Chapter 33, Problem 38GP

To determine

Power output of the star.

Answer to Problem 38GP

Power output of the star is $2.7\times {10}^{25}\text{ W}$

Explanation of Solution

Given:

The mass of a neutron star is 1.5 times that of the Sun.

Diameter of the neutron star is $20\text{ km}$ with the rotational speed of $1.0\text{ rev/s}$ .

Loss in the rotational kinetic energy is at the rate of 1 part in ${10}^9$ per day.

Formula used:

Power output is calculated as

  $\text{P = }\frac{\Delta \text{KE}}{\Delta \text{t}}$

Calculation:

Power output of the star is

  $\begin{array}{c}\text{P = }\frac{\Delta \text{KE}}{\Delta \text{t}}\\ =\frac{\frac{1}{2}\text{I}{\omega }^2(\text{fraction lost})}{\Delta \text{t}}\\ =\frac{\frac{1}{2}\frac{2}{5}{\text{MR}}^2{\omega }^2(\text{fraction lost})}{\Delta \text{t}}\\ \text{ = }\frac{1}{5}\frac{(1.5)(1.99\times {10}^{30}\text{ kg}){(1.0\times {10}^4\text{ m})}^2{(2\pi \text{ rad/s})}^2(1\times {10}^{-9})}{(1\text{d})(24\text{h/d})(3600\text{s/h})}\\ \text{ = }2.7\times {10}^{25}\text{ W}\end{array}$

Conclusion:

Thus, power output of the star is $2.7\times {10}^{25}\text{ W}$