The angular velocity of rotation of a star (of mass M and radius R) at which the matter starts to escape from its equator, is

Question

The angular velocity of rotation of a star (of mass M and radius R) at which the matter starts to escape from its equator, is (A) √(2GM/R) (B) √(2GR/M) (C) √(2GM2/R) (D) √(2GM/R3)

Tardigrade Admin · Accepted Answer

Given: Mass of star = M and radius of the star = R. We know that escape angular velocity of the star (ωe)=(ve/R)=(1/R)√(2GM/R)=√(2GM/R3). (where escape velocity (ve = √(2GM/R))

Q. The angular velocity of rotation of a star (of mass $M$ and radius $R$ ) at which the matter starts to escape from its equator, is

$\frac{2 GM}{R}$

$\frac{2 GR}{M}$

$\frac{2 G M ^{2}}{R}$

$\frac{2 GM}{R ^{3}}$

Given: Mass of star = M and radius of the star = R. We know that escape angular velocity of the star
$(ω_{e}) = \frac{v _{e}}{R} = \frac{1}{R} \frac{2 GM}{R} = \frac{2 GM}{R ^{3}} .$
(where escape velocity $(v_{e} = \frac{2 GM}{R})$

Q. The angular velocity of rotation of a star (of mass M and radius R) at which the matter starts to escape from its equator, is

R2GM​​

M2GR​​

R2GM2​​

R32GM​​