A star `2.5` times the mass of the sun and collapsed to a size of the `12 km` rotates with a speed of `1.5` rev.per second. (Extremely compact stars of this kind are known as neutron stars. Centain onserved steller objects called pulsars are believed to belong this category ). Will an object placed on its equator remain struck to its surface due to gravity ? (Mass of the sun `= 2 xx 10^(30) kg)`.

Explanation

The object will remain struck to the surface of star due to gravity, if the acceleration due to gravity is more than the centrifugal acceleration due to its rotation.
Acceleration due to gravity, `g = (GM)/(R^(2)) = (6.67 xx 10^(-11) xx 2.5 xx 2 xx 10^(30))/((12000)^(2)) = 2.3 xx 10^(12) ms^(-2)`
centrifugal acceleration `= r omega^(2) = r(2 pi nu)^(2) = 12000(2 pi xx 1.5)^(2) = 1.1 xx 10^(6) m s^(-2)`
since, `g gt r omega^(2)`, therefore the body remain struck with the surface of star.