Based on law of conservation of angular momentum, the rotational speed of the star is equal to 6,000 rev/s.
Given the following data:
First of all, we would determine the initial angular speed of the neutron star as follows:
[tex]\omega_i = \frac{1 \;Rev}{10 \;days} \\\\\omega_i = \frac{1 \;Rev}{10 \times 24 \times 60 \times 60}\\\\\omega_i = 1.157 \times 10^{-6}\;rev/s[/tex]
Mathematically, the moment of inertia of a uniform solid sphere is given by this formula:
[tex]I=\frac{2}{5} mr^2[/tex]
Where:
In order to determine the rotational speed of this neutron star, we would apply the law of conservation of angular momentum:
[tex]L_1 = L_2\\\\I_1\omega_1 = I_2\omega_2\\\\\frac{2}{5} m_1r_1^2 \omega_1 = \frac{2}{5} m_2r_2^2\omega_2\\\\m_1r_1^2 \omega_1 = m_2r_2^2\omega_2\\\\\omega_2 =\frac{r_1^2 \omega_1}{r_2^2}[/tex]
Substituting the given parameters into the formula, we have;
[tex]\omega_2 = \frac{ (7 \times 10^5 )^2 \times 1.157 \times 10^{-6}}{ (10 )^2}\\\\\\\omega_2 = \frac{ (7 \times 10^5 )^2 \times 1.157 \times 10^{-6}}{ 10 0}\\\\\omega_2 =\frac{566,930}{100}[/tex]
Final angular speed = 5,669 ≈ 6,000 rev/s.
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