Question

Suppose a star the size of our Sun, but with mass 8.0 times as great, were rotating at a speed of 1.0 revolution every 9.0 days. If it were to undergo gravitational collapse to a neutron star of radius 12 km, losing\(\frac{3}{4}\)of its mass in the process, what would its rotation speed be? Assume the star is a uniform sphere at all times. Assume also that the thrown off mass carries off either (a) no angular momentum, or (b) its proportional share\(\left( {\frac{3}{4}} \right)\)of the initial angular momentum.

Short answer

(a) The rotational speed of the star after collapsing in the absence of momentum is \(108724\;{\rm{rad/s}}\).

(b) The rotational speed of the star after collapsing when it carries a proportional share of the thrown-off mass’s angular momentum is \(81543\;{\rm{rad/s}}\).

Step-by-step solution

Identification of the given data

Let the mass of the sun be\({M_{\rm{s}}}\).

The mass of the star is\({m_{\rm{i}}} = 8{M_{\rm{s}}}\).

The radius of the sun is\({r_{\rm{i}}} = 6.96 \times {10^8}\;{\rm{m}}\)

The initial angular speed of the star is:

\({\omega _{\rm{i}}} = \frac{{1\;{\rm{rev}}}}{{9\;{\rm{days}}}} \times \frac{{2\pi \;{\rm{rad}}}}{{1\;{\rm{rev}}}} \times \frac{{1\;{\rm{day}}}}{{86400\;{\rm{s}}}} = 8.08 \times {10^{ - 6}}\;{\rm{rad/s}}\)

The radius of the neutron star is\({r_{\rm{f}}} = 12\;{\rm{km}} = 12000\;{\rm{m}}\).

The mass of the neutron star is \({m_{\rm{f}}} = \left( {1 - \frac{3}{4}} \right) \times 8{M_{\rm{s}}} = 2{M_{\rm{s}}}\).

Understanding the law of conservation of angular momentum

In rotational dynamics, if no external torque is acting in a system, the change in angular momentum is zero and the angular momentum remains conserved.

\({L_{{\rm{initial}}}} = {L_{{\rm{final}}}}\)

It can be written in terms of the moment of inertia and angular velocity as:

\({\left( {I\omega } \right)_{{\rm{initial}}}} = {\left( {I\omega } \right)_{{\rm{final}}}}\)

(a) Calculation of the angular speed of the star after collapsing when the thrown-off mass has no angular momentum

The moment of inertia of the sphere is\(I = \frac{2}{5}m{r^2}\). Applying the law of conservation of angular momentum, you get:

\(\begin{aligned}{c}{\left( {I\omega } \right)_{{\rm{initial}}}} = {\left( {I\omega } \right)_{{\rm{final}}}}\\\frac{2}{5}{m_{\rm{i}}}r_{\rm{i}}^2{\omega _{\rm{i}}} = \frac{2}{5}{m_{\rm{f}}}r_{\rm{f}}^2{\omega _{\rm{f}}}\\{m_{\rm{i}}}r_{\rm{i}}^2{\omega _{\rm{i}}} = {m_{\rm{f}}}r_{\rm{f}}^2{\omega _{\rm{f}}}\\{\omega _{\rm{f}}} = \frac{{{m_{\rm{i}}}r_{\rm{i}}^2{\omega _{\rm{i}}}}}{{{m_{\rm{f}}}r_{\rm{f}}^2}}\end{aligned}\)

Substitute the known numerical values in the above expression.

\(\begin{aligned}{c}{\omega _{\rm{f}}} = \frac{{\left( {8{M_{\rm{s}}}} \right){{\left( {6.96 \times {{10}^8}\;{\rm{m}}} \right)}^2}\left( {8.08 \times {{10}^{ - 6}}\;{\rm{rad/s}}} \right)}}{{\left( {2{M_{\rm{s}}}} \right){{\left( {12000\;{\rm{m}}} \right)}^2}}}\\ = 108724\;{\rm{rad/s}}\end{aligned}\)

(b) Calculation of the angular speed of the star after collapsing when the thrown-off mass has an angular momentum

In this case, the thrown-off mass carries its proportional share\(\left( {\frac{3}{4}} \right)\)of the initial angular momentum. Thus, the angular speed of the star after collapsing will be

\(\begin{aligned}{c}{\omega _{\rm{f}}}^\prime = \frac{3}{4}{\omega _{\rm{f}}}\\ = \frac{3}{4} \times 108724\;{\rm{rad/s}}\\ = 81543\;{\rm{rad/s}}\end{aligned}\)