Problem 86 - University Physics with Modern Physics

Question

The Crab Nebula is a cloud of glowing gas about 10 lightyears across, located about 6500 light-years from the earth (\(\textbf{Fig. P9.86}\)). 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 \(\times\) 10\(^3$$^1\) 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 0.0331 s, and this period is increasing by 4.22 \(\times 10^{-13}\) s for each second of time that elapses. (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. (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. (c) What is the linear speed of a point on the equator of the neutron star? Compare to the speed of light. (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 nucleus (about 10\(^{17}\) kg/m\(^3\)). Justify the statement that a neutron star is essentially a large atomic nucleus.

Step-by-Step Solution

Answer

Moment of inertia is calcuated via power balance; radius by sphere model. Equatorial speed and density show star's extreme nature.

Step 1: Calculating Moment of Inertia

The rate at which energy is released by the nebula is equal to the rate of energy loss of the neutron star. The rotational energy of the star is given by \( E = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. The change in rotational energy per unit time is \( \frac{dE}{dt} = I \omega \frac{d\omega}{dt} \). Since \( E = \frac{1}{2} I \omega^2 \) and \( P = 5 \times 10^{31} \text{ W} \), we equate \( P = I \omega \frac{d\omega}{dt} \) to find \( I \). Using \( \omega = \frac{2\pi}{T} \) and \( \frac{d\omega}{dt} = -\frac{2\pi \cdot 4.22 \times 10^{-13}}{0.0331^2} \), solve for \( I \).

Step 2: Determining Angular Velocity

Calculate \( \omega \), the angular velocity of the neutron star: \( \omega = \frac{2\pi}{T} = \frac{2\pi}{0.0331} \). This value will be used in the moment of inertia calculation.

Step 3: Calculating Change in Angular Velocity

Next, calculate \( \frac{d\omega}{dt} \), which is the rate of change of angular velocity: \( \frac{d\omega}{dt} = -\frac{2\pi \times 4.22 \times 10^{-13}}{0.0331^2} \).

Step 4: Solving for Moment of Inertia

We have established that the power output \( P = I \omega \frac{d\omega}{dt} \). Substitute the values calculated into this equation: \( I = \frac{P}{\omega \left| \frac{d\omega}{dt} \right|} \). Solve to find \( I \).

Step 5: Calculating Radius of Neutron Star

Model the neutron star as a solid sphere with mass \( M = 1.4 \times \text{mass of the sun} = 1.4 \times 1.989 \times 10^{30} \text{ kg} \). The moment of inertia for a sphere is \( I = \frac{2}{5} M R^2 \). Solve for the radius \( R \) using the moment of inertia obtained.

Step 6: Calculating Linear Speed at Equator

The linear speed \( v \) at the equator is given by \( v = R \omega \). Use the radius from Step 5 and angular velocity from Step 2 to calculate \( v \). Compare \( v \) with the speed of light \( c = 3 \times 10^8 \text{ m/s} \).

Step 7: Calculating Density of Neutron Star

Density \( \rho = \frac{M}{\frac{4}{3}\pi R^3} \). Use the mass and radius from previous steps to find the density. Compare this to the density of ordinary rock and an atomic nucleus.

Key Concepts

Moment of Inertia

The moment of inertia is a principle that plays a critical role in the physics of rotating objects. For a neutron star, which can be thought of as a giant spinning mass, the moment of inertia (\( I \)) is crucial in determining its rotational behavior.

• Moment of inertia is essentially the rotational equivalent of mass in linear motion.
• For a rotating body, it determines how much torque is needed for a desired angular acceleration.

In the context of a neutron star, the moment of inertia is defined by the formula:\[ I = \frac{2}{5} M R^2 \]where \( M \) is the mass and \( R \) is the radius of the neutron star. This formula assumes the star is a solid uniform sphere.Understanding moment of inertia helps us to explore how matter is organized in celestial bodies. It provides insight into the distribution of mass within the neutron star, influencing how the star responds to changes in rotation, such as those occurring after a supernova event.

Angular Velocity

Angular velocity reflects how fast an object rotates or revolves around a point or axis. For neutron stars, this determines how quickly the star completes a rotation. In the Crab Nebula, the neutron star rotates very rapidly, completing a single rotation in just 0.0331 seconds.

• Angular velocity (\( \omega \)) can be calculated using the formula: \( \omega = \frac{2\pi}{T} \), where \( T \) is the period of rotation.
• In this case, the angular velocity is derived as \( \omega = \frac{2\pi}{0.0331} \).

The concept of angular velocity is vital for understanding the dynamic state of neutron stars. It connects to the star's observable features, such as pulsation rates, and is crucial for calculating other properties like the moment of inertia or rotational kinetic energy. As a neutron star emits radiation, its rotational period slowly increases, indicating that it is losing angular momentum, a phenomenon directly linked with the star's aging and transformation.

Supernova

A supernova represents one of the universe's most powerful events. This explosion marks the death of a massive star and is essential in the formation of neutron stars.

• The Crab Nebula is a prime example of a supernova remnant. The explosion was so significant that it was noted on Earth in 1054 A.D.
• During such an explosion, the outer layers of the star are blasted into space, leaving behind a dense core, often forming a neutron star.

The material ejected from the supernova forms the glowing gases of the nebula—still visible as a remnant in the sky today. The core that remains is extremely dense, becoming a neutron star that resembles an enormous atomic nucleus in terms of density. Understanding supernovae helps astronomers explore these magnificent cosmic events and their role in the life cycle of stars.

Rotational Kinetic Energy

Rotational kinetic energy is the energy due to the rotation of an object and is a portion of the total kinetic energy.For a neutron star, which spins at staggering speeds, this concept is fundamental in understanding its energy dynamics. The rotational kinetic energy (\( E \)) of a neutron star is given by:\[ E = \frac{1}{2} I \omega^2 \]where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. The staggering rate of energy output, in turn, powers the surrounding nebula.

• This energy release occurs because the neutron star is slowing down over time—a result of losing energy to the nebula.
• The neutron star's energy output greatly surpasses other celestial phenomena, contributing extensively to the nebula's luminosity.

Understanding rotational kinetic energy provides insights into how neutron stars interact with their environment, including the radiation of energy and the emission of pulsating light, offering a glimpse into the profound physics governing such celestial entities.