With what velocity should a particle be projected so that its height becomes equal to radius of earth?

Options

A. `((GM)/(R ))^(1//2)`
B. `((8GM)/(R ))^(1//2)`
C. `((2GM)/(R ))^(1//2)`
D. `((4GM)/(R ))^(1//2)`

Answer

To find the velocity with which a particle should be projected so that its height becomes equal to the radius of the Earth, we can use the principle of conservation of mechanical energy. Here's a step-by-step solution:

### Step 1: Understand the Problem
We need to project a particle from the surface of the Earth such that it reaches a height equal to the radius of the Earth (R). Therefore, the total height (h) at which the particle will reach is:
\[ h = R \]
The total distance from the center of the Earth to the particle at its maximum height will be:
\[ d = R + h = R + R = 2R \]

### Step 2: Apply Conservation of Mechanical Energy
The mechanical energy conservation states that the initial potential energy plus the initial kinetic energy equals the final potential energy plus the final kinetic energy:
\[ U_i + K_i = U_f + K_f \]

### Step 3: Calculate Initial and Final Energies
1. **Initial Potential Energy (U_i)**:
The gravitational potential energy when the particle is on the surface of the Earth is given by:
\[ U_i = -\frac{GMm}{R} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( m \) is the mass of the particle.

2. **Initial Kinetic Energy (K_i)**:
The initial kinetic energy when the particle is projected with velocity \( v \) is:
\[ K_i = \frac{1}{2} mv^2 \]

3. **Final Potential Energy (U_f)**:
At the maximum height (2R), the potential energy is:
\[ U_f = -\frac{GMm}{2R} \]

4. **Final Kinetic Energy (K_f)**:
At the maximum height, the kinetic energy is zero because the particle momentarily stops before falling back:
\[ K_f = 0 \]

### Step 4: Set Up the Equation
Substituting the energies into the conservation of energy equation:
\[ -\frac{GMm}{R} + \frac{1}{2} mv^2 = -\frac{GMm}{2R} + 0 \]

### Step 5: Simplify the Equation
Rearranging gives:
\[ \frac{1}{2} mv^2 = -\frac{GMm}{2R} + \frac{GMm}{R} \]
\[ \frac{1}{2} mv^2 = \frac{GMm}{2R} \]

### Step 6: Solve for Velocity (v)
Now, we can cancel \( m \) 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:
\[ v = \sqrt{\frac{GM}{R}} \]

### Conclusion
The required velocity \( v \) with which the particle should be projected is:
\[ v = \sqrt{\frac{GM}{R}} \]