I am having trouble finding data on this. So far, I have found at least seven values:

1: 1.9885  e30 kg
2: 1.98855 e30 kg
3: 1.9891  e30 kg
4: 1.991   e30 kg
5: 1.989   e30 kg
6: 1.99    e30 kg
7: 2       e30 kg
  1. The first value can be found from NASA GSFC.

  2. The second value can be found from Wikipedia in (at least) two places.

  3. The third (and also the fifth) value (which, it is interesting to note, are both outside the error bound given in the second Wikipedia occurrence) can be found on this generic NASA site.

  4. The fourth value occurs in the book Handbook of Chemistry and Physics (Robert C. Weast, 1980)

  5. The fifth value apparently also occurs in the book Astrophysical Data (Kenneth R. Lang, 1990ish).

  6. The sixth value occurs in the book Physics--3rd Edition (Cutnell et al. 1995). This site lists several of the values and does a calculation to obtain the sixth value.

  7. Occurs in many places and is pretty clearly rounded.

The fifth, sixth, and seventh values (also) occur in various non-primary sources online, and I assume are rounded values of others.

My question: what is the most recent, best estimate, to the greatest precision, of the Sun's mass? Obviously, citable, primary sources only!

  • 2
    $\begingroup$ Scientists know the product $GM$$_{\odot}$ to ten decimal places. However, since scientists only know $G$ to four decimal places, four decimal places is the limit to which we can say we know the mass of the Sun. Your first two values are essentially the same. $\endgroup$ Dec 21, 2014 at 5:02
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    $\begingroup$ Bottom line: You are much better off using the product $\mu_{\odot}=GM$$_{\odot}$ than you are using numerical values for $G$ and $M_{\odot}$ and computing $GM$$_\odot$. The same applies to the eight planets and the Earth's Moon. The standard gravitational parameter for all of the above bodies is known to five decimal places or more. $\endgroup$ Dec 21, 2014 at 5:09

4 Answers 4


The mass of the Sun is determined from Kepler's laws:


Each term in this component contributes to both the value of the solar mass and our uncertainty. First, we know to very good precision that the (sidereal) year is 365.256363004 days. We have also defined the astronomical unit (AU) to be 149597870700 m. Strictly speaking, the semi-major axis of the Earth's orbit is slightly different, but by very little in the grand scheme of things (see below).

At this point, we can solve for the product $GM$, known as the gravitational parameter, sometimes denoted $\mu$. For the Sun, $$\mu_\odot=132712440018\pm9\,\mathrm{km}^3\cdot\mathrm{s}^{-2}$$

So solve for $M_\odot$, we need the gravitational constant $G$, which, as it turns out, is by far the largest contributor to the uncertainty in the solar mass. The current CODATA value is $6.67384\pm0.00080\times10^{-11}\,\mathrm{N\cdot m}^2\cdot\mathrm{kg}^{-2}$, which all combines to give $$M_\odot=1.98855\pm0.00024\times10^{30}\,\mathrm{kg}$$ where my uncertainty is purely from the gravitational constant.

The value $1.9891\times10^{30}\,\mathrm{kg}$ (and nearby values) probably come from an older value of the gravitational constant of $6.672\times10^{-11}\,\mathrm{N\cdot m}^2\cdot\mathrm{kg}^{-2}$, which is still supported by some measurements of $G$.

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    $\begingroup$ To actually solve for the Sun's standard gravitational parameter to the precision presented here, one can use the square of the Gaussian gravitational constant: $\mu_\odot = (0.01720209895)^2\,{\mathrm{AU}^3}/{\mathrm{day}^2}$. (Side note: I recommend \cdot for unit multiplication instead of periods.) $\endgroup$
    – Stan Liou
    Dec 20, 2014 at 23:15
  • $\begingroup$ That's a rather dated value for $\mu_{\odot}$, drawn from the DE405 ephemeris model (JPL's HORIZONS still uses DE405). The currently accepted value, based on DE421, is 132712440099±10 km$^3$/s$^2$ (TCB-compatible) or 132712440041±10 km$^3$/s$^2$ (TDB-compatible). Since then, the AU has been turned into a defined constant that is compatible with that value and with the Gaussian gravitational constant $k$. DE430/431 reverted to simply using $k^2$ for $\mu_{\odot}$. $\endgroup$ Dec 21, 2014 at 4:55
  • $\begingroup$ ilrs.gsfc.nasa.gov/docs/2014/196C.pdf gives the Sun's GM (for DE430/431) as 132712440041.939400 km^3/s^2 (page 49). $\endgroup$
    – user21
    Dec 22, 2014 at 23:42

http://ilrs.gsfc.nasa.gov/docs/2014/196C.pdf gives G*m for the Sun as 132712440041.939400 km^3/s^2.

http://physics.nist.gov/cgi-bin/cuu/Value?bg gives G as 6.67384*10^-11 m^3*kg^-1*s^-2

If these numbers were exact (they're not), the sun's mass (after converting m^3 to km^3 above) would be:

165890550052424250000000000000000000/83423 kg

or about

1988546924138717739712069812881.3396785059276218789 kg


1.9885469241387177397120698128813396785059276218789*10^30 kg

Of course, that level of precision is ridiculously unwarranted, but I would argue that's the most precise value you can get, if you accept NASA's definitions of the sun's Gm and the universe's G.

  • 3
    $\begingroup$ Yes, the level of precision is unwarranted; so don't write it like that. You cannot estimate it any more precisely than $G$ is determined. i.e. 6 significant figures. $\endgroup$
    – ProfRob
    Dec 23, 2014 at 12:21

The Sun is losing the equivalent of around 4.26 million metric tons per second of hydrogen, as it is converted into energy. The Sun does not have an exact mass.

  • 1
    $\begingroup$ No credit for this. At what significant figure does this affect the mass of the Sun over any human timescale... $\endgroup$
    – ProfRob
    Dec 23, 2014 at 12:17
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    $\begingroup$ The poster was interested in an EXACT figure for the mass of the Sun. $\endgroup$
    – iantresman
    Dec 23, 2014 at 12:44
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    $\begingroup$ The question was "what is the most recent, best estimate, to the greatest precision, of the Sun's mass? Obviously, citable, primary sources only!". Even if we observed for 100 years, the Sun's mass would change by less than 1 part in $10^{11}$, comparable with the changes brought about by the solar wind. As $G$ is only known to 6 significant figures, your answer is (currently) irrelevant. $\endgroup$
    – ProfRob
    Dec 23, 2014 at 12:51
  • $\begingroup$ @RobJeffries and iantresman, the title got edited by Noordung. The "exact" got added, and by the time I realized, I couldn't really revert since answers were already extant addressing that wording. $\endgroup$
    – geometrian
    Dec 23, 2014 at 23:19

The answer to your question is a very simple one but, a good one at that. The exact mass is 1.9891×1030 kg. Math and new advances in solar exploration have given us so much more information about our Sun. We have been able to see solar eruptions, (CMA's), and gained information about the Sun never known.

  • 3
    $\begingroup$ A source would be really nice, given that imallett wrote, "Obviously, citable, primary sources only!" $\endgroup$
    – HDE 226868
    Dec 22, 2014 at 23:38

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