Question

Certain neutron stars (extremely dense stars) are believed to be rotating at about 1.6 rev/s. If such a star has a radius of 48 km, what must be its minimum mass so that material on its surface remains in place during the rapid rotation?

Answer 1

Answer:

Mass, $M=4.24\times 10^9\ kg$

Explanation:

Given that,

Angular velocity of the star, $\omega=1.6\ reav/s$

Radius of the star, r = 48 cm = 0.48 m

The centripetal force acting on the on the star is caused by the gravitational force such that,

$\dfrac{GMm}{r^2}=\dfrac{mv^}{r}$

$\dfrac{GM}{r}=v^2$

Since, $v=r\omega$

$M=\dfrac{r^3\omega^2}{G}$

$M=\dfrac{(0.48)^3\times (1.6)^2}{6.67\times 10^{-11}}$

$M=4.24\times 10^9\ kg$

So, the minimum mass so that material on its surface remains in place during the rapid rotation is $4.24\times 10^9\ kg$