Question

Many pulsars radiate radio-frequency or other radiation in a periodic manner and are bound to a companion star in what is known as a binary pulsar system. In \(2003,\) a double pulsar system, \(\operatorname{PSR} J 0737-3039 \mathrm{~A}\) and \(\mathrm{J} 0737-3039 \mathrm{~B},\) was discovered by astronomers at the Jodrell Bank Observatory in the United Kingdom. In this system, both stars are pulsars. The pulsar with the faster rotation period rotates once every \(0.0230 \mathrm{~s}\), while the other pulsar has a rotation period of \(2.80 \mathrm{~s}\). The faster pulsar has a mass 1.337 times that of the Sun, while the slower pulsar has a mass 1.250 times that of the Sun. a) If each pulsar has a radius of \(20.0 \mathrm{~km},\) express the ratio of their rotational kinetic energies. Consider each star to be a uniform sphere with a fixed rotation period. b) The orbits of the two pulsars about their common center of mass are rather eccentric (highly squashed ellipses), but an estimate of their average translational kinetic energy can be obtained by treating each orbit as circular with a radius equal to the mean distance from the system's center of mass. This radius is equal to \(4.23 \cdot 10^{8} \mathrm{~m}\) for the larger star, and \(4.54 \cdot 10^{8} \mathrm{~m}\) for the smaller star. If the orbital period is \(2.40 \mathrm{~h},\) calculate the ratio of rotational to translational kinetic energies for each star.

Short answer

b) What is the ratio of rotational to translational kinetic energies for each pulsar?

Step-by-step solution

Calculate the rotational kinetic energies of the pulsars

First, we need to find the rotational kinetic energy for both pulsars. The formula for the rotational kinetic energy of a uniform sphere is given by \(E_{rot} = \frac{2}{5}I\omega^2\), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity. Moment of inertia, \(I\) for a sphere, is given by \(I = \frac{2}{5}MR^2\) with mass \(M\) and radius \(R\). Also, \(\omega\) is related to the rotation period (T) by \(\omega = \frac{2\pi}{T}\). Let's find \(\omega\) for both pulsars first: For Pulsar A with a rotation period of 0.0230 s: \(\omega_{A} = \frac{2\pi}{0.0230}\) For Pulsar B, with a rotation period of 2.80 s: \(\omega_{B} = \frac{2\pi}{2.80}\) Next, we find the rotational kinetic energies \(E_{rot,A}\) and \(E_{rot,B}\) for both pulsars: \(E_{rot,A} = \frac{2}{5}(1.337 M_{\odot})(20.0 \times 10^3 m)^2(\frac{2\pi}{0.023 s})^2\) \(E_{rot,B} = \frac{2}{5}(1.250 M_{\odot})(20.0 \times 10^3 m)^2(\frac{2\pi}{2.80 s})^2\) Where \(M_{\odot}\) is the solar mass (\(1.989 \times 10^{30} kg\)).

Calculate the translational kinetic energies of the pulsars

The formula for the translational kinetic energy of an object in circular motion is: \(E_{trans} = \frac{1}{2}Mv^2\) For objects in circular motion, we can relate velocity \(v\) to their radius and orbital period (T) by \(v = \frac{2\pi r}{T}\). Let's calculate the translational kinetic energies \(E_{trans,A}\) and \(E_{trans,B}\) for both pulsars: For Pulsar A with radius 4.23 x 10^8 m and orbital period of 2.40 h: \(E_{trans,A} = \frac{1}{2}(1.337 M_{\odot})(\frac{2\pi (4.23 \times 10^8 m)}{2.40 \times 3600 s})^2\) For Pulsar B with radius 4.54 x 10^8 m and orbital period of 2.40 h: \(E_{trans,B} = \frac{1}{2}(1.250 M_{\odot})(\frac{2\pi (4.54 \times 10^8 m)}{2.40 \times 3600 s})^2\)

Calculate the ratios of kinetic energies

We are now ready to find the required ratios: a) Ratio of rotational kinetic energies, \(\frac{E_{rot,A}}{E_{rot,B}}\) Calculate this using the calculated values of \(E_{rot,A}\) and \(E_{rot,B}\). b) Ratio of rotational to translational kinetic energies for each star: For Pulsar A, \(\frac{E_{rot,A}}{E_{trans,A}}\) Calculate this using the calculated values of \(E_{rot,A}\) and \(E_{trans,A}\). For Pulsar B, \(\frac{E_{rot,B}}{E_{trans,B}}\) Calculate this using the calculated values of \(E_{rot,B}\) and \(E_{trans,B}\).

Key concepts

Rotational Kinetic Energy

Understanding rotational kinetic energy is pivotal when studying celestial objects like pulsars in a binary system. Rotational kinetic energy is the energy possessed by an object due to its rotation. It can be thought of as the spinning equivalent of linear or translational kinetic energy, which is energy due to motion in a straight line.

In the realm of astrophysics, this concept helps us quantify the energy of rotating bodies like stars, planets, and pulsars. The formula for a spherical object, like our given pulsars, is expressed as \(E_{rot} = \frac{2}{5}I\omega^2\), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.

The moment of inertia acts as a measure of an object's resistance to changes in its rotation, and angular velocity quantifies how fast the rotation occurs. The intricate dance of pulsars in a binary system is governed by their rotational kinetic energies, which influence everything from their emissions to gravitational wave output. In our exercise, the rotational kinetic energy plays a critical role in understanding the dynamism of the binary pulsar system.

Translational Kinetic Energy

While we've delved into the spin of pulsars with rotational kinetic energy, their journey through space is equally fascinating. This movement is characterized by translational kinetic energy, which reflects the energy an object has due to its motion through space. In an astronomical ballet, celestial bodies like planets and stars move along orbits, and translational kinetic energy provides a window into this motion.

For circular orbits, the translational kinetic energy is given by the formula \(E_{trans} = \frac{1}{2}Mv^2\), where \(M\) is the mass of the object and \(v\) its velocity. Even though the pulsars in the binary system have elliptical orbits, we can approximate their motion as circular for the sake of simplicity to estimate their kinetic energy.

The comparison of rotational and translational kinetic energies offers insights into the distribution and interplay of energy within the system, revealing the delicate balance that allows the pulsars to remain in stable orbits around each other.

Moment of Inertia

The moment of inertia is a concept that can be puzzling, yet it's simply the rotational equivalent of mass in linear motion. It's a measure of an object's resistance to changes to its rotational velocity and, for our pulsars, which are treated as uniform spheres, the moment of inertia \(I\) is given by \(I = \frac{2}{5}MR^2\), where \(M\) is the mass of the sphere and \(R\) its radius.

Higher moment of inertia indicates that it's harder to spin or stop spinning the object. In the case of pulsars, this concept helps us understand how mass distribution affects their spin—and by extension, their rotational kinetic energy. Different mass distributions lead to various moments of inertia, affecting how pulsars interact with each other and their environment. Through our exercise, the moment of inertia is utilized to calculate the pulsars' rotational kinetic energy, an integral step in understanding the energy dynamics of the binary system.

Angular Velocity

Now let's look at angular velocity, a concept that might seem abstract, but it simply describes the rate of rotation of an object. In the context of a binary pulsar system, angular velocity tells us how quickly each pulsar completes a rotation around its axis.

Represented by \(\omega\), angular velocity is linked to the rotational period \(T\) by the equation \(\omega = \frac{2\pi}{T}\). So, the faster the pulsar rotates, the greater its angular velocity. A higher angular velocity translates to a larger rotational kinetic energy provided the mass and radius of the pulsar remain constant.

For objects such as pulsars, which are famed for their regular lighthouse-like beams of radiation, angular velocity is significant. It determines the frequency at which their beams sweep across Earth, allowing astronomers to measure time with astonishing precision.