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I tried to resolve $G$ from natural units (like Planck), and found that $$ G = (K \times c / r_0)^3 $$ where $ G = 6.67430 \times 10^{-11} $ is gravitational constant, $ K = 1.66053906660 \times 10^{-27} $ is unified atomic mass units (Daltons) per kilogramm, $c$ is speed of light in vacuum, and $ r_0 \approx 1.23 \times 10^{-15} $ is nuclear radius of hydrogen. When calculating I change $r_0$ from $1.23$ to $1.22725429$.

In numbers it is equals to $$ 6.67430 \times 10^{-11} = {[(1.66053906660 \times 2.99792458 / 1.22725429) \times 10^{-27+8+15} ]}^3 $$ or $$ 6.67430 \times 10 = (1.66053906660 \times 2.99792458 / 1.22725429)^3 $$

I do not think that $G$ depends on $c^3$, but it is accurate (with error about $0.7\%$) even if $r_0 = 1.23 \times 10^{-15}$ (and with error $2\times 10^{-7}\%$ if $r_0=1.22725429 \times 10^{-15}$).

That more intresting, in Planck units the planck lengh depends on both $G$ and $c^3$: $$ l_P = \sqrt{\frac{\hbar G}{c^3}} $$ In this case the equation transforms into $$ l_P = \sqrt{\frac{\hbar K^3}{r_0^3} } $$

Is there any reason to use $c^3$ in $l_P$ or in $G$?

Mikhail Ionkin
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    Planck is derived from G, not the other way around, and so is K. Natural units are derived from G, not G from natural units – Yukterez May 23 '20 at 18:59
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    Sorry, this is numerology, not physics. See https://physics.stackexchange.com/q/285829/123208 & http://physics.stackexchange.com/q/44017/2451 – PM 2Ring May 23 '20 at 19:10

2 Answers2

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Your first equation $$G=(K\cdot c/r_0)^3$$ doesn't make sense.

The left-side has dimension $\frac{m^3}{kg\cdot s^2}$. The right side has dimension $\left(\frac{kg}{s}\right)^3$. Therefore the numerical equality is just an accident caused by the choice of your units.

Thomas Fritsch
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  • In equation $F = Mm/r^2$ dimensions is not the same. So, people used $GMm/r^2$ to resolve it. In any case I can just come up with a new constant $D=1$ that will have dimension that I need, and multiple right part on $D$. – Mikhail Ionkin May 23 '20 at 19:00
  • Thanks. Seems that I underestimated probability of such accident. – Mikhail Ionkin May 23 '20 at 21:10
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If this corresponds to something physical, it shouldn't matter what units we measure these quantities in (after all, the choice of units is something that we humans have made for our own convenience, and doesn't really have much to do with the way the universe works). So let's see what happens to your calculation when we do it in, for example, CGS units.

In CGS units, we have that $K=1.66053906660\times 10^{-30}$ grams/Dalton, $c=2.99792458\times 10^{10}$ cm/s, and $r_0=1.22725429\times 10^{-13}$ cm, while $G=6.674\times 10^{-8}$ cm/(g$\cdot$s$^2$). When we repeat the calculation in these units, we find:

$$(K\times c/r_0)^3=6.6743\times 10^{-26}$$

which is now 1,000,000,000,000,000,000 times smaller than the value of $G$ in these units. Since your calculation doesn't hold up under a change of units, it's obviously unphysical.

The more general issue with this kind of reasoning is the following very important fact:

Two quantities having the same numerical value does not mean that they are related.

You can see plenty of examples of this in, for example, this webcomic: https://xkcd.com/1047/. For example, there are roughly $\frac{5}{\pi^{1/e}}$ US customary feet in 1 meter; the two quantities agree to 1 part in 4000. This does not mean that there is some connection between $e$, $\pi$, and the conversion from one arbitrary unit of length to the length of some guy's foot.

It turns out you can find mystical-looking "numerical coincidences" with basically any number you try. Someone even built a free program that will find them for you, available here: https://mrob.com/pub/ries/.

probably_someone
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