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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