correct answers to your original question. – Walter Dec 11 '14 at 8:51
• I think your question is ill-posed, as it uses the concepts of polar and equatorial diameter. However, for an irregular object these are not well-defined. – Walter Dec 11 '14 at 9:08
• @Walter: Please get off it. People know what the question means. You are mincing words to keep your point. – ThePopMachine Dec 12 '14 at 19:04

Saturn is the most oblate planet in the solar system. If the equatorial diameter is $$a$$ and the polar diameter is $$b$$ then its oblateness, $$(a-b)/a = 0.1$$.

We do not know the oblateness values for more than one or two exoplanets and even these are somewhat uncertain, but are thought to be lower than Saturn's value. For example http://adsabs.harvard.edu/abs/2010ApJ...709.1219C

The oblateness of a star depends on the ratio of it gravitational acceleration to centrifugal acceleration at the equator. So the most oblate stars ought to be the ones with big radii and fast rotation rates (i.e. types of stars that would have low surface gravities if non-rotating).

Giant stars come in two basic types - red giants, which are evolved intermediate and low-mass stars. These tend to be slowly rotating, though there is a class called FK Com variables which are fast-rotating red giants.

The other class is the blue giants, which are high mass main sequence, or slightly evolved, stars.

For maximum oblateness you want the star with the highest value of: $$f = \frac{R \omega^2}{GM/R^2} = \frac{R^3 \omega^2}{GM}$$ where $$M$$ is the stellar mass and $$\omega$$ its angular velocity. Often we don;t know $$\omega$$, but can estimate the rotation velocity at the equator $$v=R\omega$$. Hence $$f = \frac{R^3 v^2}{GMR^2} = \frac{Rv^2}{GM}$$

A possible record holder is the O-type star VTFS 102 in the Large Magellanic Cloud. This probably has a mass of $$25M_{\odot}$$ a radius of about $$10R_{\odot}$$ and has a measured rotation speed of $$\geq 600$$ km/s. Thus its value of $$f$$ is 0.75, compared with a value of $$f=0.15$$ for Saturn.

EDIT: Due to Michael B - what about "millisecond" pulsars? The shortest periods are about 1.4 ms, the radii are about 10 km and the masses are likely to be about 1.4$$M_{\odot}$$. Do the sums and you find $$f=0.1$$. So, (at least in Newtonian mechanics) they should be no more oblate than Saturn.

A further contender might by the dwarf planet Haumea that orbits at 40-50 au from the Sun. It is thought large enough to have achieved its shape via hydrostatic equilibrium between gravity and internal pressures (and rotation). Its mass is $$4\times10^{21}$$ kg, (mean) radius is 700 km and it has a rotation period of 3.91 hours. These numbers give $$f=0.25$$. It has been modelled using this, and matching its light curve, to be a triaxial ellipsoid (rather than oblate) with axis ratios 2 : 1.5 : 1 (http://adsabs.harvard.edu/abs/2006ApJ...639.1238R).

• Why are you using (a-b)/a instead of just b/a? Is this a conventional definition? b/a makes more sense to me. – ThePopMachine Dec 9 '14 at 1:49
• I started writing my own answer but realized I wasn't adding much more than this one. In general, Be stars are thought to have circumstellar disks from marginally bound material being lost from the surface. Achernar is also rotating very rapidly: I get f=0.44. – Warrick Dec 9 '14 at 6:57
• @ThePopMachine It appears to be the conventional definition of oblateness, such that a sphere has zero oblateness. – ProfRob Dec 9 '14 at 7:33
• Great answer, but what about millisecond pulsars? They are tiny, sure, but with spin frequencies of up to ~700 Hz they can be appreciably oblate. – Michael B. Dec 10 '14 at 19:24
• @MichaelB. See my edit; they are not even close to winning. – ProfRob Dec 10 '14 at 22:14

I think it is going to depend on the type of object being examined. Planetary objects tend to be more spherical than stars, which in turn tend to be more spherical than galaxies.

My vote goes toward the Heliospheric current sheet, which is thought to extend 10-20 astronomical units (about 1.5x10^9km to 3x10^10km) from the Sun, and is thought to be about 10,000 km near the orbit of the Earth. But it tends to be influenced more by electromagnetic forces, than due to spin forces.

• By object, I sort of meant planet, star, or black hole if it makes sense in this context. – ThePopMachine Dec 8 '14 at 17:48
• @ThePopMachine: Then you mean a baryonic or rigid object, as in astronomy many structures (think planetary nebulae) are considered objects. Though as an Earth dweller I understand why you are interested in rigid objects, but as you learn more about astronomy you will find that the difference is negligible. For instance, even rocky asteroids are fluid on the scales of their diameters, that is why e.g. Vesta is differentiated. The rock 'flows' at astronomical scales. – dotancohen Dec 11 '14 at 6:21
• "Planetary objects tend to be more spherical than stars," not even true in our own solar system. – ProfRob Dec 11 '14 at 12:35

The dwarf planet Haumea two equatorial diameters, it is triaxial. The longest equatorial diameter is about twice the length of the polar diameter.

According to Google and Wikipedia, the most oblate star is Achernar which has an equatorial diameter 56% larger than the polar diameter due to it rapid spin.

The discussion at http://cosmoquest.org/forum/showthread.php?80772-Oblate-Neutron-Stars says that if the oblateness is extreme enough then an oblate spheroid isn't stable and it turns into a triaxial ellipsoid and that none of the known neutron stars are oblate as that.

If you're only interested in objects devoid of non-microscopic empty spaces, then we're pretty much into the realm of stars, planets, and asteroids.

Objects of this type larger than $\sim100$km are all nearly spherical, simply because of the effect of their own gravity (the spherical shape minimises the gravitational energy). However, small asteroids can have all sort of sometimes quite irregular shapes. Of course, your concepts of polar and equatorial diameter then become somewhat ill-defined, but one can still define the three principal axes (for example of the moment of intertia) and consider their ratios.

Eros, for example, has extents 34.4 x 11.2 x 11.2 km. It is thus prolate, but the semi-minor to -major axis ratio is less than 1/3: pretty small, though not flat.