Chapter 27, Problem 87GP

To determine

The allowed energies and radii of moon.

Answer to Problem 87GP

  $E_n=-\frac{G^2{M}^2{m}^4}{2n^2{\hslash }^2}$ , $r=\frac{n^2{\hslash }^2}{GM{m}^3}$

Explanation of Solution

Given:

The known distance between the Earth and Moon, is the quantization of the energy and the apparent radius.

Formula used:

Gravitational potential energy is given as: $PE=-\frac{GMm}{r}$

Kinetic energy is given as: $KE=\frac{1}{2}m{v}^2$

Calculation:

Gravitational force can be written as: $F=\frac{GMm}{r^2}$

Centripetal force is given as: $F_c=\frac{m{v}^2}{r}$

Equate both the forces,

  $\begin{array}{l}\frac{GMm}{r^2}=\frac{m{v}^2}{r}\\ \frac{GMm}{r}=m{v}^2\mathrm{...............................}(1)\end{array}$

Now, the total energy can be given as:

  $\begin{array}{c}{E}_T=KE+PE\\ =\frac{1}{2}m{v}^2+\left(-\frac{GMm}{r}\right)\end{array}$

Using equation (1),

  $\begin{array}{c}{E}_T=\frac{1}{2}m{v}^2+\left(-\frac{GMm}{r}\right)\\ =\frac{1}{2}\left(\frac{GMm}{r}\right)-\frac{GMm}{r}\\ =-\frac{1}{2}\left(\frac{GMm}{r}\right)\mathrm{........................}(2)\end{array}$

Using the Quantization rule to find the value of $r$ ,

  $\begin{array}{l}L=mvr=n\hslash \\ v=\frac{n\hslash }{mr}\end{array}$

From equation (1),

  $\begin{array}{l}m{\left(\frac{n\hslash }{mr}\right)}^2=\frac{GMm}{r}\\ r=\frac{n^2{\hslash }^2}{GM{m}^3}\end{array}$

So, the total energy can be given as:

  $\begin{array}{l}{E}_n=-\frac{1}{2}\left(\frac{GMm}{\raisebox{1ex}{$n^2{\hslash }^2$}\!\left/ \!\raisebox{-1ex}{$GM{m}^3$}\right.}\right)\\ E_n=-\frac{G^2{M}^2{m}^4}{2n^2{\hslash }^2}\end{array}$

Conclusion:

Thus, the energy and radius can be given as: $E_n=-\frac{G^2{M}^2{m}^4}{2n^2{\hslash }^2}$ and $r=\frac{n^2{\hslash }^2}{GM{m}^3}$