A neutron star of mass 2 × 10^30kg and radius 10 km rotates with a period of 0.02 seconds. What is its rotational kinetic energy?

Rotational Kinetic Energy of a Neutron Star

To calculate the rotational kinetic energy \( K \) of the neutron star, we use the formula:

\[ K = \frac{1}{2} I \omega^2 \]

Where:

• \( I \) is the moment of inertia of the neutron star.
• \( \omega \) is the angular velocity of the neutron star.

The moment of inertia \( I \) of a sphere is given by:

\[ I = \frac{2}{5} m r^2 \]

Given:

• Mass of the neutron star, \( m = 2 \times 10^{30} \) kg
• Radius of the neutron star, \( r = 10 \) km = \( 10 \times 10^3 \) m
• Period of rotation, \( T = 0.02 \) seconds

First, let's find the angular velocity \( \omega \):

\[ \omega = \frac{2\pi}{T} \]

Substitute the given values:

\[ \omega = \frac{2\pi}{0.02} = 314.159 \, \text{rad/s} \]

Now, let's calculate the moment of inertia \( I \):

\[ I = \frac{2}{5} m r^2 \] \[ I = 0.8 \times 10^{36} \, \text{kg} \cdot \text{m}^2 \]

Now, let's plug the values into the formula for rotational kinetic energy \( K \):

\[ K = \frac{1}{2} \times (0.8 \times 10^{36}) \times (314.159)^2 \] \[ K \approx 2.501 \times 10^{42} \, \text{J} \]

So, the rotational kinetic energy of the neutron star is approximately \( 2.501 \times 10^{42} \) Joules.