Question

Suppose a star the size of our Sun, but of mass 8.0 times as great, were rotating at a speed of 1.0 revolution every 12 days. If it were to undergo gravitational collapse to a neutron star of radius 12 km, losing 3/4 of its mass in the process, what would its rotation speed be? Assume the star is a uniform sphere at all times.

(a) Assume that the thrown-off mass carries off no angular momentum, in rev/s

(b) Assume that the thrown-off mass carries off its proportional share (3/4) of the initial angular momentum, in rev/s

Answer 1

The solution to this problem is based upon the conservation of momentum, in this case 'Angular momentum'. For any system of masses of moment of inertia 'I', and angular velocity 'omega', the angular momentum 'L' is given by \[L = I \omega\]A star can be approximated to be a uniform sphere of mass 'M' and radius 'R' spinning about its axis, so we can write 'I' as \[I = \frac {2}{5} M R^2\]#For part (a) it is given that that the thrown-off mass carries off no angular momentum, so when we conserve the angular momentum of the star before and after the transition into a neutron star, we can write \[L_{1} =L_{2}\]Now it is given that the original star is the size of the sun and 8 times as massive; so if we assume the mass of the sun to be M, \[M_{1} = 8M\] and \[R_{1} = Radius Of The Sun = 695500 Kilometers =6.955\times 10 ^8 m\] Also given is the time taken to complete 1 revolution = 12 days, so \[\omega _{1}= 1 Rev/ 12 days = 1Rev/(12\times24\times60\times60 seconds) = 9.64506173\times10 ^{-7} Rev/s\]After the transition into the neutron star, the mass is reduced by 3/4 and the radius becomes 12 Km, so \[M_{2} = \frac{8M}{4}= 2M\] and \[R_{2} = 1.2\times 10^4 m\]Now plugging these values in the equation\[L_{1} =L{2}\]Using the formulae for L and I we can write, \[\frac{2}{5} M_{1}R_{1}^2\omega_{1} =\frac{2}{5} M_{2}R_{2}^2\omega_{2}\]\[\frac{2}{5}\times8M\times (6.955\times10^8)^2 \times (9.64506173\times 10^{-7}) = \frac{2}{5}\times2M\times (1.2\times10^4)^2 \times \omega_{2}\] and on further simplification we get, \[\omega_{2} = 12959.7546425 Rev/sec =1.2959\times 10^4 Rev/s\]
#For part (b) it is given that the thrown-off mass carries off its proportional share (3/4) of the initial angular momentum, so when we conserve the angular momentum of the system we get \[\frac{L_{1}}{4} = L_{2}\] following similar steps as part (a) \[\frac{1}{4}[\frac{2}{5} M_{1}R_{1}^2\omega_{1}] =\frac{2}{5} M_{2}R_{2}^2\omega_{2}\]\[\frac{1}{4}\times[\frac{2}{5}\times8M\times (6.955\times10^8)^2 \times (9.64506173\times 10^{-7})] = \frac {2}{5}\times2M\times (1.2\times10^4)^2 \times \omega_{2}\] simplifying this expression we get, \[\omega_{2} = 3.23994 \times 10^3 Rev/s\]