<p>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.</p>

CC BY-SA 3.0
2014-12-23T12:17:16.700Z
No credit for this. At what significant figure does this affect the mass of the Sun over any human timescale...
ProfRob
CC BY-SA 3.0
2014-12-23T12:44:04.803Z
The poster was interested in an EXACT figure for the mass of the Sun.
iantresman
CC BY-SA 3.0
2014-12-23T12:51:45.620Z
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.
ProfRob
CC BY-SA 3.0
2014-12-23T23:19:23.357Z
@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.
geometrian
CC BY-SA 3.0
2014-12-23T11:38:31.723Z
2014-12-23T11:38:31.723Z
iantresman
<p>The answer to your question is a very simple one but, a good one at that. The exact mass is 1.9891×10<sup>30</sup> 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.</p>

CC BY-SA 3.0
2014-12-22T23:38:02.140Z
A source would be really nice, given that imallett wrote, "Obviously, citable, primary sources only!"
HDE 226868
CC BY-SA 3.0
2014-12-22T03:28:44.640Z
2014-12-22T23:25:14.103Z
2014-12-22T23:25:14.103Z
jenni selman
<p>The mass of the Sun is determined from Kepler's laws:</p>

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

<p>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 <a href="http://hpiers.obspm.fr/eop-pc/models/constants.html" rel="noreferrer">the (sidereal) year is</a> 365.256363004 days. We have also <a href="http://www.iau.org/static/resolutions/IAU2012_English.pdf" rel="noreferrer">defined the astronomical unit (AU)</a> 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).</p>

<p>At this point, we can solve for the product $GM$, known as the <em>gravitational parameter</em>, sometimes denoted $\mu$. <a href="http://ssd.jpl.nasa.gov/?constants" rel="noreferrer">For the Sun,</a>
$$\mu_\odot=132712440018\pm9\,\mathrm{km}^3\cdot\mathrm{s}^{-2}$$</p>

<p>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 <a href="http://physics.nist.gov/constants" rel="noreferrer">current CODATA value</a> 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.</p>

<p>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 <a href="http://www.nature.com/nature/journal/v510/n7506/full/nature13507.html" rel="noreferrer">some measurements</a> of $G$.</p>

CC BY-SA 3.0
2014-12-20T23:15:17.410Z
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.)
Stan Liou
CC BY-SA 3.0
2014-12-21T04:55:18.573Z
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 $\mu_{\odot}$.
David Hammen
CC BY-SA 3.0
2014-12-20T20:48:39.787Z
2014-12-21T04:39:30.500Z
2014-12-21T04:39:30.500Z
Warrick
<p>I am having trouble finding data on this. So far, I have found at least seven values:</p>

<pre><code>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
</code></pre>

<ol>
<li><p>The first value can be found from <a href="http://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html" rel="noreferrer">NASA GSFC</a>.</p></li>
<li><p>The second value can be found from Wikipedia in (at least) two places.</p>

<ul>
<li>The <a href="http://en.wikipedia.org/wiki/Sun#cite_note-nssdc-1" rel="noreferrer">first Wikipedia occurence</a> references the GSFC page--which is erroneous since that value does not occur there.</li>
<li>The <a href="http://en.wikipedia.org/wiki/Solar_mass" rel="noreferrer">second Wikipedia occurence</a> references <a href="http://asa.usno.navy.mil/static/files/2014/Astronomical_Constants_2014.pdf" rel="noreferrer">USNO's 2014 Astronomical Constants</a>, which appear to themselves be copy-pasted from data in 2009 or 2012.  The second Wikipedia occurrence also sources <a href="http://physics.nist.gov/cgi-bin/cuu/Value?bg" rel="noreferrer">NIST</a> (presumably <a href="http://physics.nist.gov/cuu/Constants/Table/allascii.txt" rel="noreferrer">this table</a>).  I couldn't find the cited mass from either reference.</li>
</ul></li>
<li><p>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 <a href="http://solarsystem.nasa.gov/planets/profile.cfm?Object=Sun&amp;Display=Facts" rel="noreferrer">generic NASA</a> site.</p></li>
<li><p>The fourth value occurs in the book <em>Handbook of Chemistry and Physics</em> (Robert C. Weast, 1980)</p></li>
<li><p>The fifth value apparently also occurs in the book <em>Astrophysical Data</em> (Kenneth R. Lang, 1990ish).</p></li>
<li><p>The sixth value occurs in the book <em>Physics--3rd Edition</em> (Cutnell et al. 1995).  <a href="http://hypertextbook.com/facts/2000/LeonVaysburd.shtml" rel="noreferrer">This site</a> lists several of the values and does a calculation to obtain the sixth value.</p></li>
<li><p>Occurs in many places and is pretty clearly rounded.</p></li>
</ol>

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

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

CC BY-SA 3.0
2014-12-21T05:02:25.133Z
Scientists know the product $GM$$_{\odot}$ to ten decimal places. However, since scientists only know $G$ to four decimal places, four decimal places is the limit to which we can say we know the mass of the Sun. Your first two values are essentially the same.
David Hammen
CC BY-SA 3.0
2014-12-21T05:09:59.767Z
Bottom line: You are much better off using the product $\mu_{\odot}=GM$$_{\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](http://en.wikipedia.org/wiki/Standard_gravitational_parameter) for all of the above bodies is known to five decimal places or more.
David Hammen
<the-sun><mass>
What is the exact mass of the Sun?
geometrian
# What is the exact mass of the Sun?

(Asked by: username_0 on 2014-12-20)

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


The 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!

**Tags:** the-sun, mass
> **Comments:**
> **username_1**: Scientists know the product $GM$$_{\odot}$ to ten decimal places. However, since scientists only know $G$ to four decimal places, four decimal places is the limit to which we can say we know the mass of the Sun. Your first two values are essentially the same.  
> **username_1**: Bottom line: You are much better off using the product $\mu_{\odot}=GM$$_{\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](http://en.wikipedia.org/wiki/Standard_gravitational_parameter) for all of the above bodies is known to five decimal places or more.  

---

## 3 answers

### Answer by username_6 on 2014-12-20. Score: 14 | Accepted answer

The 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](http://hpiers.obspm.fr/eop-pc/models/constants.html) 365.256363004 days. We have also [defined the astronomical unit (AU)](http://www.iau.org/static/resolutions/IAU2012_English.pdf) 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,](http://ssd.jpl.nasa.gov/?constants)
$$\mu\_\odot=132712440018\pm9\,\mathrm{km}^3\cdot\mathrm{s}^{-2}$$

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](http://physics.nist.gov/constants) 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](http://www.nature.com/nature/journal/v510/n7506/full/nature13507.html) of $G$.

> **Comments:**
> **username_7**: 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.)  
> **username_1**: 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 $\mu_{\odot}$.  

---

### Answer by username_2 on 2014-12-23. Score: -1

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.

> **Comments:**
> **username_3**: No credit for this. At what significant figure does this affect the mass of the Sun over any human timescale...  
> **username_2**: The poster was interested in an EXACT figure for the mass of the Sun.  
> **username_3**: 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.  
> **username_0**: @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.  

---

### Answer by username_4 on 2014-12-22. Score: -3

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.

> **Comments:**
> **username_5**: A source would be really nice, given that imallett wrote, "Obviously, citable, primary sources only!"  

---

