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 1014 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 7.0x10^5 km (comparable to our sun); its final radius is 16 km .If the original star rotated once in 35 days, find the angular speed of the neutron star.

Answer 1

Answer:

The angular speed of the neutron star is $3976.978\ rads^{-1}$

Solution:

As per the question:

Initial radius of the star, r = $7.0\times 10^{5}\ km = 7\times 10^{8}\ m$

Final Radius, r' = 16 km

Period of rotation of the star, T = 35 days = $35\times 24\times 3600 = 3024000\ s$

Now,

To calculate the angular speed of the neutron star:

We use the principle of conservation of angular momentum:

$I\omega = I'\omega'$

where

I = Moment of inertia

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

$(7\times 10^{8})^{2}\times \frac{2\pi}{T} = (16000)^{2}\omega'$

$\omega = 3976.978\ rads^{-1}$