When a star with a mass larger than 8 times the Sun\'s mass dies, it explodes in

Question

When a star with a mass larger than 8 times the Sun's mass dies, it explodes in a supernova and leaves behind a very dense core. This core is so dense that protons and electrons have combined so that the core is just a ball of neutrons. This super-dense remnant is called a neutron star. Neutron stars pack the mass of the Sun into the size of a city. A star that was originally 9 times the mass of the Sun (M_O = 2 Times 10^ 30 kg)has a rotation velocity of 250 km/s and a radius of 4 times the radius of the Sun (R_ O = 695.700 km) How long does it take for this star to make one full rotation? What is the angular momentum of this star? Treat the star as a solid sphere. If after a supernova the star left behind a remnant with the same mass as before, but a radius of 20 km, how long would this neutron star take to make one rotation?

Explanation / Answer

a) T = 2r/v = 2 * (4 * 6.957 * 108) / (250 * 103) = 7 * 104 s

b) L = I = (2mr2/5)(v/r) = 2mvr/5 = 2 * (9 * 2 * 1030) * (250 * 103) * (4 * 6.957 * 108) / 5 = 5 * 1045 kg-m2/s

c) Using conservation of angular momentum,

I11 = I22

=> 2mv1r1/5 = 2mv2r2/5

=> v1r1 = v2r2

=> v2 = v1r1/r2 = (250 * 103) * (4 * 6.957 * 108) / (20 * 103) = 3.48 * 1010 m/s

So, new time period, T = 2 * (20 * 103) / (3.48 * 1010) = 3.6 * 10-6 s

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