Answer
See the detailed answer below.
Work Step by Step
We can use the same formula of the electric force of Bohr orbits with some editing.
We know, for the nth Bohr orbit, that
$$E_n=\dfrac{-2\pi^2Z^2e^4m_ek^2}{h^2n^2}$$
To find the energy if the gravitational force is applied, we need to replace $kZe^2$ with $Gm_Em_m$ where $G$ is the universal gravitational constant, $m_E$ is the mass of the Earth (rather than the mass of the proton), and $m_m$ is the mass of the moon (rather than the mass of the electron).
Thus,
$$E_n=\dfrac{-2\pi^2m_m( k^2Z^2e^4)}{h^2n^2}=\dfrac{-2\pi^2m_m( k Z e^2)^2}{h^2n^2}$$
Replacing;
$$E_n= \dfrac{-2\pi^2m_m( Gm_mm_E)^2}{h^2n^2}$$
$$E_n= \dfrac{-2\pi^2G^2m_m^3 m_E^2}{h^2n^2}$$
Plugging the known;
$$E_n= \dfrac{-2\pi^2(6.672\times10^{-11})^2 (7.35\times10^{22})^3 (5.98\times10^{24})^2}{(6.626\times10^{-34})^2n^2}$$
$$E_n=\dfrac{\color{red}{\bf -2.84186\times10^{165}}}{n^2}\tag 1$$
Now we need to find the radius, of the nth Bohr orbit, by the same approach;
$$r_n =\dfrac{h^2n^2}{4\pi^2 m_m(kZe^2)}$$
$$r_n =\dfrac{h^2n^2}{4\pi^2 m_m (Gm_mm_E)}$$
$$r_n =\dfrac{h^2n^2}{4\pi^2G m_m^2 m_E }\tag 2$$
Plugging the known;
$$r_n =\dfrac{(6.626\times10^{-34})^2n^2}{4\pi^2 (6.672\times10^{-11}) (7.35\times10^{22})^2 (5.98\times10^{24})}$$
$$r_n=\color{red}{\bf 5.15954\times10^{-129}}n^2$$
To find if the quantization of the energy and radius is apparent here or not, we need to plug the real distance between the Earth and the moon into (2) and solving for $n$;
$$n^2=\dfrac{4\pi^2G m_m^2 m_Er_{\rm E\;to\;m} }{h^2 } $$
$$n =\sqrt{\dfrac{4\pi^2G m_m^2 m_Er_{\rm E\;to\;m} }{h^2 } }$$
Plugging the known;
$$n =\sqrt{\dfrac{4\pi^2 (6.672\times10^{-11}) (7.35\times10^{22})^2 (5.98\times10^{24}) (3.84\times10^8)}{ (6.626\times10^{-34})^2 } }$$
$$n=2.7281\times10^{68}\approx 10^{69}$$
Therefore, the quantization of energy and radius here is not apparent.
