Question

The Crab pulsar \(\left(m \approx 2 \cdot 10^{30} \mathrm{~kg}, R=12 \mathrm{~km}\right)\) 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 \(10^{-5} \mathrm{~s} .\) 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 \(4 \cdot 10^{26}\) W.

Short answer

#tag_title# Step 4: Calculating the Rate of Energy Loss #tag_content# To find the rate of energy loss, we need to find the difference in rotational energies and divide it by the time interval (1 year): Rate of Energy Loss = \(\frac{E_\mathrm{final} - E_\mathrm{initial}}{1 \mathrm{\, year}} = \frac{8.83 \cdot 10^{44} \mathrm{~J}}{1 \mathrm{\, year}} \approx 2.8 \cdot 10^{37} \mathrm{~W}\) #tag_title# Step 5: Comparing to the Sun's Power Output #tag_content# We are given that the Sun's power output is approximately \(3.8 \cdot 10^{26} \mathrm{~W}\). Comparing this to the rate of energy loss calculated above: \(2.8 \cdot 10^{37} \mathrm{~W} \approx 7.4 \cdot 10^{10}\) times the Sun's power output. #Answer# Based on the calculations, it can be justified that the Crab pulsar is indeed losing rotational energy at a rate roughly \(7.4 \cdot 10^{10}\) times the power output of the Sun.

Step-by-step solution

Calculating the Moment of Inertia

To calculate the moment of inertia (I) of the Crab pulsar, we can assume it to be a solid sphere and use the following formula: I = \(\frac{2}{5}mR^2\) Where m is the mass of the pulsar \((2 \cdot 10^{30} \mathrm{~kg})\) and R is its radius \((12 \mathrm{~km})\). Substituting the given values, we get: I = \(\frac{2}{5}(2 \cdot 10^{30})(12 \cdot 10^3)^2 \approx 3.46 \cdot 10^{45} \mathrm{~kg\cdot m^2}\)

Determining the Angular Velocity

We know that the Crab pulsar rotates at \(30\) rotations per second, which translates to an angular velocity ω as: ω = \(2\pi \cdot 30 \mathrm{\, rad/s}\) Since the rotation period increases by \(10^{-5} \mathrm{\, s}\) each year, we can calculate the change in angular velocity: Δω = \(2\pi \cdot 30 \mathrm{\, rad/s} - 2\pi \cdot (30 + \frac{10^{-5}}{2\pi})^{-1} \mathrm{\, rad/s} \approx -1.88 \cdot 10^{-5} \mathrm{\, rad/s}\)

Computing the Rotational Energies

Using the initial angular velocity ω and the moment of inertia I, we can calculate the initial rotational energy E_initial: E_initial = \(\frac{1}{2}I\omega^2 = \frac{1}{2} (3.46 \cdot 10^{45}) (2\pi \cdot 30)^2 \approx 5.88 \cdot 10^{49} \mathrm{~J}\) Now we calculate the rotational energy E_final at the end of one year using the final angular velocity \((ω + Δω)\): E_final = \(\frac{1}{2}I(ω + Δω)^2 \approx 5.88 \cdot 10^{49} - 8.83 \cdot 10^{44} \mathrm{~J}\)

Key concepts

Moment of Inertia

The concept of moment of inertia (denoted as I) is central to understanding the rotational dynamics of objects. It's an object's resistance to changes in its rotation and is comparable to mass in linear motion. The larger the moment of inertia, the harder it is to spin or stop spinning the object.

Moment of inertia depends on the mass of the object and the distribution of that mass relative to the axis of rotation. For solid spheres like the Crab pulsar, which are uniform and symmetric, the formula \( I = \frac{2}{5}mR^2 \) nicely encapsulates how the mass (m) and radius (R) create this rotational resistance. By plugging in the pulsar's mass and radius, we can understand why it requires immense energy to influence its spin.

Angular Velocity

Angular velocity (represented by the Greek letter \(\omega\)) measures how fast an object is rotating. It is often expressed in radians per second (rad/s). For an object like the Crab pulsar, which makes full rotations, we can determine its angular velocity by the formula \( \omega = 2\pi \times \text{frequency} \), where the frequency is the number of rotations per second.

The Crab pulsar's incredibly high angular velocity contributes to its significant rotational energy. To track changes over time, such as a yearly slowdown, we consider the change in angular velocity (\(\Delta\omega\)), adding depth to our understanding of the pulsar's dynamic behavior over periods.

Rotational Kinetic Energy

Rotational kinetic energy is the energy an object possesses due to its rotation. The formula to calculate this for any rotating object is \( E = \frac{1}{2}I\omega^2 \) where E is the energy, I is the moment of inertia, and \(\omega\) represents the angular velocity. This formula shows how the moment of inertia and the angular velocity jointly determine the rotational energy of an object.

For celestial objects like the Crab pulsar, their large moment of inertia and high angular velocity mean they store an exceptional amount of rotational kinetic energy. Calculating the initial and final energy based on changes in \(\omega\) highlights the vast energy release as the pulsar's spin rate changes, a concept crucial for understanding the physics of neutron stars.

Neutron Stars

Neutron stars, like the Crab pulsar, are stellar remnants born from the explosive deaths of massive stars. These objects pack a sun's worth of mass into a city-sized sphere, resulting in some of the densest forms of matter in the universe. Their rapid rotation and strong magnetic fields produce beams of electromagnetic radiation observed as pulsars.

Neutron stars' incredible density leads to high moments of inertia, allowing them to conserve rotational energy very effectively. Their rapid spin and subsequent slowdown can emit significant amounts of energy. By comparing the stellar rotational energy loss to our sun's output, we can convey the vast power that these celestial objects are capable of outputting, dwarfing our sun's energy production.