Question

When a large star becomes a supernova, its core may be compressed so tightly that it becomes a neutron star, with a radius of about 20 km (about the size of the San Francisco area). If a neutron star rotates once every second, what is the speed of a particle on the star's equator?

Answer 1

Speed,
`v=1.25* 10^5\ m/s`

Step-by-step explanation:

Given that,

Radius of the neutron stare, R = 20 km = 20,000 m

Time taken by the Neutron star to rotate, t = 1 s

We need to find the speed of a particle on the star's equator. The total distance covered divided by total time taken is called speed of an object. Here,

`v=(2\pi R)/(t)`

`v=(2\pi * 20000)/(1)`

v = 125663.70 m/s

or

`v=1.25* 10^5\ m/s`

So, the speed of a particle on the star's equator is
`v=1.25* 10^5\ m/s`. Hence, this is the required solution.