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

Answer:

Speed,$v=1.25\times 10^5\ m/s$

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,

$v=\dfrac{2\pi R}{t}$

$v=\dfrac{2\pi \times 20000}{1}$

v = 125663.70 m/s

or

$v=1.25\times 10^5\ m/s$

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