Question

The angular velocity of rotation of star(of mass M and radius R) at which the matter start to escape equator will be
A.$\sqrt {\dfrac{{2G{M^2}}}{R}}$
B.$\sqrt {\dfrac{{2GM}}{g}}$
C.$\sqrt {\dfrac{{2GM}}{{{R^3}}}}$
D.$\sqrt {\dfrac{{2GR}}{M}}$

Answer 1

To answer this question we need to know the formula for escape velocity of earth. So using this formula we will answer this question. We will then find the angular velocity using another formula that relates radius and escape velocity.
Formula used:
$\overrightarrow E = \sqrt {\dfrac{{2GM}}{R}}$
Where $\overrightarrow E$ is the escape velocity.
G is gravitational constant,
M is mass of star
And R is the radius of the star.

Complete answer:
We know that escape velocity is given by ,
$\overrightarrow E = \sqrt {\dfrac{{2GM}}{R}}$
Let us assume that the star rotates with velocity u.
Then, velocity u is given by
$u = \dfrac{v}{R}$ (where v is escape velocity )
Or, $u = \dfrac{1}{R}\sqrt {\dfrac{{2GM}}{R}}$
Or, $u = \sqrt {\dfrac{{2GM}}{{{R^3}}}}$.

Hence the correct answer to this question will be option c.

Additional information:
Lets see what escape velocity is : The minimum speed required for a free, non-propelled object to escape from the gravitational pull of a main body and reach an infinite distance from it is known as escape velocity or escape speed in physics (particularly, celestial mechanics). It's usually expressed as an ideal speed that doesn't account for atmospheric friction. Although the term "escape velocity" is commonly used, it is more accurately described as a speed rather than a velocity because it is directionless; the escape speed increases with the primary body's mass and decreases with distance from the primary body.

Note:
In this question we need to use the formula for angular velocity, not just stop at the escape velocity formula. Also remember that the presence of escape velocity is a result of energy conservation and a finite depth energy field in this question we need to use the formula for angular velocity