### What is the exact mass of the Sun?

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 kgThe first value can be found from NASA GSFC.

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

- The first Wikipedia occurence references the GSFC page--which is erroneous since that value does not occur there.
- The second Wikipedia occurence references USNO's 2014 Astronomical Constants, which appear to themselves be copy-pasted from data in 2009 or 2012. The second Wikipedia occurrence also sources NIST (presumably this table). I couldn't find the cited mass from either reference.

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.

The fourth value occurs in the book

*Handbook of Chemistry and Physics*(Robert C. Weast, 1980)The fifth value apparently also occurs in the book

*Astrophysical Data*(Kenneth R. Lang, 1990ish).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.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!

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.

Warrick Correct answer

8 years agoThe mass of the Sun is determined from Kepler's laws:

$$\frac{4\pi^2\times(1\,\mathrm{AU})^3}{G\times(1\,\mathrm{year})^2}$$

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 $G$.

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.)

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 $\mu_{\odot}$.

http://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).

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Content dated before 7/24/2021 11:53 AM

David Hammen 8 years ago

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.