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Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a \(neutron star\). The density of a neutron star is roughly 10\(^{14}\) times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was 7.0 \(\times 10^5\) km (comparable to our sun); its final radius is 16 km. If the original star rotated once in 30 days, find the angular speed of the neutron star.

Short Answer

Expert verified
Approximately 3.26 x 10^3 radians per second.

Step by step solution

01

Understand Angular Momentum Conservation

In a collapse like this, the angular momentum of the star is conserved if no external torques are acting on it. This means that the initial angular momentum of the star must equal the final angular momentum after the collapse.
02

Define Terms and Variables

Let the initial radius of the star be \( R_i = 7.0 \times 10^5 \) km and the final radius be \( R_f = 16 \) km. The initial rotational period is 30 days, and we need to find the final angular speed \( \omega_f \) after the collapse.
03

Initial Angular Momentum

The moment of inertia for a sphere is \( I = \frac{2}{5} m R^2 \). The initial angular velocity \( \omega_i \) is connected to the period \( T_i \) by \( \omega_i = \frac{2\pi}{T_i} \). Initially, with \( T_i = 30 \times 24 \times 60 \times 60 \) seconds, \( \omega_i = \frac{2\pi}{30 \times 24 \times 60 \times 60} \) rad/s.
04

Final Angular Momentum

After the collapse, the star's radius is \( R_f \), and its angular momentum must equal the initial angular momentum, so the equation becomes \( \frac{2}{5} m R_i^2 \omega_i = \frac{2}{5} m R_f^2 \omega_f \). We can cancel \( \frac{2}{5}m \) since it's constant.
05

Solve for Final Angular Speed

We simplify \( R_i^2 \omega_i = R_f^2 \omega_f \). Solving for final angular speed, \( \omega_f = \omega_i \frac{R_i^2}{R_f^2} \). Substitute \( \omega_i \) and the radii: \( \omega_i = \frac{2\pi}{30 \times 24 \times 60 \times 60} \) and \( R_i = 7.0 \times 10^5 \) km, \( R_f = 16 \) km.
06

Calculate \( \omega_f \)

After substitution, \( \omega_f = \frac{2\pi}{30 \times 24 \times 60 \times 60} \left(\frac{(7.0 \times 10^5)^2}{16^2}\right) \). This results in \( \omega_f \approx 3.26 \times 10^3 \) rad/s.

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Key Concepts

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

Neutron Star
Neutron stars are one of the universe’s most fascinating objects. They form when a massive star exhausts its nuclear fuel and undergoes a supernova explosion. This can cause the star to collapse in on itself, squeezing its mass into a much smaller volume. As a result, the neutron star becomes incredibly dense and compact. To put it in perspective, imagine cramming something as massive as the Sun into a sphere only about 20 kilometers across.
  • Neutron stars have densities reaching around 10^{14} times that of regular matter.
  • They primarily consist of neutrons, subatomic particles found in the nucleus of atoms.
  • The gravitational pull of a neutron star is immense, due to its compactness.
Despite their small size, neutron stars possess a tremendous mass and can spin at incredibly high speeds, making them a highlight of study for both astrophysics and gravitational theory.
Moment of Inertia
The moment of inertia is a critical concept when studying rotating bodies. It represents how much torque is needed for a particular angular acceleration. For solid spheres like our star before and after collapse, the moment of inertia can be calculated using the formula:
  • \[ I = \frac{2}{5} m R^2 \]
Here, \( I \) is the moment of inertia, \( m \) is the mass, and \( R \) is the radius of the sphere. The moment of inertia depends heavily on the distribution of mass around the axis of rotation.
When the radius of the star decreases dramatically during the collapse, the moment of inertia decreases as well. Because angular momentum must remain conserved (if no external torque), any change in the moment of inertia means a change in angular speed, which explains why neutron stars often spin so quickly after forming.
Angular Speed Calculation
Calculating the angular speed of a neutron star involves understanding the principles of angular momentum conservation. When a star collapses into a neutron star, its angular momentum pre-collapse must equal its angular momentum post-collapse, assuming no external torques influence the system.
This is mathematically expressed as:
  • \[ \frac{2}{5} m R_i^2 \omega_i = \frac{2}{5} m R_f^2 \omega_f \]
Since the mass \( m \) and the factor \( \frac{2}{5} \) remain constant, they can be canceled out, simplifying to:
  • \[ R_i^2 \omega_i = R_f^2 \omega_f \]
Rearranging gives us the formula for final angular speed:
  • \[ \omega_f = \omega_i \frac{R_i^2}{R_f^2} \]
This shows that the initial angular velocity \( \omega_i \), calculated from the original star's rotational period, gets multiplied by the ratio of the initial radius squared to the final radius squared. Since the final radius is significantly smaller, this results in a markedly increased angular speed, which explains the high spin rates of neutron stars.

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