Answer:
Mass, [tex]M=4.24\times 10^9\ kg[/tex]
Explanation:
Given that,
Angular velocity of the star, [tex]\omega=1.6\ reav/s[/tex]
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,
[tex]\dfrac{GMm}{r^2}=\dfrac{mv^}{r}[/tex]
[tex]\dfrac{GM}{r}=v^2[/tex]
Since, [tex]v=r\omega[/tex]
[tex]M=\dfrac{r^3\omega^2}{G}[/tex]
[tex]M=\dfrac{(0.48)^3\times (1.6)^2}{6.67\times 10^{-11}}[/tex]
[tex]M=4.24\times 10^9\ kg[/tex]
So, the minimum mass so that material on its surface remains in place during the rapid rotation is [tex]4.24\times 10^9\ kg[/tex]
