Question

A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (Mass of the sun = 2 × 1030 kg). Question 8.23 Gravitation

Answer 1

As Fg(Gravitational force) > Fc(centrifugal force)
the body will remain stuck to the surface due to gravity .

I hope it will be helpful.

Answer 2

concept:- we know,
when centrifugal force(acting away from the centre of rotation) is less then the gravitational force .then the object will not able to  fly off. i mean it will remain stuckto the surface of stars.


e.g   mg ≥ mv²/r
          g ≥v²/r

here mass of the stars M = 2.5* mass of sun = 5* 10³⁰ kg
 radius R = 12km = 12000m 
 
now ,
g = GM/r² = 6.67*10⁻¹¹*5*10³⁰/(12*10³)²
                 = 2.3 * 10¹² m/s²



now  centripetal acceleration,
v²/r = (2πrη)²/r
      = 4π²η²r 
here, η = 1.2 rev/sec
so,  v²/r  = 4*(3.14)²*(1.2)²*12*10³
              = 0.682 * 10⁶ m/s²

here,  g≥ v²/r
so,the object will remain stuck to the star.