Question

Certain neutron stars (extremely dense stars) are believed to be rotating at about 50 rev/s. If such a star has a radius of 15 km, what must be its minimum mass so that material on its surface remains in place during the rapid rotation?

Answer 1

Answer:$M=49.95\times 10^{26} kg$

Explanation:

Given

$N=50 rev/s$

$\omega =2\pi N$

$\omega =2\pi \cdot 50=314.2 rad/s$

radius of neutron star $r=15 km$

Centripetal force on material having mass m is given by

$F_c=\frac{mv^2}{r}$

Gravitational Force between neutron star and mass m is

$F_g=\frac{GMm}{r^2}$  ,where M=mass of Neutron star

equating centripetal Force and Gravitational Pull

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

$M=\frac{v^2r}{G}$

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

$M=\frac{(314.2)^2\times (15000)^3}{6.67\times 10^{-11}}$

$M=\frac{0.4995\times 10^17}{10^{-11}}$

$M=49.95\times 10^{26} kg$