Question

The core of a star collapses during a supernova, forming a neutron star. Angular momentum of the core is conserved, and so the neutron star spins rapidly. If the initial core radius is\({\rm{5 \times 1}}{{\rm{0}}^{\rm{5}}}{\rm{ km}}\)and it collapses to\({\rm{10}}{\rm{.0 km}}\), find the neutron star’s angular velocity in revolutions per second, given the core’s angular velocity was originally\({\rm{1}}\)revolution per\({\rm{30}}\)days.

Short answer

The neutron star’s angular velocity in revolutions per second is obtained as: \({\rm{965 }}\frac{{{\rm{rev}}}}{{\rm{s}}}\).

Step-by-step solution

Angular momentum

The expression for the angular momentum is given by,

\(L = I\omega \)

Here\(L\)is the angular momentum,\(I\)is moment of inertia,\(\omega \)is the angular velocity.

The angular momentum can be conserved only when there is no external torque acting on the body.

Evaluating the neutron star’s angular velocity

Using the conservation of angular momentum, we then obtain:

\(\begin{array}{c}{{\rm{I}}_{\rm{1}}}{{\rm{\omega }}_{\rm{1}}}{\rm{ = }}{{\rm{I}}_{\rm{2}}}{{\rm{\omega }}_{\rm{2}}}\\{{\rm{\omega }}_{\rm{2}}}{\rm{ = }}\frac{{{{\rm{I}}_{\rm{1}}}}}{{{{\rm{I}}_{\rm{2}}}}}{{\rm{\omega }}_{\rm{1}}}\end{array}\)

As the moment of inertia is directly proportional to the square of the distance,

\(\begin{array}{c}{\omega _2}{\rm{ = }}{\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}{\omega _1}\\ = {\left( {\frac{{{\rm{5 \times 1}}{{\rm{0}}^{\rm{5}}}{\rm{\;km}}}}{{{\rm{10}}{\rm{.0\;km}}}}} \right)^{\rm{2}}}\frac{{{\rm{1rev}}}}{{{\rm{30\;d}}}}\\ = {\rm{8}}{\rm{.33 \times 1}}{{\rm{0}}^{\rm{7}}}\frac{{{\rm{1rev}}}}{{{\rm{1\;d}}}}\\ = {\rm{8}}{\rm{.33 \times 1}}{{\rm{0}}^{\rm{7}}}\frac{{{\rm{1rev}}}}{{{\rm{24 \times 3600\;s}}}}\\ = {\rm{965}}\,{\rm{rev/s}}\end{array}\)

Therefore, the neutron star’s angular velocity is \({\rm{965}}\,{\rm{rev/s}}\).