A particle is projected vertically upward with with velocity `sqrt((2)/(3)(GM)/(R ))` from the surface of Earth. The height attained by it is (G, M, R have usual meanings)

Options

A. 3R
B. `(R )/(2)`
C. 2R
D. `(R )/(3)`

Answer

To solve the problem of finding the height attained by a particle projected vertically upward with a specific velocity from the surface of the Earth, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Initial velocity \( v = \sqrt{\frac{2}{3} \frac{GM}{R}} \)
- Mass of the Earth \( M \)
- Radius of the Earth \( R \)
- Gravitational constant \( G \)

2. **Apply Conservation of Energy**:
The total mechanical energy at the surface of the Earth must equal the total mechanical energy at the maximum height \( h \) reached by the particle.

\[ \text{Total Energy at Surface} = \text{Total Energy at Height } h \]

This can be expressed as:
\[ \text{K.E. at surface} + \text{P.E. at surface} = \text{K.E. at height } h + \text{P.E. at height } h \]

3. **Calculate Kinetic and Potential Energy**:
- **At the surface**:
- Kinetic Energy (K.E.) = \( \frac{1}{2} mv^2 \)
- Potential Energy (P.E.) = \( -\frac{GMm}{R} \)

- **At height \( h \)**:
- K.E. = 0 (at the maximum height, the velocity is 0)
- P.E. = \( -\frac{GMm}{R+h} \)

4. **Set Up the Equation**:
\[ \frac{1}{2} mv^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h} \]

5. **Substitute the Initial Velocity**:
Substitute \( v \) into the equation:
\[ \frac{1}{2} m \left(\sqrt{\frac{2}{3} \frac{GM}{R}}\right)^2 - \frac{GMm}{R} = -\frac{GMm}{R+h} \]

Simplifying the left side:
\[ \frac{1}{2} m \cdot \frac{2}{3} \cdot \frac{GM}{R} - \frac{GMm}{R} = -\frac{GMm}{R+h} \]
\[ \frac{GMm}{3R} - \frac{GMm}{R} = -\frac{GMm}{R+h} \]

6. **Combine Terms**:
\[ \frac{GMm}{3R} - \frac{3GMm}{3R} = -\frac{GMm}{R+h} \]
\[ -\frac{2GMm}{3R} = -\frac{GMm}{R+h} \]

7. **Cancel \( GMm \)**:
Since \( GMm \) is common on both sides, we can cancel it out:
\[ \frac{2}{3R} = \frac{1}{R+h} \]

8. **Cross-Multiply**:
\[ 2(R+h) = 3R \]
\[ 2R + 2h = 3R \]

9. **Solve for \( h \)**:
\[ 2h = 3R - 2R \]
\[ 2h = R \]
\[ h = \frac{R}{2} \]

### Final Answer:
The height attained by the particle is \( h = \frac{R}{2} \).