A particle is throws vertically upwards from the surface of earth and it reaches to a maximum height equal to the radius of earth. The radio of the velocity of projection to the escape velocity on the surface of earth is

Options

A. `(1)/(sqrt2)`
B. `(1)/(2)`
C. `(1)/(4)`
D. `(1)/(2sqrt2)`

Explanation

To solve the problem, we will use the principle of conservation of energy. We need to find the ratio of the velocity of projection (V₁) to the escape velocity (Vₑ) on the surface of the Earth.

### Step-by-Step Solution:

1. **Understand the Problem**:
A particle is thrown vertically upwards and reaches a maximum height equal to the radius of the Earth (R). We need to find the ratio of the initial velocity of projection (V₁) to the escape velocity (Vₑ).

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

3. **Initial Energy (Eᵢ)**:
The initial energy of the particle when it is thrown is the sum of its kinetic energy and gravitational potential energy:
\[ Eᵢ = \frac{1}{2} m V₁^2 - \frac{GMm}{R} \]
Here, \( m \) is the mass of the particle.

4. **Final Energy (E_f)**:
At the maximum height (h = R), the kinetic energy becomes zero, and the potential energy at that height is:
\[ E_f = -\frac{GMm}{2R} \]
(since the height is equal to the radius of the Earth).

5. **Apply Conservation of Energy**:
According to the conservation of energy, the initial energy equals the final energy:
\[ Eᵢ = E_f \]
Substituting the expressions we have:
\[ \frac{1}{2} m V₁^2 - \frac{GMm}{R} = -\frac{GMm}{2R} \]

6. **Simplify the Equation**:
Rearranging the equation gives:
\[ \frac{1}{2} m V₁^2 = \frac{GMm}{R} - \frac{GMm}{2R} \]
\[ \frac{1}{2} m V₁^2 = \frac{GMm}{2R} \]
Dividing both sides by \( m \) (mass of the particle):
\[ \frac{1}{2} V₁^2 = \frac{GM}{2R} \]
Multiplying by 2:
\[ V₁^2 = \frac{GM}{R} \]

7. **Calculate V₁**:
Taking the square root gives:
\[ V₁ = \sqrt{\frac{GM}{R}} \]

8. **Calculate the Ratio**:
Now, we can find the ratio of V₁ to Vₑ:
\[ \text{Ratio} = \frac{V₁}{Vₑ} = \frac{\sqrt{\frac{GM}{R}}}{\sqrt{\frac{2GM}{R}}} \]
Simplifying this gives:
\[ \text{Ratio} = \frac{1}{\sqrt{2}} \]

9. **Final Answer**:
Therefore, the ratio of the velocity of projection to the escape velocity is:
\[ \frac{V₁}{Vₑ} = \frac{1}{\sqrt{2}} \]