Where does $\ell_{g} = \frac{\hbar^{2}}{2GMm^{2}}$ come from? $$\ell_{g} = \frac{\hbar^{2}}{2GMm^{2}}$$
This is an equation for the characteristic length scale that contains the constants $\hbar$ and $G$. 
My question is where does this equation come from and if it has a significance rather than being  just dimensionally correct?
 A: We have
$$
l_g = \frac{\hbar^{2}}{2GMm^{2}} = \frac{\left( \frac{\hbar}{mc} \right)^2}{\frac{2GM}{c^2}} = \frac{{ƛ}^2}{r_s}
$$
ie the square of the reduced Compton wavelength of a particle of mass $m$ divided by the Schwarzschild radius of a body of mass $M$.
At a guess, it's something to do with field theory on curved backgrounds, but I don't know for sure.
A: It's one half of the analogue of Bohr radius for a (quantum) particle of mass $m$ orbiting a black hole of mass $M$.
The usual Bohr radius is defined as: $$ a_0 = \frac{4 \pi \varepsilon_0 \hbar^2}{m_{\text{e}} e^2}$$
We can recognize $e^2/(4\pi \varepsilon_0)$ as a coefficient in Coulomb law for the force between electron and proton:
$$
|F_{12}|=\frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2}.$$
Now, the gravitational force for the particle orbiting a black hole (assuming Newtonian limit is applicable):
$$
|F_{12}|=\frac{G M m}{r^2}.
$$
Schrödinger equation for the nonrelativistic particle in the Newtonian field of a point mass is equivalent to equation for the hydrogen atom under substitution of $G M m $ with $e^2/(4\pi \varepsilon_0)$, so its solutions are characterized by the “gravitational Bohr radius”:
$$ a_g = \frac{\hbar^2}{GMm^2}.$$ 
OP's $\ell_g$ is one half of this value.
Applicability of Newtonian approximation could be checked by comparing $a_g$ (or $\ell_g$) with the Schwarzschild radius $r_s=2GM/c^2$. If $a_g\gg r_s$ then it is possible to have a black hole atom, quantum particle orbiting the black hole with the size of order $a_g$ described reasonably well by nonrelativistic theory.
