The condition for a uniform spherical mass `m` of a radius `r` to be a black hole is [`G` =gravitational constant and `g`=acceleration due to gravity]

Options

A. `((2Gm)/(r ))^(1//2) le c`
B. `((2gm)/(r ))^(1//2)=c`
C. `((2 Gm)/(r ))^(1//2) ge c`
D. `((gm)/(r ))^(1//2) ge c`

Explanation

To determine the condition for a uniform spherical mass \( m \) of radius \( r \) to be a black hole, we need to analyze the concept of escape velocity and how it relates to the gravitational pull of the mass.

### Step-by-Step Solution:

1. **Understanding Black Holes**: A black hole is defined as a region in space where the gravitational pull is so strong that nothing, not even light, can escape from it. This implies that the escape velocity from the surface of the mass must be equal to or greater than the speed of light \( c \).

2. **Escape Velocity Formula**: The escape velocity \( V_e \) from the surface of a spherical mass is given by the formula:
\[ V_e = \sqrt{\frac{2Gm}{r}} \]
where \( G \) is the gravitational constant, \( m \) is the mass of the object, and \( r \) is its radius.

3. **Setting Escape Velocity Equal to the Speed of Light**: For the mass to be a black hole, the escape velocity must equal the speed of light:
\[ V_e = c \]

4. **Substituting the Escape Velocity Formula**: We can substitute the escape velocity formula into the equation:
\[ \sqrt{\frac{2Gm}{r}} = c \]

5. **Squaring Both Sides**: To eliminate the square root, we square both sides of the equation:
\[ \frac{2Gm}{r} = c^2 \]

6. **Rearranging the Equation**: We can rearrange the equation to find a condition relating \( m \), \( r \), \( G \), and \( c \):
\[ 2Gm = c^2 r \]

7. **Final Condition**: Thus, the condition for a uniform spherical mass \( m \) of radius \( r \) to be a black hole is:
\[ m = \frac{c^2 r}{2G} \]