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 need to find the ratio of the velocity of projection (v) to the escape velocity (v_escape) at the surface of the Earth, given that the particle reaches a maximum height equal to the radius of the Earth (R).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A particle is thrown vertically upwards from the surface of the Earth.
- It reaches a maximum height equal to the radius of the Earth (R).
- We need to find the ratio of the initial velocity of projection (v) to the escape velocity (v_escape).

2. **Using Conservation of Mechanical Energy**:
- At the point of projection, the particle has kinetic energy (KE) and at the maximum height, all this energy is converted into gravitational potential energy (PE).
- The change in kinetic energy is equal to the change in potential energy:
\[ \Delta KE = \Delta PE \]

3. **Setting Up the Energy Equations**:
- The initial kinetic energy when the particle is projected is:
\[ KE = \frac{1}{2}mv^2 \]
- The potential energy at the surface of the Earth is:
\[ PE_{initial} = -\frac{GMm}{R} \]
- The potential energy at the maximum height (2R from the center of the Earth) is:
\[ PE_{final} = -\frac{GMm}{2R} \]

4. **Calculating the Change in Potential Energy**:
- The change in potential energy as the particle moves from the surface to the height R is:
\[ \Delta PE = PE_{final} - PE_{initial} = -\frac{GMm}{2R} + \frac{GMm}{R} = \frac{GMm}{2R} \]

5. **Equating Kinetic and Potential Energy**:
- Setting the change in kinetic energy equal to the change in potential energy:
\[ \frac{1}{2}mv^2 = \frac{GMm}{2R} \]
- Cancelling \(m\) from both sides (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:
\[ v = \sqrt{\frac{GM}{R}} \]

6. **Finding the Escape Velocity**:
- The escape velocity from the surface of the Earth is given by:
\[ v_{escape} = \sqrt{\frac{2GM}{R}} \]

7. **Calculating the Ratio**:
- Now we find the ratio of the velocity of projection to the escape velocity:
\[ \frac{v}{v_{escape}} = \frac{\sqrt{\frac{GM}{R}}}{\sqrt{\frac{2GM}{R}}} \]
- Simplifying the ratio:
\[ \frac{v}{v_{escape}} = \frac{\sqrt{1}}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]

8. **Final Answer**:
- The ratio of the velocity of projection to the escape velocity is:
\[ \frac{v}{v_{escape}} = \frac{1}{\sqrt{2}} \]