A body is projected vertically upwards from the surface of the earth with a velocity equal to half of escape velocity of the earth. If `R` is radius of the earth, maximum height attained by the body from the surface of the earth is

Options

A. `(R)/(6)`
B. `(R)/(3)`
C. `(2R)/(3)`
D. `R`

Explanation

To solve the problem of finding the maximum height attained by a body projected vertically upwards with a velocity equal to half of the escape velocity of the Earth, we can use the principle of conservation of mechanical energy.

### Step-by-step Solution:

1. **Identify Escape Velocity**:
The escape velocity \( v_{\text{escape}} \) from the surface of the Earth is given by the formula:
\[ v_{\text{escape}} = \sqrt{2gR} \]
where \( g \) is the acceleration due to gravity and \( R \) is the radius of the Earth.

2. **Determine Initial Velocity**:
Since the body is projected with a velocity equal to half of the escape velocity, we have:
\[ v_0 = \frac{1}{2} v_{\text{escape}} = \frac{1}{2} \sqrt{2gR} \]

3. **Apply Conservation of Mechanical Energy**:
The total mechanical energy at the surface of the Earth (initial) is equal to the total mechanical energy at the maximum height (final). The initial kinetic energy (KE) and gravitational potential energy (PE) at the surface are:
\[ \text{Initial KE} = \frac{1}{2} m v_0^2 \]
\[ \text{Initial PE} = -\frac{GMm}{R} \]
At maximum height \( h \), the kinetic energy becomes zero and the potential energy is:
\[ \text{Final PE} = -\frac{GMm}{R+h} \]

4. **Set Up the Energy Conservation Equation**:
The conservation of energy gives us:
\[ \frac{1}{2} m v_0^2 - \frac{GMm}{R} = -\frac{GMm}{R+h} \]

5. **Substituting Values**:
Substitute \( v_0 \) into the equation:
\[ \frac{1}{2} m \left(\frac{1}{2} \sqrt{2gR}\right)^2 - \frac{GMm}{R} = -\frac{GMm}{R+h} \]
Simplifying the left side:
\[ \frac{1}{2} m \cdot \frac{1}{4} \cdot 2gR - \frac{GMm}{R} = -\frac{GMm}{R+h} \]
\[ \frac{mgR}{4} - \frac{GMm}{R} = -\frac{GMm}{R+h} \]

6. **Using \( g = \frac{GM}{R^2} \)**:
Substitute \( g \) in terms of \( G \) and \( R \):
\[ \frac{m \cdot \frac{GM}{R^2} \cdot R}{4} - \frac{GMm}{R} = -\frac{GMm}{R+h} \]
\[ \frac{GMm}{4R} - \frac{GMm}{R} = -\frac{GMm}{R+h} \]

7. **Combine Terms**:
Combine the left side:
\[ \left(\frac{1}{4} - 1\right) \frac{GMm}{R} = -\frac{GMm}{R+h} \]
\[ -\frac{3}{4} \frac{GMm}{R} = -\frac{GMm}{R+h} \]

8. **Cross Multiply and Solve for \( h \)**:
\[ \frac{3}{4} (R+h) = R \]
\[ 3R + 3h = 4R \]
\[ 3h = R \implies h = \frac{R}{3} \]

### Final Answer:
The maximum height attained by the body from the surface of the Earth is:
\[ h = \frac{R}{3} \]