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 8.0×105 km (comparable to our sun); its final radius is 16 km. the original star rotated once in 35 days, find the angular speed of the neutron star.

Answer 1

Answer:

The angular speed wf of the neutron star is calculated to be $5.19*10^{3}$ rad/s

Explanation:

The reason for such a rapid spin-rate is due to the principle of angular momentum. The angular momentum of a system can be given as:

Angular Momentum (L) = Mass * Radius^2 * Angular Velocity (w)

Applying this principle to our context, we would say that the angular momentum of the star before and after collapsing is constant. In order to not break this principle, we know that the mass of the star did not change but the radius shrank by a significant amount after collapsing, and so in order to keep the angular momentum (L) constant after collapse, the star had to increase it's angular velocity, which is evident in our answer.

The calculations of the answer are as follows:

Star's Initial Radius Ri = $8.0 * 10^{5}$ km ( $8.0 * 10^{8}$ m)

Star's Initial Angular Velocity wi = $\frac{2\pi} {35 days * 24 hrs * 3600 sec}$ = $2.077 * 10^{-6}$ rad/sec

Star's final radius Rf = $16 * 10^{3}$ m

Now, we can equate the initial and final states of the star i.e. the angular momentum of star before and after the collapse as following:

Li = Lf (where i and f denote initial and final state)

Solving of Final Angular Velocity we have:

wf = wi * (Ri / Rf) ^ 2

Plugging in our known values:

wf = $2.077 * 10^{-6}$ x $(\frac{8 * 10^{8}}{16 * 10^{3}} )^{2}$ = 5.19 * 10^3 rad/s