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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 \(2004,\) a double pulsar system, PSR J0737-3039A and J0737-3039B, 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 \(0.023 \mathrm{~s}\), while the other pulsar has a rotation period of \(2.8 \mathrm{~s}\). 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 \(2.4 \mathrm{~h},\) calculate the ratio of rotational to translational kinetic energies for each star.

Step by step solution

01

a) Finding the ratio of rotational kinetic energies

For a rotating object, the rotational kinetic energy (\(K_{rot}\)) is given by the formula: \(K_{rot} = \frac{1}{2} I \omega^2\), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity. For a uniform sphere, the moment of inertia is given by \(I=\frac{2}{5}mR^2\), where \(m\) is the mass and \(R\) is the radius of the sphere. The angular velocity can be found by dividing \(2\pi\) by the rotation period: \(\omega = \frac{2\pi}{T}\). Now let's find \(K_{rot}\) for each pulsar: Pulsar A: \(I_A = \frac{2}{5} m_A R^2 = \frac{2}{5} \cdot 1.337 M_{sun} \cdot (20 \times 10^3 m)^2\), \(\omega_A = \frac{2 \pi}{0.023 s}\), \(K_{rot}^A = \frac{1}{2} I_A \omega_A^2\) Pulsar B: \(I_B = \frac{2}{5} m_B R^2 = \frac{2}{5} \cdot 1.250 M_{sun} \cdot (20 \times 10^3 m)^2\), \(\omega_B = \frac{2 \pi}{2.8 s}\), \(K_{rot}^B = \frac{1}{2} I_B \omega_B^2\) Now, find the ratio \(\frac{K_{rot}^A}{K_{rot}^B}\) Next, let's move to part (b), in which we need to find the ratio of rotational to translational kinetic energies for each star.

02

b) Calculating the ratio of rotational to translational kinetic energies

For a moving object, the translational kinetic energy (\(K_{trans}\)) is given by the formula: \(K_{trans} = \frac{1}{2} mv^2\), where \(m\) is the mass and \(v\) is the linear velocity. To find the linear velocity from the mean distance (\(r\)) and the orbital period (\(T\)), we can use the formula: \(v = \frac{2 \pi r}{T}\). Now let's find \(K_{trans}\) for each pulsar: Pulsar A: \(v_A = \frac{2 \pi (4.23 \times 10^8 m)}{2.4 h}\) \(K_{trans}^A = \frac{1}{2} m_A v_A^2\) Pulsar B: \(v_B = \frac{2 \pi (4.54 \times 10^8 m)}{2.4 h}\) \(K_{trans}^B = \frac{1}{2} m_B v_B^2\) Now, find the ratios \(\frac{K_{rot}^A}{K_{trans}^A}\) and \(\frac{K_{rot}^B}{K_{trans}^B}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rotational Kinetic Energy

Rotational kinetic energy is the energy possessed by an object due to its rotation. For a pulsar, or any rotating body, the rotational kinetic energy (\(K_{rot}\)) is calculated using the formula:

• \(K_{rot} = \frac{1}{2} I \omega^2\)

where \(I\) represents the moment of inertia and \(\omega\) refers to angular velocity. The moment of inertia signifies how mass is distributed with respect to the axis of rotation and varies based on the shape of the object. For a uniform sphere, like a pulsar, this is expressed as:

• \(I = \frac{2}{5} m R^2\)

This formula shows that the rotational kinetic energy of a pulsar depends significantly on both its mass and radius, squared, as well as the square of its angular velocity.
In the context of our exercise, we apply the rotational kinetic energy formula separately to each pulsar and then determine their ratio. This step helps compare their energies by considering their mass, radius, and the speed at which they spin.

Translational Kinetic Energy

Translational kinetic energy is the energy that an object possesses due to its motion through space. It is different from rotational kinetic energy, which pertains to spinning motion.

• The formula for translational kinetic energy is \(K_{trans} = \frac{1}{2} m v^2\)

where \(m\) is the mass of the object and \(v\) is its velocity. In this celestial binary pulsar system, the translation kinetic energy reflects the energy as the pulsars orbit around their common center of mass.
To calculate this energy, we need to find the velocity (\(v\)) of each pulsar in their orbit. The velocity depends on the distance from the system's center of mass and the orbital period:

• \(v = \frac{2 \pi r}{T}\)

Here, \(r\) is the average distance from the center of mass, and \(T\) is the orbital period. With these calculations, we derive the translational kinetic energy for each pulsar to interpret the relative movement of these stellar giants in a binary system.

Moment of Inertia

The moment of inertia is a critical concept in understanding rotational motion, often dubbed as the rotational equivalent of mass. It quantifies how hard it is to change the rotational motion of an object. For a uniform solid sphere, this is given by:

• \(I = \frac{2}{5} m R^2\)

where \(m\) represents the mass and \(R\) the radius of the object. The larger these values, the greater the moment of inertia, indicating more energy is needed to alter its rotation.
In pulsar systems like PSR J0737-3039A and J0737-3039B, calculating the moment of inertia allows us to comprehend the energy requirements to maintain or alter their rapid spins. As both pulsars are modeled as uniform spheres, their moments of inertia depend directly on their masses and the square of their radii.
These calculations are fundamental not only in determining rotational kinetic energies but also in comparing them across different systems, highlighting the unique dynamics of binary pulsars.