A body is projected vertically up from surface of the earth with a velocity half of escape velocity. The ratio of its maximum height of ascent and radius of earth is

Options

A. `1:1`
B. `1:2`
C. `1:3`
D. `1:4`

Explanation

To solve the problem, we need to find the ratio of the maximum height \( h \) attained by a body projected vertically upwards from the surface of the Earth with a velocity equal to half of the escape velocity, to the radius of the Earth \( R \).

### Step-by-Step Solution:

1. **Understand Escape Velocity**:
The escape velocity \( V_e \) from the surface of the Earth is given by the formula:
\[ V_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Initial Velocity**:
The body is projected with a velocity \( V_1 \) which is half of the escape velocity:
\[ V_1 = \frac{1}{2} V_e = \frac{1}{2} \sqrt{\frac{2GM}{R}} = \sqrt{\frac{GM}{2R}} \]

3. **Conservation of Mechanical Energy**:
The total mechanical energy at the surface (initial state) is equal to the total mechanical energy at the maximum height \( h \) (final state). The total energy consists of kinetic energy (KE) and potential energy (PE).

- **At the Surface**:
- Kinetic Energy (KE) at surface:
\[ KE = \frac{1}{2} mv_1^2 = \frac{1}{2} m \left(\sqrt{\frac{GM}{2R}}\right)^2 = \frac{1}{2} m \cdot \frac{GM}{2R} = \frac{mGM}{4R} \]
- Potential Energy (PE) at surface:
\[ PE = -\frac{GMm}{R} \]
- Total Energy at surface:
\[ E_{initial} = KE + PE = \frac{mGM}{4R} - \frac{GMm}{R} = \frac{mGM}{4R} - \frac{4mGM}{4R} = -\frac{3mGM}{4R} \]

- **At Maximum Height \( h \)**:
- Kinetic Energy (KE) at maximum height:
\[ KE = 0 \quad (\text{at maximum height, velocity is zero}) \]
- Potential Energy (PE) at height \( h \):
\[ PE = -\frac{GMm}{R+h} \]
- Total Energy at height \( h \):
\[ E_{final} = 0 - \frac{GMm}{R+h} = -\frac{GMm}{R+h} \]

4. **Setting Energies Equal**:
From conservation of energy, we set the total energy at the surface equal to the total energy at height \( h \):
\[ -\frac{3mGM}{4R} = -\frac{GMm}{R+h} \]

5. **Canceling Common Terms**:
We can cancel \( -GMm \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{3}{4R} = \frac{1}{R+h} \]

6. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 3(R + h) = 4R \]
Simplifying this:
\[ 3R + 3h = 4R \implies 3h = 4R - 3R \implies 3h = R \implies h = \frac{R}{3} \]

7. **Finding the Ratio**:
Now we find the ratio of maximum height \( h \) to the radius of the Earth \( R \):
\[ \frac{h}{R} = \frac{R/3}{R} = \frac{1}{3} \]

### Final Answer:
The ratio of the maximum height of ascent to the radius of the Earth is:
\[ \frac{h}{R} = \frac{1}{3} \]