Is there any relation between Planck constant and Gravitational constant?

Why is the Gravitational constant about $10^{23}$ times of the Planck constant in SI-units? Is there any relation between them? I mean Planck constant is about $6.6\times 10^{-34}$ $Js$ and Gravitational constant is about $6.6×10^{−11} \frac{N·m^2}{kg^2}$.

• Comment to the question (v2): If one only uses two significant figures for $G$, then one should round up to $6.7 \times 10^{−11} \frac{N·m^2}{kg^2}$. – Qmechanic Jun 2 '14 at 19:42
• Short answer: it's a coincidence. – Harry Johnston Jun 2 '14 at 21:40

$G$ is not exactly larger than $h$ by a factor of $10^{23}$ in SI units, as you are probably aware (just making sure). There is also no expected numerical relationship between the two that has a physical interpretation. You have to understand that these constants are mostly just due to our (to some extent) arbitrary choices of units. These are, of course, motivated by everyday convenience. But this doesn't mean that the commonly used SI units have any physical significance. In fact, there are several other unit systems. One particularly interesting one that is quite popular among physicists doing fundamental research is known as the Planck unit system.

In terms of Planck units, both $G$ and $\hbar$ are equal to $1$, as well as $c$, $k_B$ and $4\pi\epsilon_0$, the speed of light, Boltzmann's constant and the inverse of the Coulomb constant respectively. The Planck unit system attempts to eliminate the arbitrary choices due to the perspective of humans, which just so happen to live on certain energy, length, etc. scales. This is done by defining the units of measurement only in terms of fundamental constants of nature. The idea is that these constants are really what 'nature measures in', so setting their numerical value to $1$ makes sense. Related is the concept of a natural unit system, of which several exist. These all attempt to formulate things in a 'natural' sense (which, among other things, depends on the field of study).

• Setting the Coulomb constant itself to $1$ is equivalent. – Double AA Jun 2 '14 at 14:05
• @DoubleAA I talked to Jamal about this as well ;) his point was just that it's called the inverse Coulomb constant. Mathematically, it's irrelevant, and he realizes that too :) – Danu Jun 2 '14 at 14:39
• Ahh I hadn't seen the edit history. Why not just list it as $c, k_B$ and $\frac{1}{4 \pi \epsilon _0}$ and call it the Coulomb constant? – Double AA Jun 2 '14 at 14:40
• @DoubleAA edit your heart out ;) – Danu Jun 2 '14 at 16:18
• @eqb setting $\pi$ to 1 is not 'ok'. It is a dimensionless constant already. – Danu Jun 3 '14 at 6:11

Planck's constant divided by the gravitational constant is 9.93x10^-24 kg s^2/m. This is not exactly 10^-23, and even if it was there is nothing special about 10^-23 kg s^2/m, since the number is dependent in the conventions used to define the units.

Express the Planck constant as $2.34\times 5^{-48}$ and the gravitational constant as $2.01\times 5^{-15}$ and the coincidence goes away. Physics doesn't care that we have ten fingers.

There is nothing special about the number 10 (nor about the number 23, though not everybody agrees about that)

• His emphasis was on 6.6 not 23. – user28737 Jun 2 '14 at 13:56

You know that SI units are arbitrary and not based on physical observations. I mean, kilogram, meter and second are arbitrary quantities based on what was reasonable in the 1700's. Their exact definition has been updated, but the amount their represent is arbitrary and bares no significance in nature.

So there nothing special about any quantity defined in SI units. Now use, plank time, plank mass and speed of light to compare quantities and you might have something there.

The $G$ constant can be related to Planck's various constants,

$$G=\frac{\ell_P^3}{m_P\cdot t_P^2}$$

where $\ell_P$ is the Planck length, $m_P$ is the Planck mass, and $t_P$ the Planck time.

G = 6.673x10^-8 cm^3/g sec^2

volume per mass acceleration

Its an interesting correlation......

• That's not an interesting correlation at all, given that the Planck units are defined by using $G$, $c$, $\hbar$, $k_e$, and $k_B$. – Kyle Kanos Aug 8 '14 at 16:52