The angular velocity of rotation of star (of mass M and radius R ) at which the matter start to escape from its equator will be

Options

A. `sqrt((2GM^(2))/R)`
B. `sqrt((2GM)/g)`
C. `sqrt((2GM)/R^(2))`
D. `sqrt((2GR)/M)`

Answer

To find the angular velocity of a star at which matter starts to escape from its equator, we can use the concept of gravitational force and centripetal force.

### Step-by-Step Solution:

1. **Understanding the Forces**:
At the equator of the star, the gravitational force must equal the centripetal force required to keep the matter in circular motion. The gravitational force acting on a mass \( m \) at the surface of the star is given by:
\[ F_g = \frac{GMm}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the star, and \( R \) is the radius of the star.

2. **Centripetal Force**:
The centripetal force required to keep the mass \( m \) moving in a circular path at the equator with angular velocity \( \omega \) is given by:
\[ F_c = mR\omega^2 \]

3. **Setting Forces Equal**:
For the matter to just start escaping, the gravitational force must equal the centripetal force:
\[ \frac{GMm}{R^2} = mR\omega^2 \]

4. **Canceling Mass**:
We can cancel the mass \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{GM}{R^2} = R\omega^2 \]

5. **Rearranging the Equation**:
Rearranging the equation gives:
\[ \omega^2 = \frac{GM}{R^3} \]

6. **Taking the Square Root**:
Taking the square root of both sides, we find the angular velocity \( \omega \):
\[ \omega = \sqrt{\frac{GM}{R^3}} \]

### Final Answer:
The angular velocity of rotation of the star at which the matter starts to escape from its equator is:
\[ \omega = \sqrt{\frac{GM}{R^3}} \]

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