Consider the earth–sun system as a gravitational analog to the hydrogen atom.

(a) What is the potential energy function (replacing Equation 4.52)? (Let be the mass of the earth, and M the mass of the sun.)

$\mathit{V}\left(r\right)\mathbf{=}\mathbf{-}\frac{{\mathbf{e}}^{\mathbf{2}}}{\mathbf{4}\mathbf{\pi }{\displaystyle \stackrel{\mathbf{,}}{{\mathbf{0}}_{\mathbf{0}}}}}\frac{\mathbf{1}}{\mathbf{r}}$

(b) What is the “Bohr radius,”${\mathit{a}}_{\mathbf{g}}$,for this system? Work out the actual number.

(c) Write down the gravitational “Bohr formula,” and, by equating ${\mathbf{E}}_{\mathbf{n}}$to the classical energy of a planet in a circular orbit of radius ${\mathit{r}}_{\mathbf{0}}$, show that $\mathit{n}\mathbf{=}\sqrt{{\mathbf{r}}_{\mathbf{0}}\mathbf{/}{\mathbf{a}}_{\mathbf{g}}}$.From this, estimate the quantum number n of the earth.

(d) Suppose the earth made a transition to the next lower level$\left(n-1\right)$ . How much energy (in Joules) would be released? What would the wavelength of the emitted photon (or, more likely, gravitation) be? (Express your answer in light years-is the remarkable answer a coincidence?).

A potential function is a function of the position of an object. It can be defined only for conservative forces. A forces is conservative if the work it does on n object only on the initial and final position of the object and net on the path. The gravitational force is a conservative force.

The potential is given by:

$V\left(r\right)=-\frac{e^2}{4\pi {ϵ}_0}\frac{1}{r}$

Consdier the Earth-sun system as a gravitational analog to the hydrogen atom. The potential is:

$V\left(r\right)=-G\frac{Mm}{r}m/r.$

Where G is the gravitational constant, $G=6.673\times {10}^{-11}{m}^{3}k{g}^{-1}{s}^{-2}$

translates hydrogen results to the gravitational analogs.

The Bohr radius for the Earth can be found by replacing$e^2$ by$mM$ and$1/4{\mathrm{\pi \epsilon }}_0$ by G ,so we get:

$a_g=\frac{{\hslash }^2}{GM{m}^2}$

Substitute with the numerical values to get:

$$\begin{gathered} a_g=\frac{{\left(1.0546\times {10}^{-34}Js\right)}^2}{\left(6.6726\times {10}^{-11}m/kg.s^2\right)\left(1.9892\times {10}^{30}kg\right){\left(5.98\times {10}^{24}kg\right)}^2} \\ {a}_g=2.34\times {10}^{-138}m \end{gathered}$$

Equation 4.70 (the allowed energies) is given by:

$\Rightarrow E_n=-\left[\frac{m}{2{ħ}^2}(GMm{)}^2\right]\frac{1}{n^2}$

There are many items that could be altered to test the recreation of another. These changing quantities are called variables. A variables is any factor, trait, or condition that can exist in differing amounts or types. An experiment usually amounts or types, An experiment usually has three kinds of variables: independent, dependent, and controlled.

$$\begin{gathered} E_c=\frac{1}{2}m{v}^2-G\frac{Mm}{r_o} \\ G\frac{Mm}{r_o^2}=\frac{m{v}^2}{r_o} \\ \Rightarrow \frac{1}{2}m{v}^2 \\ =\frac{GMm}{2r_o} \end{gathered}$$

So,

$c{E}_c=-\frac{GMm}{2r_o}$

$$\begin{gathered} =-\left[\frac{m}{2{\hslash }^2}(GMm{)}^2\right]\frac{1}{n^2} \\ \Rightarrow n^2=\frac{GM{m}^2}{{\hslash }^2}{r}_o \\ =\frac{r_o}{a_g} \end{gathered}$$

$\Rightarrow n=\sqrt{\frac{r_o}{a_g}}.$

$$\begin{gathered} r_o=\mathrm{earth}-\mathrm{sundistance}=1.496\times {10}^{11}m \\ \Rightarrow n=\sqrt{\frac{1.496\times {10}^{11}}{2.34\times {10}^{-138}}} \\ =2.53\times {10}^{74}{r}_0 \\ =\mathrm{earth}-\mathrm{sun}\mathrm{distance} \\ =1\mathrm{m}. \end{gathered}$$

The wavelength of the emitted photon

$∆E=-\left[\frac{G^2{M}^2{m}^3}{2{ħ}^2}\right]\left[\frac{1}{{\left(n+1\right)}^2}-\frac{1}{n^2}\right].\frac{1}{{\left(n+1\right)}^2}=\frac{1}{n^2\left(1+1/n^2\right)}\approx \frac{1}{n^2}\left(1-\frac{2}{n}\right)$

So,

role="math" localid="1658389122151" $$\begin{gathered} \left[\frac{1}{(n+1{)}^2}-\frac{1}{n^2}\right]\approx \frac{1}{n^2}\left(1-\frac{2}{n}-1\right)=-\frac{2}{n^3}; \\ \Delta E=\frac{G^2{M}^2{m}^3}{{\hslash }^2{n}^3} \\ ∆E=\frac{{\left(6.67\times {10}^{-11}\right)}^2{\left(1.99\times {10}^{30}\right)}^2{\left(5.98\times {10}^{24}\right)}^3}{{\left(1.055\times {10}^{-34}\right)}^2{\left(2.53{\times }^{74}\right)}^3} \\ =2.09\times {10}^{-41}J \end{gathered}$$

$$\begin{gathered} E_p=∆E=hv=\frac{hc}{\lambda } \\ \lambda =\left(3\times {10}^8\right)\left(6.63\times {10}^{-34}\right)/\left(2.09\times {10}^{-41}\right) \\ =9.52\times {10}^{15}m \end{gathered}$$

$$\begin{gathered} \mathrm{But}1ly=9.46\times {10}^{15}m.that\lambda \approx 1ly,n^2=GM{m}^2{r}_o/{\hslash }^2 \\ \lambda =\frac{ch}{∆E} \\ =c2\mathrm{\pi ħ}\frac{{\mathrm{ħ}}^2{\mathrm{n}}^3}{{\mathrm{G}}^2{\mathrm{M}}^2{\mathrm{m}}^3} \\ =\mathrm{c}\frac{2{\mathrm{\pi ħ}}^2}{{\mathrm{G}}^2{\mathrm{M}}^2{\mathrm{m}}^3}{\left(\frac{{\mathrm{GMm}}^2{\mathrm{r}}_0}{{\mathrm{ħ}}^2}\right)}^{3/2} \\ =\mathrm{c}\left(2\mathrm{\pi }\sqrt{\frac{{\mathrm{r}}_0^3}{\mathrm{GM}}}\right) \end{gathered}$$

$\mathrm{But}(\mathrm{from}(\mathrm{c}\left)\right)$

$v=\sqrt{GM/r_o}=2\pi r_o$, where T is the period of the orbit (in this case one year),

so $T=2\pi \sqrt{r_o^3/GM}$and hence ${}^{\lambda =cT}$(one light year).