For a \(1.5 \mathrm{M}_{\odot}\) neutron star with \(R=15 \mathrm{km},\) rotating 100 times per second, compare the gravitational force on an object at the surface with the force required to produce the circular motion for that object (at the star's equator). In other words, compare the weight of an object with the centripetal force on it.

Gravitational force is the attraction between two masses. The formula to calculate this force is given by Newton's law of universal gravitation:
\[ F_g = \frac{G M m}{R^2} \]
Here, \( G \) is the gravitational constant, \( M \) is the mass of the neutron star, \( m \) is the mass of the object, and \( R \) is the radius of the neutron star.
The gravitational force is always attractive, meaning it pulls objects toward the center of mass. For a neutron star, which is incredibly dense, this force is immense at the surface.
To use the formula, plug in:

• \( G \) = \(6.67 \times 10^{-11} \, \text{N} \, \text{m}^2 \text{kg}^{-2}\)
• \( M \) = \(1.5 \times 1.989 \times 10^{30} \, \text{kg}\)
• \( R \) = \( 15 \times 10^3 \, \text{m}\)

In this problem, the gravitational force calculated is \( F_g \approx 8.84 \times 10^{11} m \), showing a significant force acting on any object on the neutron star's surface.

Centripetal force (\( F_c \)) is the force required to keep an object moving in a circular path. This force acts towards the center of the circle, maintaining the circular motion.
The formula for centripetal force is:
\[ F_c = \frac{m v^2}{R} \]
Here,

• \( m \) is the mass of the object
• \( v \) is the tangential velocity
• \( R \) is the radius of the circular path

For an object at the equator of a neutron star rotating 100 times per second, we can determine the velocity as:
\[ v = 2 \times \text{π} \times R \times f \]
where \( f \) is the frequency of rotation. Substituting the given values:
\( R = 15 \times 10^3 \, \text{m} \) and \( f = 100 \, \text{Hz} \)
we find \( v \approx 9.42 \times 10^5 \, \text{m/s} \).
Using this velocity in the centripetal force formula, we find \( F_c \approx 59.21 \times 10^6 m \). This demonstrates how strong the centripetal force needs to be to keep an object in circular motion on such a rapidly rotating neutron star.

Neutron stars are the remnants of massive stars that have undergone supernova explosions. These stars are incredibly dense, with masses about 1.4 times that of our Sun but diameters of only about 20 kilometers.
Key characteristics include:

• Extreme density: A single teaspoon of neutron star material would weigh billions of tons on Earth.
• Strong gravity: Their surface gravity is about \( 2 \times 10^{11} \) times that of Earth's.
• Intense magnetic fields: Hundreds of millions to a billion times stronger than Earth's magnetic field.

These factors result in extreme conditions that greatly affect the physics involved. When examining forces on a neutron star, both the gravitational and centripetal forces are of a magnitude far beyond anything experienced on Earth.

Rotational motion refers to an object spinning around an axis. For neutron stars, this involves extremely rapid rotation rates, sometimes hundreds of times per second.
Key concepts in rotational motion include:

• Angular velocity (\( \omega \)): The rate of change of the angle, often given in radians per second.
• Tangential velocity (\( v \)): The linear speed of a point on the rotating object, found using \( v = \text{ω} \times R \).

For a neutron star rotating 100 times per second, the angular velocity (\( \omega \)) would be:
\[ \omega = 2 \times \text{π} \times f = 2 \times \text{π} \times 100 \, \text{radians/s} \]
resulting in \( \omega \approx 628 \, \text{radians/s} \).
This rotational motion contributes to the centripetal force required to keep an object at the neutron star's surface moving in a circle. The interplay between these forces elucidates the extreme and fascinating nature of neutron stars.