The escape velocity of a particle of a particle from the surface of the earth is given by

Options

A. `(gR)^(1//2)`
B. `(2gR)^(1//2)`
C. `(3gR)^(1//2)`
D. `(gR//2)^(1//2)`

Explanation

To find the escape velocity of a particle from the surface of the Earth, we can follow these steps:

### Step 1: Understand the formula for escape velocity
The escape velocity (\( V_{escape} \)) from the surface of a celestial body is given by the formula:
\[ V_{escape} = \sqrt{\frac{2GM}{R}} \]
where:
- \( G \) is the universal gravitational constant (\( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)),
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth.

### Step 2: Relate gravitational acceleration to escape velocity
We know that the gravitational acceleration (\( g \)) at the surface of the Earth can be expressed as:
\[ g = \frac{GM}{R^2} \]
From this, we can rearrange the equation to express \( GM \):
\[ GM = gR^2 \]

### Step 3: Substitute \( GM \) into the escape velocity formula
Now, we can substitute \( GM \) into the escape velocity formula:
\[ V_{escape} = \sqrt{\frac{2gR^2}{R}} = \sqrt{2gR} \]

### Step 4: Final expression for escape velocity
Thus, the escape velocity from the surface of the Earth can be simplified to:
\[ V_{escape} = \sqrt{2gR} \]

### Conclusion
Therefore, the escape velocity of a particle from the surface of the Earth is given by:
\[ V_{escape} = \sqrt{2gR} \]