Question

Certain neutron stars (extremely dense stars) are believed to be rotating at about$1\text{rev}/\text{s}$. If such a star has a radius of$20\text{\hspace{0.17em}km}$, what must be its minimum mass so that material on its surface remains in place during the rapid rotation?

Short answer

Mass of the material on the surface of the neutron star is$5\times {10}^{24}\text{\hspace{0.33em}kg}$

Step-by-step solution

The given data 

1) Frequency of rotation of neutron star is

$f=1\frac{rev}{s}$

2)Radius of neutron star is$R=20000\text{\hspace{0.33em}m}$

Understanding the concept of Differential acceleration equation 

We use the acceleration difference equation to find the mass of the material on the surface of the neutron star.

Formula:

Acceleration difference equation due to revolution,$g=a_g-{\omega }^{2}R$ ... (i)

Calculation of the mass of the material 

From equation (i), we can get

$a_g=\frac{GM}{R^2}$

To stay stationary on the surface of the neutron star, forces should be balanced so that, g= 0.

Hence substituting g=0 in equation (i), we get

$\begin{array}{c}0=\frac{GM}{R^2}-{\omega }^{2}R\\ \frac{GM}{R^2}={\omega }^{2}R\\ M=\frac{{\omega }^2{R}^3}{G}\end{array}$

Substituting the values of$\omega =2\pi f\text{and\hspace{0.33em}}R=20000\text{\hspace{0.33em}m}$we get,

$\begin{array}{c}M=\frac{{\left(2\pi \times 1\frac{\text{rev}}{\text{s}}\right)}^2{\left(20000\text{\hspace{0.17em}\hspace{0.17em}m}\right)}^3}{6.67\times {10}^{-11}{\text{\hspace{0.17em}\hspace{0.17em}Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}}\\ =4.7\times {10}^{24}\\ \approx 5\times {10}^{24}\text{\hspace{0.33em}kg}\end{array}$

The mass of the material on the neutron star should be$5\times {10}^{24}\text{\hspace{0.33em}kg}$.