A star 2.5 times the mass of the sun which collapsed to a radius of 12 km rotates with a speed of 1.5 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain observed stellar objects called pulsars are believed to belong to this category). Will an object placed on its equator remain struck to its surface due to gravity ? (Mass of the sun = 2 xx10^30 kg).

A star 2.5 times the mass of the sun which collapsed to a radius of 12

1.	A star 2.5 times the mass of the sun which collapsed to a radius of 12 km rotates with a speed of 1.5 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain observed stellar objects called pulsars are believed to belong to this category). Will an object placed on its equator remain struck to its surface due to gravity ? (Mass of the sun = 2 xx10^30 kg).
Answer» Solution :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(2pi V)^2 = 12000 (2pi xx 1.5)^2 = 1.1 xx 10^6 ms^(-2)` SINCE `g GT r omega^2` , the body will remain struck with the surface of star.