I know that earth has a radius $R_E = 6371\,km$ and gold has the following properties:

  • $\rho_{Au} = 1.932 \cdot 10^4 \frac{kg}{m^3}$
  • $M_{Au} = 196.9666 \cdot u \quad (u = 1.66054*10^{-27}\,kg)$

where $\rho_{Au}$ is the density of gold and $M_{Au}$ is the mass of a gold atom.

Assuming the equator is a perfect circle, we can determine its circumference $$ \begin{align*} U_E &= 2\pi R_E \\ &= 2\pi \cdot 6371 \, km \\ &\approx 40\,030 \,km \end{align*} $$ Let's take circle and cut it open at one point, so we can stretch it out to a straight line with length $U_E$. (please correct me here if that's wrong)

To calculate how many atoms fit around the equator, we can divide the circumference $R_E$ by the diameter of a single gold atom, let's call it $D_{Au}$.

Two problems:

(1) Is it actually possible to treat the circumference of the circle as a simple straight line, given that the ring of atoms around the equator has height $1$.

(2) How can I find the diameter/radius of a gold atom across a line? I tried viewing a gold atom as a simple sphere and calculating its radius using the formula $$ V_{Au} = \frac{4}{3} \pi R_{Au}^3 $$ Which won't give any value of the atomic radii that you can find on the internet, since they all define the atomic radius as the distance between $2$ atoms in a given packing in a volume or on an area. But I need to get the radius in the context of a line, right?

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    $\begingroup$ Hints: 1) you know the density and molar mass of gold, so you can calculate the dimensions of a cube that contains one mole of gold atoms; 2) that cube has the same width, height, and depth. Multiplying the number of atoms in the cube's height, width, and depth gives one mole. This means that the cube root of the Avagadro number is the number of gold atoms down the side of the cube, assuming that each gold atom is just touching its neighbors. $\endgroup$ Commented Oct 19, 2021 at 19:25

1 Answer 1


An isolated atom of gold doesn't have a well defined radius. For the purpose you your hypothetical calculation you would have to make an assumption about how closely the atoms would pack if placed in a line. off the top of my head I can't think why the spacing of a single line of gold atoms would be significantly different to a 2d or 3d array of them, so I would go with the interatomic distance figures given on the internet.

As for how to model the equator... if you treat it as a circle and ignore the extra height of the individual gold atoms, you will underestimate the number of atoms by just over three. However, that will be a trivial correction, vastly dwarfed by the effect of assuming that the equator is a circle, which ignores all the changes in sea level and the height of land over which the equator passes.

  • $\begingroup$ Yeah I can assume it's a perfect circle. Why would I underestimate the number of atoms by 3? $\endgroup$ Commented Oct 19, 2021 at 20:05
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    $\begingroup$ Because if you sit the atoms on the equator, the line joining their centres will form a circle with a radius equal to the radius of the Earth plus the radius of a gold atom, so the length of that line will be longer than the equator by 2𝜋 times the radius of a gold atom, which means it will accommodate an extra 𝜋 atoms. $\endgroup$ Commented Oct 19, 2021 at 20:16

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