neutron stars and supernova remnants. The crab nebula is a bloud of glowing gas

Question

neutron stars and supernova remnants. The crab nebula is a bloud of glowing gas about 10 ly across located about 6500 light years from the earth. It is the remanant of a star that uderwent a super nova,seen on earth in 1054 A.D. Energy is released by the crab nebula at a rate of about 5*10^31W, about 10^5 time the rate at which the sun radiates energy.The crab nebula obtains its energy from the rotational kinetic energy of a rapidly spinning neutron star at its center. The object rotates once every .03331 s, and this period is increasing by 4.22*10^-13 for each second of the time that elapses. A)If the rate at which energy is lost by the neutron star is equal to the rate which energy is released by the nebula, find the moment of inertia of the neutron star.B)Theories of the supernovae predict that the neutron star in the crab nebula has a mass about 1.4 times that of the sun. Modeling the neutron star as a solid uniform sphere,calculate its radius in km C)what is its linear speed of a point on the equator of the neutron star?compare to the speed of light.D)Assume that the neutron star is uniform calculate its density. compare to the density of a ordinary rock (3000kg/m^3)and to the density of an atomic nucleus(about 10^17 kg/m^3).Justify the statement that the neutron star is essentially a large atomic nucleus.

Explanation / Answer

A)
Pn=Power Release by the Nebula=5*1031 Watts
I=rotational inertia
w=angular frequency
T=period=.0331 s
T'=dT/dt=4.22E-13
w=2*pi/T

E=rotational energy = I2 = 4*pi2*I*T-2
Ps=Power radiate by the star=dE/dt=E'=4*pi2*I*(-2)*T-3*T'=-8...

Ps = -Pn

4*pi2*I*(-2)*T-3*T' = 5*1031

I = 1.1×1038 Km*m2

B)
I=(2/5)M*r2, moment of inertia of a solid sphere
M=1.4 Solar mass

r=9.9 KM

C)

v=*r
=2*pi/T  

v=1.9*106 m/s

D)

Density = mass/volume
mass = 1.4 solar mass
volume=(4/3) pi r^3
p=6.9*1017 kg/m3

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