The earth is assumed to be a sphere of raduis `R`. A plateform is arranged at a height `R` from the surface of the `fv_(e)`, where `v_(e)` is its escape velocity form the surface of the earth. The value of `f` is

Options

A. `sqrt(2)`
B. `(1)/(sqrt(2))`
C. `(1)/(3)`
D. `(1)/(2)`

Answer

To solve the problem, we need to find the value of \( f \) when a platform is arranged at a height \( R \) from the surface of the Earth, where \( R \) is the radius of the Earth and \( v_e \) is the escape velocity from the surface of the Earth.

### Step-by-Step Solution:

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

2. **Position of the Platform**:
The platform is at a height \( R \) from the surface of the Earth. Therefore, the total distance from the center of the Earth to the platform is:
\[ d = R + R = 2R \]

3. **Escape Velocity at Height \( R \)**:
The escape velocity \( v' \) at this height can be derived from the gravitational potential energy and kinetic energy considerations. The total mechanical energy must be zero for the object to escape:
\[ \text{Kinetic Energy} + \text{Potential Energy} = 0 \]
The kinetic energy at height \( R \) is:
\[ KE = \frac{1}{2} m v'^2 \]
The potential energy at height \( 2R \) is:
\[ PE = -\frac{GMm}{2R} \]
Setting the total energy to zero:
\[ \frac{1}{2} m v'^2 - \frac{GMm}{2R} = 0 \]

4. **Solving for \( v' \)**:
From the equation:
\[ \frac{1}{2} m v'^2 = \frac{GMm}{2R} \]
We can cancel \( m \) (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v'^2 = \frac{GM}{2R} \]
Multiplying both sides by 2:
\[ v'^2 = \frac{GM}{R} \]
Taking the square root:
\[ v' = \sqrt{\frac{GM}{R}} \]

5. **Relating \( v' \) to \( v_e \)**:
We know:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
Now, we can express \( v' \) in terms of \( v_e \):
\[ v' = \sqrt{\frac{GM}{R}} = \sqrt{\frac{1}{2}} \cdot \sqrt{\frac{2GM}{R}} = \frac{1}{\sqrt{2}} v_e \]
Therefore, we can write:
\[ v' = f \cdot v_e \]
where \( f = \frac{1}{\sqrt{2}} \).

6. **Final Result**:
Thus, the value of \( f \) is:
\[ f = \frac{1}{\sqrt{2}} \]