Question

Certain neutron stars (extremely dense stars) are believed to be rotating at about 1 rev/s. If such a star has a radius of \(20 \mathrm{~km}\) (a typical value), \((a)\) what is the speed of a point on the equator of the star, and \((b)\) what is the centripetal acceleration of this point?

Short answer

The speed of a point on the equator of the star is \(125600 m/s\) and the centripetal acceleration of this point is \(7.96 * 10^7 m/s^2\).

Step-by-step solution

Understanding the given information

The problem gives certain values: the neutron star is rotating at 1 rev/s (revolution per second), and the radius of the star is 20 km. We need to convert the radius into meters for easier calculation and better understanding. So, the radius \( r = 20 * 10^3 = 20000 m \). The rotation speed is \( w = 1 rev/s \). To calculate the speed of a point on the star, we need to understand that one revolution is \(2\pi\) radians.

Calculating the speed at the equator of the star

The formula to calculate speed (v) from rotational speed is \( v = w * r \). To use it directly, we need to convert the rotational speed \( w \) from rev/s to rad/s. As one revolution is \(2\pi\) radians, then \( w = 1 rev/s = 2\pi rad/s = 2*3.14 rad/s \). Substituting the given rotation speed and radius into the formula, we find \( v = 2*3.14 rad/s * 20000 m = 125600 m/s \). This is the speed at the equator of the star.

Calculating the centripetal acceleration at the equator

Centripetal acceleration \( a \) at a point in rotational motion is given by the formula \( a = w^2 * r \). Plugging in the values we have, \( a = (2*3.14 rad/s)^2 * 20000 m = 7.96*10^7 m/s^2 \). This is the centripetal acceleration at the point on the equator of the neutron star.

Key concepts

Rotational Motion

Rotational motion is the movement of an object in a circular path around a fixed point or axis. In the context of a neutron star, this fixed point would be the star's center of mass. An object undergoing rotational motion experiences a continuous change in direction that results in a velocity vector that is always tangent to the circular path.

For a neutron star rotating at 1 revolution per second, it's helpful to visualize the equator as a point moving in a circle with a radius of 20 kilometers. The rotational speed, or angular velocity, determines how fast this point travels along the circular path. Since the distance covered in one complete revolution is the circumference of the circle, we can determine the linear speed at the equator by relating the angular velocity to the radius. In essence, the faster the neutron star spins, the greater the speed at any point along its equator. It's akin to a child on a merry-go-round; the further from the center, the faster they seem to go.

Centripetal Acceleration

Centripetal acceleration is the acceleration that is required to keep an object moving in a circular path and is directed towards the center of the circle. This type of acceleration is necessary for any type of rotational motion because, without it, the object would move off in a straight line due to inertia. For a massive object like a neutron star, the force providing this acceleration is gravity.

The formula to calculate centripetal acceleration is \( a = w^2 * r \) which relates directly to the angular velocity (\(w\)) squared and the radius of the rotation (\(r\)). The extraordinary speed of the neutron star's rotation results in a massive centripetal acceleration, much greater than we experience on Earth. If an object on the equator of the neutron star were to suddenly lose this acceleration, it would be propelled straight out into space at an incredible speed due to the immense force that was keeping it in circular motion.

Angular Velocity

Angular velocity is a measure of the rate of rotation. It specifies how quickly an object rotates or revolves around an axis. The angular velocity is not the same as the linear velocity (speed along a straight line) but is related to it through the radius of the circular path. For the neutron star spinning at 1 revolution per second, the angular velocity is 2π radians per second, because there are 2π radians in one complete revolution.

Expressing the angular velocity in radians allows us to calculate various other rotational parameters, such as linear speed and centripetal acceleration. It's important to understand that although all points on the neutron star have the same angular velocity (since the star rotates as a rigid body), the linear speed of a point depends on its distance from the axis of rotation—the larger the radius, the higher the linear speed at the equator. This concept is much like a record player, with parts closer to the edge traveling faster than those near the center, despite all parts completing one revolution in the same amount of time.