Chapter 8, Problem 74GP

To determine

To Find: The rotation speed of the star.

Answer to Problem 74GP

  $1.9\times {10}^5\text{ rad/s}$

Explanation of Solution

Given data:

Mass of star, $m_{star}=8\times m_{sun}$

Speed of $1\text{revolution}$ every $12\text{days}$ .

Radius of neutron star, $r=11\text{km}$

Formula used:

According to Angular momentum conservation,

  $I\omega =I_1{\omega }_1=I_2{\omega }_2$

  ${\omega }_2=\left(\frac{I_1}{I_2}\right){\omega }_1$

Where,

I₁, I₂− Moment of inertia before and after collapse,

${\omega }_{1,}{\omega }_2-$ Angular speed before and after collapse.

Calculation:

According to Angular momentum conservation,

  $I_1{\omega }_1=I_2{\omega }_2$

  ${\omega }_2=\left(\frac{I_1}{I_2}\right){\omega }_1$

  $\begin{array}{l}{M}_1=M_{sun}\\ M_2=M_{star}\end{array}$

  $M_2=\frac{M_{sun}}{8}$

  $R_1=R_{sun}=6.96\times {10}^5\text{km}$

  $R_2=R_{star}=11\text{ km}$

  $\begin{array}{l}{\omega }_1=\frac{2\pi }{t_1}\text{ rad/s}\\ {\omega }_1=\frac{2\pi }{12\text{days}}=\frac{2\pi }{12\times 24\times 3600\text{s}}\\ \end{array}$

  ${\omega }_1=0.00000606\text{ rad/s}$

Now, substituting the values in the equation of ${\omega }_2$ ,

  ${\omega }_2=\left(\frac{I_1}{I_2}\right){\omega }_1\text{ rad/s}$

  $I=\frac{2}{5}M{R}^2$ , similarly, for I₁, I₂

  ${\omega }_2=\frac{M_1{R}_1{}^2}{M_2{R}_2{}^2}$

Substituting all values of M₁, M₂ and R₁, R₂

  ${\omega }_2=8\left(\frac{6.96\times {10}^5}{11}\right)\times 0.00000606\text{ rad/s}$

  ${\omega }_2=1.9\times {10}^5\text{ rad/s}$

Conclusion: 

The rotational speed of the neon star is $1.9\times {10}^5\text{ rad/s}$ .