The Crab pulsar (m=2.00x10^30 kg) is a neutron star located in the Crab Nebula. The rotation rate of the Crab pulsar is currently about 30.0 rotations per second, or 60.0pi rad/s. The rotation rate of the pulsar, however, is decreasing; each year, the rotation period increases by 1.00x10^-5 s. Justify the following statement: The loss in rotational energy of the pulsar to 1.00*10^5 times the power output of the Sun. ( The total power radiated by the Sun is about 4.00*10^26 W.)

Question

The Crab pulsar (m=2.00x10^30 kg) is a neutron star located in the Crab Nebula. The rotation rate of the Crab pulsar is currently about 30.0 rotations per second, or 60.0pi rad/s. The rotation rate of the pulsar, however, is decreasing; each year, the rotation period increases by 1.00x10^-5 s. Justify the following statement: The loss in rotational energy of the pulsar to 1.00*10^5 times the power output of the Sun. ( The total power radiated by the Sun is about 4.00*10^26 W.)
P/Ps=_________

bobpursley · Answer

rotational energy loss= 1/2 I (wi^2-wf^2)

= 1/2 I (wi+wf)(wi-wf)
now the moment of inertia for a solid sphere is 2/5 mr^2

= 1/2 * 2/5*2E30 * (wi*r-wf*r)(wi*r+wi*r)

now period= 1/freq=1/30
so w=2PI/period

wi=2PI/30
wf=2PI/(30-1E-5)

so, finally, (wi^2-wf^2)=
= (2PI)^2 (1/30^2 -1/(30-1E-5)^2
= (2PI)^2 ( 900-60E-5-1E-10 -900)/(30^3*(30-1E-5)^2

= appx (2PI)^2 ( 1/15 * 1E-5)

then, power loss is appx
1/5 *2E30*(2PI)^2 (1/15 E-5)
= 2.6E24
so recheck my work.