# How much gold is there in our sun?

XKCD 1944 claims that there is "more gold in the sun than water in the oceans". Is this really true?

• Who here would dare contradict xkcd.com ? :-) Jan 19, 2018 at 9:49
• 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." Jan 19, 2018 at 16:25
• It was already explained on explainxkcd.com/wiki/index.php/1944
– vsz
Jan 19, 2018 at 18:16
• @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. Jan 21, 2018 at 17:45
• Also discussed in the xkcd forum. Jan 22, 2018 at 10:26

The mass of the sun is 1.989 × 1030 kg.

Abundance in the Sun of the elements gives a percentage 1 × 10-7 % for gold *, so that leaves you with a mass of 1.989 × 1021 kg of gold.

HowStuffWorks states that there is 1.26 × 1021 kg water on Earth, of which 98% is in the oceans, i.e. 1.235 × 1021 kg.

This would mean the XKCD statement is true: there is 1.6 times as much gold in the sun as there is water in the oceans.

* They cite WolframAlpha as their source. Executing SolarAbundance "Gold" there confirms this (mass) percentage.

• There are at least two problems with that calculation: (1) The source for abundance doesn't say whether the percentage is of mass or of number of atoms; (2) if is it percentage of mass, 1 × 10<sup>-7</sup> % means somewhere between 0.5 × 10<sup>-7</sup> % and 1.5 × 10<sup>-7</sup> %, so the proportion could be as low as 0.8, which is less than 1. Jan 19, 2018 at 12:29
• How did that gold get there? I am under the impression that, as a main sequence star, the sun cannot create its own gold through element synthesis. So I am guessing that the gold in the sun was present when the sun first started burning, and I guess it must come from older generation supernovae? Jan 19, 2018 at 17:11
• @ChocolateAndCheese Correct, virtually all of the elements heavier than Helium in the Sun (and the rest of the solar system) are the remains of older stars. Jan 19, 2018 at 19:27
• @ Peter Taylor, similarly, the XKCD cartoon doesn't state whether it is more by mass or more by number of atoms/molecules. Jan 19, 2018 at 22:37
• @Octopus or by volume. 2 lbs of gold is quite a bit smaller than 1 lb of sea-water. Jan 20, 2018 at 6:18

"Element Abundances in the Sun - The Elements Handbook", KnowledgeDoor 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).

• 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: par.nsf.gov/servlets/purl/10036398 Jan 22, 2018 at 14:57
• @DavePhD, so in other words this is an under-estimate, strengthening the conclusion? Jan 23, 2018 at 17:13
• yes, an underestimate Jan 23, 2018 at 17:37