If `M_(e )` and `R_(e )` be the mass and radius of earth, then the escape velocity will be _________.

Answer

To find the escape velocity from the surface of the Earth, we can use the principle of conservation of mechanical energy. The escape velocity is the minimum velocity required for an object to escape the gravitational pull of the Earth without any additional propulsion.

### Step-by-Step Solution:

1. **Understanding Escape Velocity**:
Escape velocity is the minimum speed needed for an object to break free from the gravitational attraction of a celestial body, in this case, Earth.

2. **Conservation of Mechanical Energy**:
The total mechanical energy (kinetic energy + potential energy) at the surface of the Earth must equal the total mechanical energy at a point far away from the Earth (where gravitational potential energy is zero).

3. **Kinetic Energy at the Surface**:
At the surface of the Earth, the kinetic energy (K.E) of the object can be expressed as:
\[ K.E = \frac{1}{2} m v_e^2 \]
where \( m \) is the mass of the object and \( v_e \) is the escape velocity.

4. **Potential Energy at the Surface**:
The gravitational potential energy (P.E) at the surface of the Earth is given by:
\[ P.E = -\frac{G M_e m}{R_e} \]
where \( G \) is the universal gravitational constant, \( M_e \) is the mass of the Earth, and \( R_e \) is the radius of the Earth.

5. **Total Energy at the Surface**:
The total mechanical energy at the surface of the Earth is:
\[ E_{initial} = K.E + P.E = \frac{1}{2} m v_e^2 - \frac{G M_e m}{R_e} \]

6. **Total Energy Far Away**:
At a point far away from the Earth (as \( r \to \infty \)), the kinetic energy will be zero and the potential energy will also be zero:
\[ E_{final} = 0 \]

7. **Setting Initial Energy Equal to Final Energy**:
By conservation of energy:
\[ \frac{1}{2} m v_e^2 - \frac{G M_e m}{R_e} = 0 \]

8. **Solving for Escape Velocity**:
Rearranging the equation gives:
\[ \frac{1}{2} m v_e^2 = \frac{G M_e m}{R_e} \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v_e^2 = \frac{G M_e}{R_e} \]
Multiplying both sides by 2:
\[ v_e^2 = \frac{2 G M_e}{R_e} \]
Taking the square root:
\[ v_e = \sqrt{\frac{2 G M_e}{R_e}} \]

### Final Answer:
The escape velocity from the surface of the Earth is:
\[ v_e = \sqrt{\frac{2 G M_e}{R_e}} \]