Chapter 33, Problem 37GP

To determine

Change in the rotational kinetic energy when a star collapses to a neutron star.

Answer to Problem 37GP

Final rotational kinetic energy is $4.9\times {10}^9$ times the initial kinetic energy.

Explanation of Solution

Given:

Final angular velocity of the star is $4.9\times {10}^9\text{ rev/month}$

Formula used:

Rotational kinetic energy is calculated as

  $KE=\frac{1}{2}I{\omega }^2$

Calculation:

Change in rotational kinetic energy when a star collapses to a neutron star is

  $\begin{array}{c}\frac{K{E}_{final}}{K{E}_{initial}}=\frac{\frac{1}{2}{I}_{final}{\omega }_{final}^2}{\frac{1}{2}{I}_{initial}{\omega }_{initial}^2}\\ =\frac{\frac{2}{5}M{R}_{final}^2{\omega }_{final}^2}{\frac{2}{5}M{R}_{initial}^2{\omega }_{initial}^2}\\ ={\left(\frac{R_{final}{\omega }_{final}}{R_{initial}{\omega }_{initial}}\right)}^2\\ \frac{K{E}_{final}}{K{E}_{initial}}\text{=}{\left(\frac{(1\times {10}^4\text{ m})(4.9\times {10}^9\text{ rev/month})}{(7\times {10}^8\text{ m})(1\text{ rev/month})}\right)}^2\\ =4.9\times {10}^9\end{array}$

Conclusion:

Thus, final rotational kinetic energy is $4.9\times {10}^9$ times the initial kinetic energy.