Question

Given the following information, what would the star's rotation speed be?

Suppose a star the size of our Sun, but with mass 3.0 times as great, were rotating at a speed of 1.0 revolution every 11 days. If it were to undergo gravitational collapse to a neutron star of radius 15 km, losing three-quarters of its mass in the process, what would its rotation speed be

Assume that the star is a uniform sphere at all times, and that the lost mass carries off no angular momentum.

Answer 1

`I_1\omega_1 = I_2\omega_2 \qquad => \omega_2 = (I_1/I_2)\omega_1`
`\qquad \quad \omega_2= (8.6118\times10^9)\times(6.611\times10^{-6}) = 5.693\times10^{4} (rad)/s`

Explanation:

Angular Momentum Conservation: In the absence of an external torque, the net angular momentum of a system remains conserved.

`L = I\omega = const \qquad => I_1\omega_1 = I_2\omega_2`

`I_1, I_2` - Moment of inertia before and after collapse.
`\omega_1, \omega_2` - Angular speed before and after collapse.
`T_1, T_2` - Rotational time period before and after collapse.

The Moment-of-Inertia of a sphere of mass `M` and radius `R` spinning along an axis passing through its centre is :
`I=2/5MR^2`

`I_1/I_2 = (2/5 M_1R_1^2) / (2/5 M_2R_2^2) = (M_1/M_2)(R_1/R_2)^2`
`\qquad = 4((6.96\times10^5\quad km)/(15 \quad km))^2 = 8.6118\times10^9`

`M_1 = M_{sun}; \qquad M_2 = M_{sun}/4; \qquad`
#R_1 = R_{sun} = 6.96\times10^5\quad km; \qquad R_2 = 15\quad km;#
`\omega_1 = (2\pi)/T_1 = (2\pi)/(11 \quad days)= ((2\pi)/(11\times24\times3600\quad s))`
`\qquad =6.611\times10^{-6}\quad rad.s^{-1}`

`I_1\omega_1 = I_2\omega_2 \qquad => \omega_2 = (I_1/I_2)\omega_1`
`\qquad \quad \omega_2= (8.6118\times10^9)\times(6.611\times10^{-6}) = 5.693\times10^{4} (rad)/s`

`T_2 = (2\pi)/omega_2 = 0.11\times10^{-3} \quad s = 0.11 \quad milli-seconds`

A milli-second Pulsar?