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

Question

During most of its lifetime, a star maintains an equilibrium size in which the inward force of gravity on each atom is balanced by an outward pressure force due to the heat of the nuclear reactions in the core. But after all the hydrogen "fuel" is consumed by nuclear fusion, the pressure force drops and the star undergoes a gravitational collapse until it becomes a neutron star. In a neutron star, the electrons and protons of the atoms are squeezed together by gravity until they fuse into neutrons. Neutron stars spin very rapidly and emit intense pulses of radio and light waves, one pulse per rotation. These "pulsing stars" were discovered in the 1960s and are called pulsars. Part A A star with the mass (M=2.0×1030kg) and size (R=3.5×108m) of our sun rotates once every 35.0 days. After undergoing gravitational collapse, the star forms a pulsar that is observed by astronomers to emit radio pulses every 0.200 s . By treating the neutron star as a solid sphere, deduce its radius. Express your answer with the appropriate units. SubmitMy AnswersGive Up Part B 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 be accurately modeled as a solid sphere. Express your answer with the appropriate units. SubmitMy AnswersGive Up

Explanation / Answer

As the star collapses, it's angular mometnum remains conserved.

Li = Lf

Ii*wi = If*wf

M*Ri^2*wi = M*Rf^2*wf^2

wi = 2*pi*fi = 2*pi/Ti

M*Ri^2*2*pi/Ti^2 = M*Rf^2*2*pi/Tf^2

Rf/Ri = sqrt (Tf/Ti)

35 days = 35*24*3600

Rf = 3.5*10^8*sqrt (0.2/(35*24*3600))

Rf = 90010.28 m = 90.01 km

B.

v = Rf*wf = 2*pi*Rf/Tf

v = 90.01*10^3*2*3.14/0.2

v = 2.83*10^6 m/sec

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