Relatively recent measurements indicate that the Sun is nearly the roundest object ever measured. If scaled to the size of a beach ball, it would be so round that the difference between the widest and narrow diameters would be much less than the width of a human hair.

I do appreciate that above result is just one measurement , and I looked for confirmation of the result. However, Wikipedia accepts its validity:

By this measure, the Sun is a near-perfect sphere with an oblateness estimated at about 9 millionths, which means that its polar diameter differs from its equatorial diameter by only 10 kilometres (6.2 mi).

enter image description here

Two questions on this subject:

  • To me at least, it is a completely counter-intuitive result. Can anyone explain from what causes this symmetry emerged? Is it a combination of a slow rotation rate combined with a highly isotropic central gravitational field? I  thought there would be an equatorial bulge, even though the rotation rate is slow.

  • Does this result, for just one ordinary, as far as I know, star indicate that asymmetrical stellar collapses are much less likely than may have been previously envisaged? Admitted, it is just one star out of countless billions, but on the other hand, as it is a random sample, it may well be indicative of many more similar "extremely" (if I can use that word) spherical objects.

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    $\begingroup$ This was such a surprise to me, I thought you must have made a mistake! But no, your figures check out. Thank you for enlightening me. $\endgroup$
    – TonyK
    Sep 21, 2015 at 23:17
  • $\begingroup$ @TonyK you are welcome, the link at the top of the post gives more details, and Rob gives the answer. I would have lost a bet if anybody asked me "what is the roundest thing in the solar system?" Thanks and regards $\endgroup$
    – user81619
    Sep 21, 2015 at 23:24
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    $\begingroup$ That image has the strangest optical illusion. It seems like the outer gaseous part shrinks in. $\endgroup$
    – camden_kid
    Sep 22, 2015 at 15:44
  • $\begingroup$ Thank for the questions and the remarks. I only object, as a devil's advocate, to the Sun being a “random sample”. There just might be something in its symmetry that was a slight tiny bit more conducive to life in one of its planets, and so to the presence of us observing it. $\endgroup$
    – DaG
    Sep 23, 2015 at 8:05
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    $\begingroup$ To me at least, it is a completely counter-intuitive result. I apologize in advance for my apparently quite different intuition. But for me, I see nothing counter intuitive whatsoever in the sun being very, very round. Just like a drop of water in free space would be very very round. This question doesn't say why this is weird and counterintuitive. How come this apparently is such a surprise? $\endgroup$
    – Steeven
    Sep 23, 2015 at 20:15

3 Answers 3


The symmetry of the Sun has got very little to do with any symmetry in its formation.

The Sun has had plenty of time to reach an equilibrium between its self gravity and its internal pressure gradient. Any departure from symmetry would imply a difference in pressure in regions at a similar radius but different polar or azimuthal angles. The resultant pressure gradient would trigger fluid flows that would erase the asymmetry.

Possible sources of asymmetry in stars could include rapid rotation or the presence of a binary companion, both of which break the symmetry of the effective gravitational potential, even if the star were spherically symmetric. The Sun has neither of these (the centrifugal acceleration at the equator is only about 20 millionths of the surface gravity, and Jupiter is too small and far away to have an effect) and simply relaxes to an almost spherically symmetric configuration.

The relationship between oblateness/ellipticity and rotation rate is treated in some detail here for a uniform density, self-gravitating spheroid and the following analytic approximation is obtained for the ratio of equatorial to polar radius $$ \frac{r_e}{r_p} = \frac{1 + \epsilon/3}{1-2\epsilon/3}, $$ where $\epsilon$, the ellipticity is related to rotation and mass as $$\epsilon = \frac{5}{4}\frac{\Omega^2 a^3}{GM}$$ and $a$ is the mean radius, $\Omega$ the angular velocity.

Putting in numbers for the Sun (using the equatorial rotation period), I get $\epsilon=2.8\times10^{-5}$ and hence $r_e/r_p =1.000028$ or $r_e-r_p = \epsilon a = 19.5$ km. Thus this simple calculation gives the observed value to a small factor, but is obviously only an approximation because (a) the Sun does not have a uniform density and (b) rotates differentially with latitude in its outer envelope.

A final thought. The oblateness of a single star like the Sun depends on its rotation. You might ask, how typical is the (small) rotation rate of the Sun that leads to a very small oblateness? More rapidly rotating sun-like (and especially more massive) stars do exist; very young stars can rotate up to about 100 times faster than the Sun, leading to significant oblateness. However, Sun-like stars spin-down through a magnetised wind as they get older. The spin-down rate depends on the rotation rate and this means that single (or at least stars that are not in close, tidally locked binary systems) stars converge to a close-to-unique rotation-age relationship at ages beyond a billion years. Thus we expect (it remains to be proven, since stellar ages are hard to estimate) that all Sun-like stars with a similar age to the Sun should have similar rotation rates and similarly small oblateness.

  • $\begingroup$ Thanks for that, I just did not realise before reading an article how really spherical it was. $\endgroup$
    – user81619
    Sep 21, 2015 at 22:09
  • $\begingroup$ @count_to_10 Nor I, but I guess 2 km/s at the equator is an acceleration only 20 millionths that of the surface gravity. $\endgroup$
    – ProfRob
    Sep 21, 2015 at 22:17
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    $\begingroup$ Bit of a tangent but interesting; does this answer imply that the asymmetry of other spherical bodies e.g. the Earth, is caused by its speed of rotation and the presence of the moon? Makes sense I guess... $\endgroup$ Sep 22, 2015 at 10:59
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    $\begingroup$ @mike-source It's about the ratio between the gravitational force and the centrifugal force, basically. The Sun rotates fast, but not quite fast enough to offset its massive gravity. The Earth is much less massive, and as such, the rotational distortion is much more pronounced. Also, Earth has solids - this is much less important than the rotation on average, but it does give some interesting local maxima and minima. The Moon's gravity is much too low to matter - the acceleration is tiny; you might have been misled by the usual (and wrong) explanation of how tides work. $\endgroup$
    – Luaan
    Sep 22, 2015 at 11:35
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    $\begingroup$ @yo' Some timescales - the thermal timescale of the Sun $\sim 10^{7}$ years; the thermal timescale of the convective envelope of the Sun $\sim 10^{5}$ years; thermal timescale of convective cells near the photosphere $\sim 1$ day; diffusive timescale in the envelope $\sim 1$ year. I would expect non-sphericity to be eliminated on some mixture of these timescales, all of which are much smaller than its age. $\endgroup$
    – ProfRob
    Sep 22, 2015 at 15:30

I) In this answer we will only discuss the equilibrium shape. Recall that when we discussed the shape of Earth in this Phys.SE post, the gravitational quadrupole moment was important. Unlike the Earth, from a surface perspective, it is a very good approximation to assume that all the mass of the Sun sits in the center, cf. below graph.

Moreover, the Newton's shell theorem helps out here. We conclude that it is enough to consider the gravitational monopole field

$$\tag{1} g(r)~=~\frac{GM}{r^2}$$

of the Sun. From Wikipedia, we get that

$$\tag{2} G~=~ 6.674\cdot 10^{−11} {\rm Nm}^2/{\rm kg}^2\quad\text{and}\quad M~=~(1.98855 \pm 0.00025)\cdot 10^{30} {\rm kg}. $$ The equatorial radius and period are $$\tag{3} r_e~=~(696342\pm 65)~{\rm km} \quad\text{and}\quad T_e~=~25.05 ~{\rm days}, $$

respectively. The equatorial speed is

$$\tag{4} v_e~=~\omega_e r_e~=~\frac{2\pi r_e}{T_e}~\approx~2.02 ~{\rm km/s}.$$

The equatorial surface gravity is then

$$\tag{5} g_e~=~\frac{GM}{r_e^2}~\approx~274~{\rm m/s^2}. $$

Repeating Mark Eichenlaub's monopole argument for the Sun, the height difference between the equatorial and polar radius becomes

$$\tag{6} h~:=~r_e-r_p~=~\frac{v_e^2}{2g_e}~\approx~7.5 ~{\rm km}, $$

leading to a flattening

$$\tag{7} f~=~\frac{h}{r_e}~\approx~ 11 \cdot 10^{-6} .$$

This estimate overshoots by 20% the actual observed flattening, which is only $9 \cdot 10^{-6}$.

II) In the remainder of this answer, we would like to argue that the 20% difference in eq. (7) is mainly due to the fact that the Sun does not spin as a rigid body, which we implicitly assumed in Section I. The polar period

$$\tag{8} T_p~=~34.4 ~{\rm days} $$

is slower than the equatorial period (3). To proceed, let us for simplicity assume that the square $T^2$ of the period $T$ depends on the polar angle $\theta$ in the following way$^1$

$$\tag{9} T^2~=~T_p^2 + s (T_e^2-T_p^2) , \qquad s~\equiv~\sin^2\theta,\qquad \omega~\equiv~ \frac{2\pi}{T}. $$

Analogously, define for later convenience the quantity

$$\tag{10} A~:=~\frac{GM}{\omega^2}~=~ A_p + s A^{\prime}, \qquad A^{\prime}~:=~A_e-A_p~<~0, $$

which is proportional to $T^2$. The centrifugal acceleration is $$\tag{11} a_{\rm cf}~=~\omega^2 r\sin\theta.$$

Using arguments similar to my Phys.SE answer here, the total force should be perpendicular to the surface

$$\tag{12} \left(g -a_{\rm cf} \sin \theta \right)\mathrm{d}r -a_{\rm cf} \cos \theta ~r\mathrm{d}\theta~=~0.$$

The differential (12) is inexact. After multiplying with an integrating factor, we have

$$ \tag{13} \mathrm{d}U~=~\lambda(u) \left[\left(\frac{A}{r^2}-sr\right)\mathrm{d}r -\frac{r^2}{2}\mathrm{d}s\right], $$


$$ \tag{14} \lambda(u)~:=~\exp\left(\frac{2}{3}A^{\prime} u^3 \right) , \qquad u~\equiv~\frac{1}{r}. $$

The potential becomes

$$ \tag{15} U~=~ -A_p \int_0^u \! du^{\prime} ~ \lambda(u^{\prime}) - s\frac{\lambda(u)}{2u^2} . $$

The difference between equatorial and polar potential should be zero:

$$\tag{16} 0~=~U_e-U_p~=~ A_p \int_{u_e}^{u_p} \! du~\lambda(u) -\frac{\lambda(u_e)}{2u_e^2} , $$

or equivalently,

$$ \frac{1}{2 A_p u_e^2} ~\stackrel{(16)}{=}~ \int_{u_e}^{u_p} \! du~ \exp\left(\frac{2}{3}A^{\prime} (u^3-u_e^3) \right)~$$ $$ \tag{17}\approx~ \int_{u_e}^{u_p} \! du~ e^{2 A^{\prime} (u-u_e) u_e^2} ~=~\frac{e^{2 A^{\prime} (u_p-u_e) u_e^2}-1}{2A^{\prime} u_e^2}.$$

The height difference becomes

$$\tag{18} h~:=~r_e-r_p~\approx~\frac{u_p-u_e}{u_e^2}~\stackrel{(17)}{\approx}~\frac{r_e^4}{2A^{\prime}} \ln \left(1 + \frac{A^{\prime}}{A_p}\right)~\approx~5.3~{\rm km},$$

leading to a flattening

$$\tag{19} f~=~\frac{h}{r_e}~\approx~ 8 \cdot 10^{-6} ,$$

which is 10% below the observed flattening. Anyway, the above simple model demonstrates that it is important to take into account the non-rigid differential rotation of the Sun.


$^1$ Besides fulfilling the correct boundary conditions, the ansatz (9) is admittedly chosen to make the integrating factor (14) simple (rather than being based on observations or astrophysical models).


I think that distance lends a false image to appear as a sphere... At a closer look, you would actually see the ups and downs, rugged exterior... Maybe, the outer gaseeous cover, which is light, turns even on the entire surface of the Sun due to strong gravity giving it the round shape...


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