<p><a href="http://www.knowledgedoor.com/2/elements_handbook/element_abundances_in_the_sun.html" rel="noreferrer">"Element Abundances in the Sun - The Elements Handbook", <em>KnowledgeDoor</em></a> claims that the base 10 log of the number of atoms of gold in the Sun for every $10^{12}$ atoms of hydrogen is $1.01 \pm 0.15$. If I'm reading their references correctly, that's from <strong>Abundances of the Elements: Meteoritic and Solar</strong>, Anders, Edward, and Nicolas Grevesse, <em>Geochimica et Cosmochimica Acta</em>, volume 53, number 1, 1989, pp. 197–214, doi:10.1016/0016-7037(89)90286-X</p>

<p>The atomic mass of gold is $197$ times the atomic mass of hydrogen (more precise figures are available, but irrelevant given the accuracy of the atomic proportions). So $2020$ kg of gold for every $10^{12}$ kg of hydrogen, meaning that ignoring all other elements and running with $1.99 \times 10^{30}$ kg for the mass of the Sun, it contains $4 \times 10^{21}$ kg of gold. Taking other elements into account - helium is actually significant - reduces that value to $3 \times 10^{21}$ kg.</p>

<p>This is about twice as much as the mass of the ocean, which corresponds to 2 standard deviations ($\log_{10} 2 \approx 0.3$ vs the standard deviation of $0.15$ in the log 10 value of the abundance).</p>

CC BY-SA 3.0
2018-01-22T14:57:05.333Z
The 10^1.01 number is the abundance *in the photosphere* relative to H=10^12 .  Elements other than H are expected to be gravitationally depleted in the photosphere, so this number is not representative of the entire sun. See section 2.2.1.2.2 here: https://par.nsf.gov/servlets/purl/10036398
DavePhD
CC BY-SA 3.0
2018-01-23T17:13:43.663Z
@DavePhD, so in other words this is an under-estimate, strengthening the conclusion?
Peter Taylor
CC BY-SA 3.0
2018-01-23T17:37:55.870Z
yes, an underestimate
DavePhD
CC BY-SA 3.0
2018-01-19T12:42:05.953Z
2018-01-21T18:19:26.057Z
2018-01-21T18:19:26.057Z
Peter Taylor
<p><a href="https://xkcd.com/1944/" rel="noreferrer">XKCD 1944</a> claims that there is "more gold in the sun than water in the oceans". Is this really true? </p>

<p><a href="https://i.sstatic.net/1xFzK.png" rel="noreferrer"><img src="https://i.sstatic.net/1xFzK.png" alt="The end of the rainbow"></a></p>

CC BY-SA 3.0
2018-01-19T09:49:18.097Z
Who here would dare contradict xkcd.com ? :-)
StephenG - Help Ukraine
CC BY-SA 3.0
2018-01-19T16:25:16.337Z
Alt text for people that don't want to click through to XKCD: _"The retina is the exposed surface of the brain, so if you think about a pot of gold while looking at a rainbow, then there's one at BOTH ends."_
JPhi1618
CC BY-SA 3.0
2018-01-19T18:16:24.040Z
It was already explained on http://www.explainxkcd.com/wiki/index.php/1944
vsz
CC BY-SA 3.0
2018-01-21T17:45:57.220Z
@StephenG Much as I hate to contradict XKCD, that's not how rainbows work. You don't get a partial rainbow if there's simply a cloud in the right position. It has to be raindrops.
Mr Lister
CC BY-SA 3.0
2018-01-22T10:26:40.293Z
Also discussed in the [xkcd forum](http://fora.xkcd.com/viewtopic.php?f=7&t=124122#p4298000).
Raidri
<the-sun>
How much gold is there in our sun?
Jakub Šturc
# How much gold is there in our sun?

(Asked by: username_0 on 2018-01-19)

[XKCD 1944](https://xkcd.com/1944/) claims that there is "more gold in the sun than water in the oceans". Is this really true?

[![The end of the rainbow](https://i.sstatic.net/1xFzK.png)](https://i.sstatic.net/1xFzK.png)

**Tags:** the-sun

> **Comments:**
> **username_1**: Who here would dare contradict xkcd.com ? :-)  
> **username_2**: Alt text for people that don't want to click through to XKCD: _"The retina is the exposed surface of the brain, so if you think about a pot of gold while looking at a rainbow, then there's one at BOTH ends."_  
> **username_3**: It was already explained on http://www.explainxkcd.com/wiki/index.php/1944  
> **username_4**: @username_1 Much as I hate to contradict XKCD, that's not how rainbows work. You don't get a partial rainbow if there's simply a cloud in the right position. It has to be raindrops.  
> **username_5**: Also discussed in the [xkcd forum](http://fora.xkcd.com/viewtopic.php?f=7&t=124122#p4298000).  

---

## 1 answers

### Answer by username_6 on 2018-01-19. Score: 31

["Element Abundances in the Sun - The Elements Handbook", *KnowledgeDoor*](http://www.knowledgedoor.com/2/elements_handbook/element_abundances_in_the_sun.html) claims that the base 10 log of the number of atoms of gold in the Sun for every $10^{12}$ atoms of hydrogen is $1.01 \pm 0.15$. If I'm reading their references correctly, that's from **Abundances of the Elements: Meteoritic and Solar**, Anders, Edward, and Nicolas Grevesse, *Geochimica et Cosmochimica Acta*, volume 53, number 1, 1989, pp. 197–214, doi:10.1016/0016-7037(89)90286-X

The atomic mass of gold is $197$ times the atomic mass of hydrogen (more precise figures are available, but irrelevant given the accuracy of the atomic proportions). So $2020$ kg of gold for every $10^{12}$ kg of hydrogen, meaning that ignoring all other elements and running with $1.99 \times 10^{30}$ kg for the mass of the Sun, it contains $4 \times 10^{21}$ kg of gold. Taking other elements into account - helium is actually significant - reduces that value to $3 \times 10^{21}$ kg.

This is about twice as much as the mass of the ocean, which corresponds to 2 standard deviations ($\log\_{10} 2 \approx 0.3$ vs the standard deviation of $0.15$ in the log 10 value of the abundance).

> **Comments:**
> **username_7**: The 10^1.01 number is the abundance *in the photosphere* relative to H=10^12 .  Elements other than H are expected to be gravitationally depleted in the photosphere, so this number is not representative of the entire sun. See section 2.2.1.2.2 here: https://par.nsf.gov/servlets/purl/10036398  
> **username_6**: @username_7, so in other words this is an under-estimate, strengthening the conclusion?  
> **username_7**: yes, an underestimate  

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