I'm searching for the "official" mass of the sun as a unit in astrophysics. The mass of the sun can be calculated by: $M_{\odot}=\frac{4\pi^2\times(1 \ \text{ua})^3}{G\times(1\ \text{year})^2}$

So in this formula:

$\pi = 3.1415926535898...$

$1 \ \text{ua} = 149597870700 \ \text{m}$ (it's an exact definition according to the IAU 2012 resolution http://www.iau.org/static/resolutions/IAU2012_English.pdf)

$G = 6.67384\times10^{-11} \ \text{m}^3\text{kg}^{-1}\text{s}^{-2}$ according to CODATA (http://physics.nist.gov/cgi-bin/cuu/Value?bg) but $G = 6.67428\times10^{-11} \ \text{m}^3\text{kg}^{-1}\text{s}^{-2}$ according to IAU 2009 (http://maia.usno.navy.mil/NSFA/IAU2009_consts.html), so which one do we choose?

Finally, for the year here, which year to choose? Is it the same year as for the light speed: $1 \ \text{year}=365.25\times24\times3600=31557600 \ \text{s}$ (Julian year)? Or the tropical year $1 \ \text{year}=365.2421897\times24\times3600=31556925.2 \ \text{s}$ (tropical year)?

For some of the preceding values we obtain:

$M_{\odot}=\frac{4\pi^2\times(149597870700)^3}{6.67384\times10^{-11}\times(31557600)^2}=1.988622... \ \text{kg}$

which is not what gives wikipedia (http://en.wikipedia.org/wiki/Solar_mass) or google (https://www.google.fr/search?q=solar+mass)

So my question is: what are the correct preceding values, and consequently what is the current good value for the sun mass? (as some units may have been redefined since some measurements it's not that simple)

  • $\begingroup$ For what it's worth, WolframAlpha reports a value of 1.988435×10^30 kg (kilograms) $\endgroup$
    – FrankH
    Sep 17, 2012 at 4:48
  • 1
    $\begingroup$ Worth noting that $GM_\odot$ is known far more precisely than $G$ or $M_\odot$ separately. Thus if you ever need to use $GM_\odot$ in a calculation, computing it from $G$ and $M_\odot$ separately amounts to doing extra effort for a less precise answer. $\endgroup$
    – user10851
    Sep 8, 2013 at 19:47
  • $\begingroup$ Google says the mass is "1.989 × 10^30 kg". Google is always right. $\endgroup$
    – Aloha
    Dec 15, 2015 at 9:58

2 Answers 2


The two values for the gravitational constant are both from CODATA. Given the choice of using the old 2009 CODATA value, or current value, use the latter.

There is no official "mass of the sun" in kilograms. However there is an official "solar mass parameter" $ \mathrm{GM}_\mathrm{s}$ (previously known as the heliocentric gravitation constant). This value has units of $ \frac{L^3}{M T^2}$, dividing it by the gravitational constant $G$, gives the solar mass in kilograms.

Wikipedia's value for the solar mass was out of date (2004), so I updated it. The current best estimate is $ 1.98855 \times 10^{30} \ \mathrm{kg}$.

Expressions for the solar mass:
$ M_{\odot}=\frac{4\pi^2(1 \ \text{AU})^3}{G(1 \ \mathrm{Gaussian \ year})^2} = \frac{k^2(1 \ \text{AU})^3}{G (1 \ \text{day})^2}$

where k is the Gaussian gravitational constant. A Gaussian year is defined as:
$ \mathrm{Y}_{\mathrm{Gauss}} = \frac{ 2 \pi (1 \ \mathrm{day}) }{k} $


I don't belive there's a generally accepted definition - if so, I've never heard of it. Just take a value from the literature you like best, eg the solar mass at J2000.

Personally, I think that as the astronomical unit now has a definite value, it would make sense to use the formula you gave above as a definition. If one should always use the most precise value for $G$ available or just take the current one is up to debate - I could argue either way.

If you do this, of course the sun will no longer have solar mass, but it's not a fixed quantity anyway and that's not stopped us before (case in point: julian year and astronomical unit).


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