Question

Consider a neutron star with a mass equal to the sun, a radius of 10 kmkm, and a rotation period of 1.0 ss. What is the speed of a point on the equator of the star

Answer 1

The speed of a point on the equator of a neutron star with a radius of 10 km and a rotation period of 1.0 second is 62.83 km/s, which is below the speed of light, thereby not violating Einstein's theory of relativity.

Step-by-step explanation:

To calculate the speed of a point on the equator of a neutron star with a mass equal to the Sun and a radius of 10 km, rotating with a period of 1.0 second, we use the formula for the circumference of a circle and the definition of speed (velocity = distance / time).

The circumference (C) of the neutron star is given by 2πR, where R is the radius of the neutron star. Substituting R with 10 km, we get:

C = 2π × 10 km = 62.83 km.

Given the rotational period (T) is 1.0 second, the speed (v) of a point on the equator is:

v = C / T = 62.83 km / 1.0 s = 62.83 km/s.

This speed must be less than the speed of light to ensure that the pulsar model does not violate Einstein's theory of relativity. Since the speed of light is approximately 300,000 km/s, a rotational speed of 62.83 km/s is well below this threshold, thus consistent with relativity.

Answer 2

6.28 * 10⁴ m/s

Step-by-step explanation:

Using the relation :

Given :

Radius, R = 10km

Period, T = 1.0 s

v = (2πR) / T

Where v = velocity of a point

R in meters ;

1km = 1000m = 10³m

10km = 10 * 10³m = 1 * 10^4 m

v = (2π*10^4m) / 1 s

v = 6.2831853 * 10⁴ m/s

v = 6.28 * 10⁴ m/s