A typical neutron star (what is left behind when a star goes supernova) has a ma

Question

A typical neutron star (what is left behind when a star goes supernova) has a mass of 1.4 solar masses and a radius of roughly 12 km. Just after formation, a neutron star can be spinning at a rate of 500 turns per second or so. Estimate the total rotational kinetic energy of the neutron star (describe any approximations or assumptions you make). If this energy were somehow converted to other forms of energy at the same rate as the sun radiates energy (which is roughly 3.9x10^26 J/s), how long would it take to bring the neutron star to zero angular speed?

Explanation / Answer

mass,M = 1.4*1.9891 × 10^30 = 2.785*10^30 kg
radius,r = 12 km = 12000 m
angular speed,w = 500 round per second = 500*2pi = 1000 rad/s

moment of inertia of a solid sphere,I = (2/5)Mr^2 = 1.6042*10^38 kg.m^2

rotational KE = 0.5*Iw^2 = 7.92*10^44 J

if rate of energy dissipation = 3.9*10^26 J/s

then time taken for complete energy dissipation = time taken to bring neutron star to zero angular speed = 7.92*10^44 / 3.9*10^26 = 2.031*10^18 s = 6.44*10^10 years Ans

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