A particle is given a velocity `(v_(e ))/(sqrt(3))` in a vertically upward direction from the surface of the earth, where `v_(e )` is the escape velocity from the surface of the earth. Let the radius of the earth be R. The height of the particle above the surface of the earth at the instant it comes to rest is :

Options

A. 3R
B. 2R
C. `(3R)/(4)`
D. `(R )/(2)`

Explanation

To solve the problem, we will use the principle of conservation of mechanical energy. The total mechanical energy (kinetic energy + potential energy) at the initial position (when the particle is projected) will be equal to the total mechanical energy at the final position (when the particle comes to rest).

### Step-by-step Solution:

1. **Identify Initial Conditions:**
- The initial velocity of the particle is given as \( v = \frac{v_e}{\sqrt{3}} \), where \( v_e \) is the escape velocity.
- The initial kinetic energy \( K_1 \) can be calculated using the formula:
\[ K_1 = \frac{1}{2} m v^2 = \frac{1}{2} m \left( \frac{v_e}{\sqrt{3}} \right)^2 = \frac{1}{2} m \frac{v_e^2}{3} \]

2. **Calculate Escape Velocity:**
- The escape velocity \( v_e \) is given by:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
- Therefore, substituting this into the kinetic energy expression:
\[ K_1 = \frac{1}{2} m \frac{1}{3} \left( \frac{2GM}{R} \right) = \frac{mGM}{3R} \]

3. **Potential Energy at Initial Position:**
- The potential energy \( U_1 \) at the surface of the Earth (height = 0) is:
\[ U_1 = -\frac{GMm}{R} \]

4. **Total Initial Energy:**
- The total initial energy \( E_1 \) is:
\[ E_1 = K_1 + U_1 = \frac{mGM}{3R} - \frac{GMm}{R} = \frac{mGM}{3R} - \frac{3mGM}{3R} = -\frac{2mGM}{3R} \]

5. **Final Conditions:**
- At the maximum height \( h \), the particle comes to rest, so its kinetic energy \( K_2 = 0 \).
- The potential energy \( U_2 \) at height \( h \) is given by:
\[ U_2 = -\frac{GMm}{R + h} \]

6. **Total Final Energy:**
- The total final energy \( E_2 \) is:
\[ E_2 = K_2 + U_2 = 0 - \frac{GMm}{R + h} = -\frac{GMm}{R + h} \]

7. **Setting Initial Energy Equal to Final Energy:**
- Using the conservation of energy:
\[ E_1 = E_2 \]
\[ -\frac{2mGM}{3R} = -\frac{GMm}{R + h} \]

8. **Solving for \( h \):**
- Cancel \( -GMm \) from both sides:
\[ \frac{2}{3R} = \frac{1}{R + h} \]
- Cross-multiplying gives:
\[ 2(R + h) = 3R \]
\[ 2h = 3R - 2R = R \]
\[ h = \frac{R}{2} \]

### Final Answer:
The height of the particle above the surface of the Earth at the instant it comes to rest is \( \frac{R}{2} \).

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