Problem 39 - University Physics with Modern Physics

Question

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 \(\left.4 \cdot 10^{26} \mathrm{~W} .\right)\)

Step-by-Step Solution

Answer

The loss in rotational energy of the Crab pulsar is approximately 59% of the Sun's power output.

Step 1: Calculate the moment of inertia of the pulsar

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. \(I = \frac{2}{5} \times 2 \cdot 10^{30} \mathrm{~kg} \times (5 \cdot 10^{3} \mathrm{~m})^2 = \frac{2}{5} \times 2 \cdot 10^{30} \mathrm{~kg} \times 25 \cdot 10^{6} \mathrm{m^2}\) \(I = 2 \cdot 10^{37} \mathrm{~kg \cdot m^2}\)

Step 2: Determine the current angular velocity of the pulsar

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

Step 3: Compute the current rotational energy of the pulsar

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

Step 4: Calculate the change in angular velocity per year

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: \(\Delta \omega = \frac{2\pi \mathrm{rad}}{\mathrm{s}}\left(\frac{10^{-5} \mathrm{s}}{\mathrm{yr}}\right) \Rightarrow \Delta \omega = 2\pi \cdot 10^{-5} \mathrm{rad/(s\cdot yr)} \Rightarrow \Delta \omega \approx 6.283 \cdot 10^{-5} \mathrm{rad/(s\cdot yr)}\)

Step 5: Calculate the loss in rotational energy per year

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

Step 6: Calculate the power loss and compare with the power output of the Sun

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.

Key Concepts

Neutron Stars

Neutron stars are some of the most fascinating and dense objects in the universe. They are born from the remnants of a supernova explosion, a dramatic end to some stars' life cycles. After a massive star has exhausted its nuclear fuel, its core collapses under gravity, and if it's not massive enough to become a black hole, it forms a neutron star. Despite their small size, usually around 20 kilometers in diameter, neutron stars can have a mass up to twice that of the Sun.
The incredible density of neutron stars means their gravity is tremendously strong, making them perfect laboratories for some of the most extreme physics in the universe. The huge mass compacted in such a small volume leads to pressures so high that even atomic nuclei are squished closer together, forming neutrons. This process gives these stars their name. When it comes to studying phenomena like gravitational waves and neutron degeneracy pressure, neutron stars are ideal objects.

Rotational Energy

Rotational energy is the energy due to the spinning motion of an object. For neutron stars, this is a critical aspect since they often rotate at extremely high speeds. The rotational energy can be calculated using the formula: \[E = \frac{1}{2} I \omega^2\] where \(E\) is the rotational energy, \(I\) is the moment of inertia, and \(\omega\) is the angular velocity. The faster the star spins, the more rotational energy it has.
For neutron stars like the Crab pulsar, which spins about 30 times per second, the rotational energy is colossal. This energy gradually decreases as the star loses speed over time, primarily due to the emission of electromagnetic radiation and particles. Understanding these energy dynamics is crucial, as it helps us understand the life cycle of these enigmatic objects.

Moment of Inertia

The moment of inertia is a physical quantity that measures how resistant an object is to rotational acceleration about an axis. In the case of neutron stars, their moment of inertia is huge due to their massive mass collapsed into a small size. It's essential for calculating the rotational energy of a neutron star. The moment of inertia \(I\) for a solid sphere is computed as: \[I = \frac{2}{5} m R^2\] where \(m\) is the mass and \(R\) is the radius of the star. The moment of inertia plays a significant role in the star's rotational dynamics, influencing how the star reacts to forces and torques.
For neutron stars, this property helps scientists study different physical phenomena, including pulsar emissions and rotational decay. When observing changes in neutron star rotation, understanding the moment of inertia helps interpret how energy and momentum are distributed in such extreme conditions.

Angular Velocity

Angular velocity is a vector quantity that represents the rate of rotation of an object. In astrophysics, it is used to describe how fast astronomical objects like neutron stars rotate around their axis. For neutron stars, these rates can be astoundingly high; the Crab Pulsar, for example, rotates over 30 times each second, equating to an angular velocity of approximately \(60 \pi \ \text{rad/s}\).
Angular velocity is derived from an object's rotational period, which is the time it takes to complete one full rotation. Formulaically, it's expressed as: \[\omega = \frac{2\pi}{T}\] where \(\omega\) is the angular velocity and \(T\) is the period. As neutron stars age, their angular velocity typically decreases due to energy loss and other astrophysical factors, altering their physical characteristics and emissions significantly.
Understanding angular velocity helps astrophysicists track changes in a neutron star's behavior, providing insights into its energy output and rotational dynamics.