he Crab Nebula is a cloud of glowing gas about 10 light-years across, located ab

Question

he Crab Nebula is a cloud of glowing gas about 10 light-years across, located about 6500 light years from the earth (the figure (Figure 1) ). It is the remnant of a star that underwent a supernova explosion, seen on earth in 1054 a.d. Energy is released by the Crab Nebula at a rate of about 5*10^31 W , about 10^5 times 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. This object rotates once every 0.0343 s, and this period is increasing by 4.16*10^(-13) s for each second of time that elapses.

Part A
If the rate at which energy is lost by the neutron star is equal to the rate at which energy is released by the nebula, find the moment of inertia of the neutron star.

Part B
Theories of 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 kilometers.

Part C
What is the linear speed of a point on the equator of the neutron star? Compare to the speed of light.

Part D
Assume that the neutron star is uniform and calculate its density. compare to the density of ordinary rock (3000 kg/m^3) and to the density of an atomic nucles (about 10^17 kg/m^3)

Explanation / Answer

a)

E = 0.5 I w^2

==> E = 0.5 I (2pi/T)^2

==> dE/dt = 0.5 I (2pi)^2 (-2/T^3) (dT/dt)

==> dE/dt = -I (2pi)^2 (1/T^3) (dT/dt)

==> 5e31 = -I (2*3.1416)y2 * (1/0.0343y3) * (4.16e-13)

==> I = 1.2286e38

==> I = 1.23 x 10^38 Kg.m2

b)

I = (2/5) M R^2

==> R = sqrt((5/2) (I/M)) = sqrt((5/2)*(1.2286e38/(1.4*1.989e30))) = 10503 = 10.5 km

c)

v = R w = R (2pi/T) = 10503 * (2*3.1416/0.0343) = 1.92 x 10^6 m/s

Compare to the speed of light:

v/c = (1.92e6)/(3e8) = 6.41 x 10^-3

d)

density = rho = M/(4/3 pi R^3) = (1.4*1.989e30)/(4/3*3.1416*10503y3) = 5.74 x 10^17 Kg/m3

compare to the density of ordinary rock (3000 kg/m^3):

(5.74e17)/(3000) = 1.91 x 10^14

compare to the density of an atomic nucles (about 10^17 kg/m^3)

(5.74e17)/(1e17) = 5.74

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