The Crab pulsar \(\left(m \approx 2 \cdot 10^{30} \mathrm{~kg}, R=5 \mathrm{~km}\right)\) is a neutron star located in the Crab Nebula. The rotation rate of the Crab pulsar is currently about 30 rotations per second, or \(60 \pi \mathrm{rad} / \mathrm{s} .\) The rotation rate of the pulsar, however, is decreasing; each year, the rotation period increases by \(10^{-5}\) s. Justify the following statement: The loss in rotational energy of the pulsar is equivalent to 100,000 times the power output of the Sun. (The total power radiated by the Sun is about \(\left.4 \cdot 10^{26} \mathrm{~W} .\right)\)

Moment of inertia (I) can be found using the formula for a solid sphere: \(I = \frac{2}{5}mR^2\) where m is the mass and R is the radius of the pulsar. \(I = \frac{2}{5} \times 2 \cdot 10^{30} \mathrm{~kg} \times (5 \cdot 10^{3} \mathrm{~m})^2 = \frac{2}{5} \times 2 \cdot 10^{30} \mathrm{~kg} \times 25 \cdot 10^{6} \mathrm{m^2}\) \(I = 2 \cdot 10^{37} \mathrm{~kg \cdot m^2}\)

The problem states that the pulsar's rotation rate is 30 rotations per second, which can be converted to radians per second as follows: \(\omega = 60 \pi \mathrm{rad/s}\)

Rotational energy (E) can be found using the formula: \(E = \frac{1}{2}I\omega^2\) \(E = \frac{1}{2} \times 2 \cdot 10^{37} \mathrm{~kg \cdot m^2} \times (60 \pi \mathrm{rad/s})^2\) \(E \approx 1.13 \cdot 10^{45} \mathrm{~J}\)

The problem states that the rotation period is increasing by \(10^{-5} \mathrm{s}\) per year. Let's determine the change in angular velocity per second within a year: \(\Delta \omega = \frac{2\pi \mathrm{rad}}{\mathrm{s}}\left(\frac{10^{-5} \mathrm{s}}{\mathrm{yr}}\right) \Rightarrow \Delta \omega = 2\pi \cdot 10^{-5} \mathrm{rad/(s\cdot yr)} \Rightarrow \Delta \omega \approx 6.283 \cdot 10^{-5} \mathrm{rad/(s\cdot yr)}\)

Using the previous results, we can determine the loss in rotational energy per year, according to the formula: \(\Delta E = -I\omega\Delta\omega\) \(\Delta E \approx -2 \cdot 10^{37} \mathrm{~kg \cdot m^2} \times 60 \pi \mathrm{rad/s} \times 6.283 \cdot 10^{-5} \mathrm{rad/(s\cdot yr)}\) \(\Delta E \approx -7.49 \cdot 10^{33} \mathrm{~J/yr}\)

Power loss (P) in watts is given by the negative of the rate of change of energy per second. To convert the energy loss per year to watts, we have: \(P = -\frac{\Delta E}{\mathrm{yr}} = -\frac{7.49 \cdot 10^{33} \mathrm{~J/yr}}{(365.25 \times 24 \times 60 \times 60) \mathrm{s/yr}} = -\frac{7.49 \cdot 10^{33} \mathrm{~J/yr}}{3.156 \cdot 10^7 \mathrm{s/yr}} = -2.37 \cdot 10^{26} \mathrm{~W}\) The given total power radiated by the Sun is \(4 \cdot 10^{26} \mathrm{~W}\). Let's compare the power loss of the pulsar to the power output of the Sun: \(\frac{|P_{pulsar}|}{P_{Sun}} = \frac{2.37 \cdot 10^{26} \mathrm{~W}}{4 \cdot 10^{26} \mathrm{~W}} \approx 0.59\) This result indicates that the loss in rotational energy of the pulsar is not equivalent to 100,000 times the power output of the Sun, as it is approximately 59% of the Sun's power output. Therefore, the given statement is not justified.