Chapter 8, Problem 63P

To determine

The rotation rate of Sun. Also, the final kinetic energy in terms of initial kinetic energy.

Answer to Problem 63P

The new rotation rate of Sun is 129.6 s

Final KE of Sun is 20000 times the original kinetic energy.

Explanation of Solution

Given:

Original mass of Sun, M

New mass of Sun, $\frac{M}{2}$

Original radius, R

New radius, 0.01R

Original rotation period, $T_1=30days$

Formula used:

Moment of inertia of spherical body, $I=\frac{2}{5}M{R}^2$

Angular velocity, $\omega =\frac{2\pi }{T}$

Law of conservation of angular momentum,

  $\begin{array}{l}{L}_1=L_2\\ I_1{\omega }_1=I_2{\omega }_2\\ I_1\frac{2\pi }{T_1}=I_2\frac{2\pi }{T_2}\\ \frac{I_1}{T_1}=\frac{I_2}{T_2}\end{array}$

Kinetic energy of rotating body, $K=\frac{1}{2}I{\omega }^2=\frac{L^2}{2I}$

Calculation:

Substituting values for moment of inertia,

  $\begin{array}{l}{I}_1=\frac{2}{5}M{R}^2\\ $ I_2=\frac{2}{5}\left(\frac{M}{2}\right){(0.01R)}^2=\frac{0.0001M{R}^2}{5}$\end{array}$

Substituting values in law of conservation of angular momentum formula,

  $\begin{array}{l}\frac{\frac{2}{5}M{R}^2}{30}=\frac{\frac{0.0001M{R}^2}{5}}{T_2}\\ T_2=0.0015days\\ T_2=(0.0015)(86400)=129.6s\end{array}$

By law of conservation of angular momentum, $L_1=L_2=L$

Substituting values for kinetic energy,

  $K{1}_{}=\frac{L^2}{2\left(\frac{2}{5}M{R}^2\right)}=\frac{5L^2}{4M{R}^2}$

  $K_2=\frac{L^2}{2\left(\frac{0.0001M{R}^2}{5}\right)}=\frac{5L^2}{0.0002M{R}^2}$

Taking ratio of kinetic energies,

  $\begin{array}{l}\frac{K_2}{K_1}=\frac{\frac{5L^2}{0.0002M{R}^2}}{\frac{5L^2}{4M{R}^2}}\\ \frac{K_2}{K_1}=20,000\\ K_2=20000K_1\end{array}$

Conclusion:

The new rotation rate of Sun is 129.6 s

Final KE of Sun is 20000 times the original kinetic energy.