Most stars maintain an equilibrium size by balancing two forces - an inward gravitational force and an outward force resulting from the star's nuclear reactions. When the star's fuel is spent, there is no counterbalance to the gravitational force. Whatever material is remaining collapses in on itself. Stars about the same size as the Sun become white dwarfs, which glow from leftover heat. Stars that have about three times the mass of the Sun compact into neutron stars. And a star with a mass greater than three times the Sun's mass collapses into a single point, called a black hole. In most cases, protons and electrons are fused together to form neutrons-this is the reason for the name neutron star. Neutron stars rotate very fast because of the conservation of angular momentum. Imagine a star of mass \(5.00 \cdot 10^{30} \mathrm{~kg}\) and radius \(9.50 \cdot 10^{8} \mathrm{~m}\) that rotates once in 30.0 days. Suppose this star undergoes gravitational collapse to form a neutron star of radius \(10.0 \mathrm{~km} .\) Determine its rotation period.

Step by step solution

01

Recall the equation of angular momentum

The equation for angular momentum (L) is given by: $$L = Iω$$ where I is the moment of inertia and ω is the angular velocity.

02

Calculate the initial angular momentum

Before the collapse, the star has mass \(M_1 = 5.00 \cdot 10^{30} \mathrm{~kg}\), radius \(R_1 = 9.50 \cdot 10^{8} \mathrm{~m}\), and rotation period of \(P_1 = 30.0 \mathrm{~days}\). The moment of inertia for a sphere is given by: $$I_1 = \dfrac{2}{5}M_1R_1^2$$ To find the initial angular velocity, we can use the formula: $$ω_1 = \dfrac{2π}{P_1}$$ Note that we need to convert the period from days to seconds. The initial angular momentum, \(L_1\), is given by: $$L_1 = I_1ω_1$$

03

Calculate the final moment of inertia and angular momentum

After the collapse, the star has a new radius, \(R_2 = 10.0 \mathrm{~km}\). We assume that the mass remains the same during the collapse, so \(M_2 = M_1\). The final moment of inertia, \(I_2\), is given by: $$I_2 = \dfrac{2}{5}M_2R_2^2$$ Using the conservation of angular momentum, we know that the final angular momentum, \(L_2\), is equal to the initial angular momentum, \(L_1\) i.e. $$L_2 = L_1$$

04

Calculate the final angular velocity and rotation period

Now, we can find the final angular velocity, \(ω_2\), using the final angular momentum and moment of inertia: $$ω_2 = \dfrac{L_2}{I_2}$$ To find the final rotation period, \(P_2\), we use the equation: $$P_2 = \dfrac{2π}{ω_2}$$ Finally, we can convert the rotation period from seconds back to days.

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