Question

An extremely small and dense neutron star of mass $M$ and radius $R$ is rotating at an angular frequency $\omega$. If an object is placed at its equator, it will remain stuck to it due to gravity if

• A. $M>\frac{R\omega }{G}$
• B. $M>\frac{R^2{\omega }^2}{G}$
• C. $M>\frac{R^3{\omega }^2}{G}$
• D. $M>\frac{R^2{\omega }^3}{G}$

Answer 1

Answer: (C)

$M>\frac{R^3{\omega }^2}{G}$

If we are working from frame of reference of star(rotating around its own axis) then they are two forces acting on the body.
1 gravitational force (due to mass) towards the centre of star
2 Centrifugal force (pseudo force ) away from the centre (we need pseudo force because we are applying newtons laws of motion in non-inertial frame) so it will remain stuck to it if force towards centre is greater than force away from the centre.
An object of mass $m$, placed at the equator of the star, will experience two forces: (i) an attractive force due to gravity towards the centre of the star and (ii) an outward centrifugal force due to the rotation of the star. The centrifugal force arises because the object is in a rotating (non-inertial) frame; this force is equal to the inward centripetal force but opposite in direction. Force on object due to gravity is
$F_g=\frac{GmM}{R^2}$
Centrifugal force on the object is
$F_c=mR{\omega }^2$
The object will remain stuck to the star and not fly off if
$F_g>F_c$
or $\frac{GmM}{R^2}>mR{\omega }^2$ or $M>\frac{R^3{\omega }^2}{G}$
Hence the correct choice is (c).