During most of its lifetime, a star maintains an equilibrium sizein which the in

Question

During most of its lifetime, a star maintains an equilibrium sizein which the inward force of gravity on each atom is balanced by anoutward pressure force due to the heat of the nuclear reactions inthe core. But after all the hydrogen "fuel" is consumed by nuclearfusion, the pressure force drops and the star undergoes agravitational collapse until it becomes a neutronstar. In a neutron star, the electrons and protons of theatoms are squeezed together by gravity until they fuse intoneutrons. Neutron stars spin very rapidly and emit intense pulsesof radio and light waves, one pulse per rotation. These "pulsingstars" were discovered in the 1960s and are calledpulsars.

During most of its lifetime, a star maintains an equilibrium sizein which the inward force of gravity on each atom is balanced by anoutward pressure force due to the heat of the nuclear reactions inthe core. But after all the hydrogen "fuel" is consumed by nuclearfusion, the pressure force drops and the star undergoes agravitational collapse until it becomes a neutronstar. In a neutron star, the electrons and protons of theatoms are squeezed together by gravity until they fuse intoneutrons. Neutron stars spin very rapidly and emit intense pulsesof radio and light waves, one pulse per rotation. These "pulsingstars" were discovered in the 1960s and are calledpulsars. A star with the mass (M = 2. 0 times 1030 kg) and size (R = 3. 5 times 108 m) of our sun rotates once every 30.0 days. After undergoinggravitational collapse, the star forms a pulsar that is observed byastronomers to emit radio pulses every 0.200 s. By treating the neutron star as a solidsphere, deduce its radius. What is the speed of a point on the equator of the neutron star?Your answer will be somewhat too large because a star cannot beaccurately modeled as a solid sphere.

Explanation / Answer

Radio pulse rate = 1 at every 0.2 sec

Rotation rate = 0.2 sec

Omega = 2*pi/t = 10 pi = 31.416 rad/s

Initial omega = 2*pi/(30*24*3600) = 2.424*10^-6 rad/s

When the star collapse, the angular momentum will remain constant (No external torque)

So, L initial = L final

I initial * w initial = I final * w final

R*R * 2.424*10^-6 = r*r * 31.416

r = 97222.1 m

b) V = omega * R = 3.054*10^6 m/s

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