# Number of Atoms on Earth's Surface (Question based on Vsauce video which involves fractal dimension)

In a Vsauce video titled "How much of the Earth can you see at once" they try to do a calculation to estimate the number of atoms on Earth's surface.

The first part is easy:

1) Calculate surface area of the Earth (let's call it 'A')

2) Divide by cross sectional area of an atom (let's call it 'a')

3) The total number of atoms is: N = A/a

The next part is what got me thinking and I just can't seem to understand how they got to it. They explain that Earth has surface roughness which is consistent with a fractal dimension of 2.3.

They then say that by considering that this applies from the scale of a human hair to that of a mountain, the total number of atoms that could fit on the Earth is 1000x greater than what is estimated from the previous calculation.

Since they never explained their calculations, would anyone be willing to guess and perhaps share how they would do this?

Cheers!

P.S. If you can go step by step I would GREATLY appreciate it!

• " cross sectional area of an atom (let's call it 'a') " I think you need use distance between atoms; and th number of atoms on square meter of Eath surface depends on type of atomic lattice and type of atomic/molecule. – nouret May 9 at 17:13

Let's say the width of a human hair is 0.1mm and a mountain is 10km wide. That's a difference in scale by a factor of $$10^8$$. (I don't know what numbers they used in the video, but I'll go with that.)
Say we have some number $$x$$ of surface atoms on an average 0.1mm$$\times$$0.1mm region. Assuming no fractal dimension (only smooth, two-dimensional surfaces), the number of atoms on a 10km$$\times$$10km area (the scale of a mountain) becomes $$10^8\cdot 10^8x$$, since that's how many 0.1mm$$\times$$0.1mm squares you can fit in a 10km$$\times$$10km square. Note that $$10^8\cdot 10^8x = (10^8)^2x = 10^{16}x$$ See that $$^2$$ there? That's where the dimension comes in. If we had only cared about the number of atoms along a line instead of on a surface, the result would've been $$(10^8)^1x$$, and if we cared about the number of atoms in a volume, the result would've been $$(10^8)^3x$$. In other words, "dimension" in this case is a measure of how much more you get of something as you increase the scale. Distances, areas and volumes have dimensions of $$1, 2$$ and $$3$$ respectively. Fractal dimension just means that this number is not an integer, and that's all there is to it.
Now we assume a fractal dimension of $$2.3$$ all the way between these two scales. That means that if there are $$x$$ surface atoms on an average 0.1mm$$\times$$0.1mm region, there are $$(10^8)^{2.3}\approx 2.511\cdot 10^{18}x$$ surface atoms on an average 10km$$\times$$10km region.
We see that using this fractal dimension, we got an answer which was $$251$$ times larger. Of course, if you assume that the difference between a human hair and a mountain is larger than what I have used here, then the difference will be larger as well. For each factor of $$10$$ in the difference between a hair and a mountain the effect of using fractal dimension doubles (the difference in dimension is $$2.3-2 = 0.3$$, and we have $$10^{0.3}\approx 2$$). So if they actually got $$1000$$ times more as their answer, then they are working with a scale difference of $$10^{10}$$ (which corresponds to 0.01mm to 100km)