We know all that the gravitational constant is $$G=6.67428±0.00067\times 10^{-11}\mathrm{m^{3} \:kg^{-1} s^{-2}}.$$
But can we calculate it theoretically?
[Physics] Can the gravitational constant $G$ be calculated theoretically
gravitynewtonian-gravityphysical constants
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There is, in fact, a standard way of defining the gravitational constant in higher dimensions. We reason as follows:
We might try to generalize the formula for the gravitational force given by Newton's Law of Gravitation as you have written above, but this doesn't lead to natural generalization because it's unclear how the power of $r$ should be generalized to higher dimensions. On the other hand, we can rewrite this law in the form of a Poisson equation: \begin{align} \Delta \Phi = 4\pi G \rho. \end{align} Where $\Delta$ is the three-dimensional laplacian, $\Phi$ is the gravitational potential and $\rho$ is the mass density. Now, to generalize to higher dimensions, we simply assert that the Poisson equation still describes Newtonian gravitation in higher dimensions. This can be motivated by taking general relativity in higher dimensions and then taking the weak field limit. Then the gravitational constant is defined by the higher-dimensional Poisson equation; \begin{align} \Delta^{(D-1)}\Phi^{(D)} = 4\pi G^{(D)}\rho^{(D)} \end{align} My notation here is that $D$ is the number of spacetime dimensions, so $G^{(4)}$ is the standard gravitational constant. In particular, I write $\Delta^{(D-1)}$ to emphasize that there are only spatial derivatives in the laplacian; \begin{align} \Delta^{(D-1)} = \partial_1^2 + \cdots + \partial_{D-1}^2. \end{align}
The Laplacian has units of one over length squared in all dimensions, and $\Phi^{(D)}$ has units of energy over mass in all dimensions, so the units of the left hand side are dimension-invariant. This means that the units of the right hand side must also be dimension-invariant; \begin{align} [G^{(4)}\rho^{(4)}] = [G^{(D)}\rho^{(D)}] \end{align} But the units of $\rho^{(D)}$ are mass per unit spatial volume, $M/L^{D-1}$ in any spactime dimension $D$, so we get the following relationship between the $D$-dimensional gravitational constant and the $4$-dimensional gravitational constant: \begin{align} [G^{(D)}] = [G^{(4)}]\frac{[\rho^{(4)}]}{[\rho^{(D)}]} =[G^{(4)}]\frac{ML^{D-1}}{ML^3} = [G^{(4)}]L^{D-4}. \end{align} So, for example, the gravitational constant in $5$ spacetime dimensions has units of length times that the of gravitational constant in $4$ spacetime dimensions.
There is a nice discussion of this with more detail in section 3.8 of
A First Course in String Theory, Zwiebach (2nd Ed.)
Zwiebach also has a discussion of how the numerical vale of $G$ changes when you add extra compact spatial dimensions. For example, he demonstrates that with one extra compact dimension of length $\ell_C$, the five-dimensional gravitational constant becomes \begin{align} G^{(5)} = \ell_C G^{(4)} \end{align} Generally speaking, the value of the gravitational constant in higher dimensions depends on the sizes of these extra (compact) dimensions. If the extra dimensions are non-compact, I'm not exactly sure how one would proceed because one needs an extra characteristic length scale for each extra dimension.
$G$ 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).
Best Answer
You can't calculate the numerical value of Newton's constant from the first principle because it is a dimensionful constant – it has units – so the numerical value depends on the magnitude of the units. And because e.g. the kilogram is defined as the mass of a platinum prototype hosted by a French chateau (the kilogram has the "least objective" definition so far), it's clear that a "pure calculation" can't know how large the kilogram is, which also means that it can't determine the numerical value of Newton's constant which depends on the definition of a kilogram.
In other units, e.g. Planck units, people often set $G=1$ or $G=1/8\pi$. In that case, the constant may be calculated – I just did it. If one uses such units, there are other – dimensionless, and therefore potentially calculable from the first principles – constants of Nature such as the electron mass (in the unit of the Planck mass). String theory is the only framework in physics that allows one to calculate all these continuous dimensionless universal constants of physics. One may prove that for a given (stabilized) compactification of string theory, all of these constants are fully determined. In practice, physicists can't do that yet because they don't know how to choose the right compactification (which is just a discrete amount of currently uncertain information that must be inserted to the calculation).