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That is, it's twice the radius where the radius is from the centre of the sun to some edge. But what is that edge?

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  • $\begingroup$ In addition to the other great answers here, you can use total solar eclipses and Baily's Beads to measure solar diameter: poyntsource.com/Richard/Solareclipse.htm $\endgroup$
    – user21
    Jan 4, 2016 at 21:04

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Fusion reactions taking place inside the core of the star produce a huge amount of energy, most of which becomes heat. These reactions are not evenly distributed through the star and so there are phenomena such as sun spots and solar flares, however the total amount of energy produced tends to be reasonably constant.

I would say that the edge is defined by the average point where the gravity reaches equilibrium with the pressure of the star's super-heated gases (as a result of internal fusion).

See the picture of the Sun on Wikipedia

That edge/balance will change when the sun begins to run low on hydrogen. At this time, the reactions inside the star will change causing it to become become a giant red star.

I guess you could compare it with the surface of sea water on Earth. It's technically not still and stable, but we can calculate an average value of the sea level. And it is because it's an average value that we can rely on that to determine altitude and earth radius as well.

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    $\begingroup$ Thank you Donald.McLean for completing my answer with more scientific terms. $\endgroup$
    – Thibault
    Dec 5, 2013 at 15:55
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    $\begingroup$ This is a very confusing answer, it defines the edge as the point of hydrostatic equilibrium... but stable stars tend to be at least near hydrostatic equilibrium everywhere, so there is no apparent reason why this definition would unambiguously pick out the "edge". $\endgroup$
    – Stan Liou
    Mar 15, 2014 at 12:45
  • $\begingroup$ The whole of the Sun is in an equilibrium between pressure (gradient) and gravity. $\endgroup$
    – ProfRob
    Dec 18, 2014 at 19:56
  • $\begingroup$ As said above, all of the Sun is in hydrostatic equilibrium. The physical principle that defines the surface is really the transition from optically thick to optically thin material. $\endgroup$
    – Warrick
    Dec 22, 2014 at 6:44
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Most literature will define the diameter of the Sun up to the photosphere, the layer of the solar atmosphere you would see if you were to observe the Sun in white light.

The base of the Photosphere is defined as the region where the optical depth is around 2/3, or the region where the plasma becomes transparent to most optical light wavelengths.

Of course the true edge of the solar atmosphere could be considered as the heliopause, where the direct influence of the Sun's magnetic field and solar wind end and interstellar space begins.

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I thought I'd contribute an answer because there's a very recent paper on the subject:

Measuring the solar radius from space during the 2012 Venus Transit

It appeared in my RSS feeds this morning! A related writeup is online at the HMI website.

To answer the question, this measurement uses the transit of Venus to fit the limb-darkening law of the Sun. That is, the Sun is a bit fainter the further from the centre that you look. As you reach the optically thinner layers near the "surface", the brightness falls off rapidly, towards zero in the vacuum of space. The inflection point of the curve (as a function of distance from the centre of the disk) is a reasonable estimate of the "radius". As pointed out elsewhere, the value changes depending on which wavelength you use, but only by a few hundred km, compared to the Sun's overall radius of about 700 000km (actually more like 695 946 km), so the uncertainty is at or below the 0.1% level. Phil Plait wrote about a similar measurement (by the same team, I believe) that used the transits of Mercury in 2003 and 2006.

Finally, the team also used the limb-darkening (I think) to measure how round the Sun is. i.e. the diameter from top-to-bottom versus left-to-right. Answer: the Sun is very very round, with the radii differing by a few parts per million.

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The radius of the Sun depends highly on which wavelength you are looking (well, taking a photo). In each of them, you'll have a well defined sharp boundary as explained by Zsbán Ambrus in his answer, but it is not the same: it varies with the wavelenght.

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Look at the Sun. You shouldn't do this directly with the naked eye, but you can do so through a very dark filter, or project a suitably dark image through a pinhole. You can even find photos of the Sun on the internet.

What you see is a disk, uniformly bright and with a sharp boundary, surrounded by a comparatively much darker sky. The bright region is the part we consider the Sun, and that's how we get the radius.

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  • $\begingroup$ A sensibly experimentalist answer. $\endgroup$
    – Stan Liou
    Mar 15, 2014 at 12:45
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    $\begingroup$ Except you would get a wavelength dependent answer. $\endgroup$
    – ProfRob
    Dec 18, 2014 at 19:58
  • $\begingroup$ It's not uniformly bright at all. The limb darkening accelerates as you get close to the edge, it sort-of goes down hill. As @RobJeffries points out it's wavelength dependent limb-darkening. $\endgroup$
    – uhoh
    Jul 24, 2017 at 19:14

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