We can estimate the quantized energy levels of the Earth's orbit by analogy with the hydrogen atom since both are inverse square forces--just with different constants. For hydrogen: $$E_n = -\frac{m_e}{2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2\frac{1}{n^2\hbar^2}$$ Replacing $m_e$ with the mass of Earth ($m$) and the parenthesized expression with the corresponding expression from the gravitational force ($GMm$, where $M$ is the mass of the sun and $G$ is the gravitational constant) to get $$E_n = -\frac{m}{2}\left(GMm\right)^2\frac{1}{n^2\hbar^2}$$ Setting this equal to the total orbital energy $$E_n = -\frac{m}{2}\left(GMm\right)^2\frac{1}{n^2\hbar^2} = -\frac{GMm}{2r}$$ Solving for $n$ and plugging in values gives: $$n = \frac{m}{\hbar}\sqrt{GMr} = 2.5\cdot 10^{74}$$ The fact that Earth's energy level is at such a large quantum number means that any energy transition (which are proportional to $1/n^2$) will be undetectably small.
Solving for $r$: $$r = n^2\left(\frac{\hbar}{m}\right)^2\frac{1}{GM}$$ An increase in the principal quantum number ($n$) by one results in a change in orbital distance of \begin{align} \Delta r &= \left[(n+1)^2 - n^2\right]\left(\frac{\hbar}{m}\right)^2\frac{1}{GM} \\ &= \left[2n + 1\right]\left(\frac{\hbar}{m}\right)^2\frac{1}{GM} \\ &= 1.2\cdot 10^{-63}\ \textrm{meters} \end{align} Again, way too small to measure.