# If you covered Canada with 1 mole's worth of golf balls, how thick would the layer of golf balls be? [closed]

I know moles are a chemistry thing, but only really the quantity matters in this question.

Canada, without counting the water, has an area of $$9,093,507 \text{ km}^2$$. A golf ball (pretending it is square) has a width/height of $$4.318 \times 10^{-5}$$ km. If I covered Canada with 1 mole ($$6.022 \times 10^{23}$$) of golf balls, how thick would the pile of golf balls be?

• See also: a mole of moles Commented Oct 31, 2022 at 20:01

The answer will depend on how the golf balls are packed. A random packing of equally sized spheres has packing density $$0.6400$$, that is, it would occupy $$64.00\%$$ of the space that contains it (see MathWorld: Sphere Packing). A hexagonal close packing, on the other hand, is far more efficient and has packing density $$0.7405$$. If there are $$n$$ many balls, each of volume $$v$$, packed together with density $$\rho$$, they will have a total volume that satisfies $${\rho}=\frac{nv}{V}$$. If those packed balls are amassed up to an approximately level height $$h$$ on a surface with area $$A$$, then they will approximately form a prism (with an irregular base) of volume $$V=hA$$. It follows from elementary algebra that $$h=\frac{nv}{\rho A}$$. In your case, a random packing (hence, density $$\rho=0.6400$$) of $$n=6.0\times10^{23}$$ approximately spherical balls of diameter $$4.3\,\mathrm{cm}$$ (hence, volume $$v=41\,\mathrm{cm}^3$$) on a surface of area $$9\times10^{6}\,\mathrm{km}^2$$ will have height $$4\times10^{3}\,\mathrm{km}$$.
• This estimate is consistent with the XKCD estimate for a mole (amount) of moles (animals) on the earth, where the packed objects have equal number and volume of the same order of magnitude ($75\, \mathrm{g}\cdot 1\,\mathrm{cm}^3/\mathrm{g}$ [moles are smaller than I realized!]), and they are spread out over an area $+2$ orders of magnitude greater in extent ($510\times10^6\,\mathrm{km}^2$, the surface area of the Earth). So we should expect in that case to have a height $\approx -2$ orders of magnitude less, as indeed we do ($80 \mathrm{km}$).
• With that much height, you should take into account the fact that the Earth is not flat. If we take the Earth to be a sphere of radius $R$, covering area $A$ of the surface to height $h$ produces a volume of $$A ((R+h)^3 - R^3))/(3R^2)= A(h + h^2/R + h^3/(3 R^2)$$ Commented Oct 31, 2022 at 21:00
• For completeness: A cubic packing, with an efficiency of $\frac{\pi}{6} \approx 0.5236$, would require a depth of 5332 km. In contrast, a completely efficient packing (if balls can be deformed or melted while preserving their volume) would need only 2792 km.
• It's not the area of the base, it's the height: $4 \times 10^3$ km is nearly $63\%$ of the radius of the Earth. Commented Nov 1, 2022 at 5:08