A particle of mass ‘m’ is kept at rest at a height 3R from the surface of earth, where ‘R’ is radius of earth and ‘M’ is mass of earth. The minimum speed with which it should be projected, so that it does not return back, is (g is acceleration due to gravity on the surface of earth)

Options

A. `((GM)/R)^(1/2)`
B. `((GM)/(2R))^(1/2)`
C. `((gR)/4)^(1/4)`
D. `((2g)/R)^(1/2)`

Answer

To solve the problem, we need to find the minimum speed with which a particle of mass 'm' should be projected from a height of \(3R\) above the Earth's surface so that it does not return back. Here, \(R\) is the radius of the Earth, and \(M\) is the mass of the Earth. We will use the concepts of gravitational potential energy and kinetic energy.

### Step-by-Step Solution:

1. **Identify the height from the center of the Earth**:
- The particle is at a height of \(3R\) above the Earth's surface.
- Therefore, the distance from the center of the Earth to the particle is:
\[ d = R + 3R = 4R \]

2. **Calculate the initial gravitational potential energy (PE)**:
- The gravitational potential energy at a distance \(d\) from the center of the Earth is given by:
\[ PE = -\frac{GMm}{d} \]
- Substituting \(d = 4R\):
\[ PE_1 = -\frac{GMm}{4R} \]

3. **Calculate the final gravitational potential energy (PE) at infinity**:
- As the particle escapes to infinity, the potential energy approaches zero:
\[ PE_2 = 0 \]

4. **Use the conservation of energy**:
- The kinetic energy (KE) of the particle when projected must equal the change in potential energy:
\[ KE = PE_2 - PE_1 \]
- Therefore:
\[ KE = 0 - \left(-\frac{GMm}{4R}\right) = \frac{GMm}{4R} \]

5. **Express kinetic energy in terms of velocity**:
- The kinetic energy is also given by:
\[ KE = \frac{1}{2} mv^2 \]
- Setting the two expressions for kinetic energy equal:
\[ \frac{1}{2} mv^2 = \frac{GMm}{4R} \]

6. **Solve for velocity \(v\)**:
- Cancel \(m\) from both sides (assuming \(m \neq 0\)):
\[ \frac{1}{2} v^2 = \frac{GM}{4R} \]
- Multiply both sides by 2:
\[ v^2 = \frac{GM}{2R} \]
- Taking the square root gives:
\[ v = \sqrt{\frac{GM}{2R}} \]

7. **Relate \(g\) to \(GM\)**:
- The acceleration due to gravity \(g\) at the surface of the Earth is given by:
\[ g = \frac{GM}{R} \]
- Thus, we can express \(GM\) as:
\[ GM = gR \]

8. **Substitute \(GM\) back into the velocity equation**:
- Replace \(GM\) in the velocity equation:
\[ v = \sqrt{\frac{gR}{2R}} = \sqrt{\frac{g}{2}} \cdot \sqrt{R} \]

9. **Final expression for the minimum speed**:
- Therefore, the minimum speed \(v\) required is:
\[ v = \sqrt{\frac{gR}{2}} \]

### Conclusion:
The minimum speed with which the particle should be projected so that it does not return back is:
\[ v = \sqrt{\frac{gR}{2}} \]