Question

Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star. The density of a neutron star is roughly 10^14 times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was 7.0×10^5 km (comparable to our sun); its final radius is 18km .

Answer 1

Answer:

The angular speed of the neutron star is 3130.5 rad/s.

Explanation:

Given that,

Initial radius $r_{1}=7\times10^{5}\ km$

Final radius $r_{2}=18 km$

Density of a neutron $\rho= 10^{14}$

Equal masses of two stars $m_{1}=m_{2}$

Suppose, If the original star rotated once in 35 days, find the angular speed of the neutron star

Time period of original star T =  35 days = 3024000 s

We need to calculate the initial angular speed of original star

Using formula of angular star

$\omega=\dfrac{2\pi}{T}$

Put the value into the formula

$\omega_{1}=\dfrac{2\pi}{3024000}$

$\omega_{1}=0.00207\times10^{-3}\ rad/s$

Let the initial moment of inertia of the star is

$I_{1}=m_{1}r_{1}^2$

Final moment of inertia of the star is

$I_{2}=m_{2}r_{2}^2$

From the conservation of angular momentum

$I_{1}\omega_{1}=I_{2}\omega_{2}$

$\omega_{2}=\dfrac{I_{1}\omega_{1}}{I_{2}}$

$\omega_{2}=\dfrac{m_{1}r_{1}^2\omega_{1}}{m_{2}r_{2}^2}$

Put the value into the formula

$\omega_{2}=\dfrac{(7.0\times10^{5})^2\times0.00207\times10^{-3}}{18^2}$

$\omega_{2}=3130.5\ rad/s$

Hence, The angular speed of the neutron star is 3130.5 rad/s.