In a famous talk (from 14:30 to 15:00) Nima Arkani-Hamed tells us that what "fixes the size of the earth" is the equilibrium between the gravitational pressure against the atomic pressure of the atoms in it. He sais

The radius of the Earth divided the radius of the atom is precisely the square root of the ratio of the electric force to the gravitational force between a pair of protons

Thus, if i calculate

$$ F_q/F_g=\frac{r^2}{Gm_p^2}\frac{q^2}{4\pi\epsilon r^2}\sim1,2\cdot10^{36} $$ I can easily find $$ R_\oplus^2\approx F_q/F_g\cdot\langle r_{atom}\rangle^2 $$

My problem is that in order to obtain $R_\oplus^2\approx4\cdot10^{13}$ m i must have $\langle r_{atom}\rangle^2\sim3,3\cdot10^{-23}$ m which leads to $\langle r_{atom}\rangle\approx6\cdot10^{-10}$ m. Now to me 600 pm are nowhere near what i was expecting (something between tens and a hundred of pm)... am i confused on that or is there something i am missing?

  • $\begingroup$ Do all atoms have the same radius? $\endgroup$ – D. Halsey Jan 2 at 0:39
  • $\begingroup$ of course not but i was expecting something avarage and as far as i remember no one up to uranium has more than 300 pm. $\endgroup$ – Pietro Oliva Jan 2 at 0:42
  • $\begingroup$ Just ignore the word “precisely”. Arkani-Hamed didn’t mean it literally. You’re getting a number in the right ballpark. $\endgroup$ – G. Smith Jan 2 at 0:53
  • $\begingroup$ If you try this with the radius of Mars, Venus or any other planet it demonstrates that equating these two ratios is ridiculous. And you can tweak the $r$ between two protons and the $\langle r_{atom} \rangle$ more or less arbitrarily to get any result you want. Equilibrium of forces - yes, the rest just silliness. $\endgroup$ – StephenG Jan 2 at 2:54
  • $\begingroup$ Exactly! i actually found a radius jupiter-like at my first guess, but i could easly reach some dwarf star radius just fooling around. I think what he intended was just to justify the macroscopic "elementary" objects tout court with no intention to be accurate, just within few orders of magnitude... $\endgroup$ – Pietro Oliva Jan 2 at 19:09

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