A particle is situated at a height `3 R` from the earth surface . The velocity with which it should be projected vertically upward so that it does not return to earth is

Options

A. `sqrt((GM)/(R ))`
B. `sqrt((2GM)/(R ))`
C. `sqrt((GM)/(2R ))`
D. `sqrt((GM)/(4R ))`

Answer

To solve the problem of determining the velocity with which a particle should be projected vertically upward from a height of \(3R\) above the Earth's surface so that it does not return to Earth, we can follow these steps:

### Step 1: Understand the Problem
We need to find the escape velocity for a particle located at a height of \(3R\) from the Earth's surface. The total distance from the center of the Earth will be \(4R\) (since the radius of the Earth is \(R\) and the particle is at a height of \(3R\)).

### Step 2: Apply Conservation of Mechanical Energy
The principle of conservation of mechanical energy states that the total mechanical energy (kinetic + potential) at the initial position must equal the total mechanical energy at the final position (which is at infinity, where potential energy is zero).

### Step 3: Write the Expressions for Potential and Kinetic Energy
1. **Initial Potential Energy (U_initial)** at height \(3R\):
\[ U_{\text{initial}} = -\frac{GMm}{r} \]
Here, \(r = 4R\) (the distance from the center of the Earth), so:
\[ U_{\text{initial}} = -\frac{GMm}{4R} \]

2. **Initial Kinetic Energy (K_initial)** when projected with velocity \(V_e\):
\[ K_{\text{initial}} = \frac{1}{2}mV_e^2 \]

3. **Final Potential Energy (U_final)** at infinity:
\[ U_{\text{final}} = 0 \]

4. **Final Kinetic Energy (K_final)** at infinity:
\[ K_{\text{final}} = 0 \]

### Step 4: Set Up the Energy Conservation Equation
Using the conservation of energy:
\[ K_{\text{initial}} + U_{\text{initial}} = K_{\text{final}} + U_{\text{final}} \]
Substituting the values:
\[ \frac{1}{2}mV_e^2 - \frac{GMm}{4R} = 0 \]

### Step 5: Solve for Escape Velocity \(V_e\)
Rearranging the equation:
\[ \frac{1}{2}mV_e^2 = \frac{GMm}{4R} \]
Dividing both sides by \(m\) (mass of the particle):
\[ \frac{1}{2}V_e^2 = \frac{GM}{4R} \]
Multiplying both sides by 2:
\[ V_e^2 = \frac{GM}{2R} \]
Taking the square root gives us the escape velocity:
\[ V_e = \sqrt{\frac{GM}{2R}} \]

### Final Answer
The velocity with which the particle should be projected vertically upward so that it does not return to Earth is:
\[ V_e = \sqrt{\frac{GM}{2R}} \]