Return to Answer

The orbital energy of the Earth around the Sun is quantized. Measuring this quantization directly is infeasible, as I'll show below, but other experiments with bouncing neutrons show that motion in a classical gravity field is subject to energy quantization.

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 $1/n^3$) will be undetectably small.

In fact, to transition to the next energy level, Earth would have to absorb: $$\Delta E_{n \to n+1} = m\left(GMm\right)^2\frac{1}{n^3\hbar^2} = 2\cdot 10^{-41}\ \textrm{J} = 1\cdot 10^{-22}\ \textrm{eV}$$ For a sense of how little this energy is, a photon of this energy has a wavelength of $10^{16}$ meters--or, one light-year.

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.

The orbital energy of the Earth around the Sun is quantized. Measuring this quantization directly is infeasible, as I'll show below, but other experiments with bouncing neutrons (Nature paper) show that motion in a classical gravity field is subject to energy quantization.

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 $1/n^3$) will be undetectably small.

In fact, to transition to the next energy level, Earth would have to absorb: $$\Delta E_{n \to n+1} = m\left(GMm\right)^2\frac{1}{n^3\hbar^2} = 2\cdot 10^{-41}\ \textrm{J} = 1\cdot 10^{-22}\ \textrm{eV}$$ For a sense of how little this energy is, a photon of this energy has a wavelength of $10^{16}$ meters--or, one light-year.

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.

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 $1/n^3$) will be undetectably small.

In fact, to transition to the next energy level, Earth would have to absorb: $$\Delta E_{n \to n+1} = m\left(GMm\right)^2\frac{1}{n^3\hbar^2} = 2\cdot 10^{-41}\ \textrm{J} = 1\cdot 10^{-22}\ \textrm{eV}$$ For a sense of how little this energy is, a photon of this energy has a wavelength of $10^{16}$ meters--or, one light-year.

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.

The orbital energy of the Earth around the Sun is quantized. Measuring this quantization directly is infeasible, as I'll show below, but other experiments with bouncing neutrons show that motion in a classical gravity field is subject to energy quantization.

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 $1/n^3$) will be undetectably small.

In fact, to transition to the next energy level, Earth would have to absorb: $$\Delta E_{n \to n+1} = m\left(GMm\right)^2\frac{1}{n^3\hbar^2} = 2\cdot 10^{-41}\ \textrm{J} = 1\cdot 10^{-22}\ \textrm{eV}$$ For a sense of how little this energy is, a photon of this energy has a wavelength of $10^{16}$ meters--or, one light-year.

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.

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.

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 $1/n^3$) will be undetectably small.

In fact, to transition to the next energy level, Earth would have to absorb: $$\Delta E_{n \to n+1} = m\left(GMm\right)^2\frac{1}{n^3\hbar^2} = 2\cdot 10^{-41}\ \textrm{J} = 1\cdot 10^{-22}\ \textrm{eV}$$ For a sense of how little this energy is, a photon of this energy has a wavelength of $10^{16}$ meters--or, one light-year.

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.