A body projected from the surface of the earth attains a height equal to the radius of the earth. The velocity with which the body was projected is

Options

A. `sqrt((GM_(e))/(R))`
B. `sqrt((2GM_(e))/(R))`
C. `sqrt((5)/(4)(GM_(e))/(R))`
D. `sqrt((3GM_(e))/(R))`

Explanation

To solve the problem of finding the velocity with which a body must be projected from the surface of the Earth to reach a height 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
The body is projected from the surface of the Earth and reaches a height equal to the radius of the Earth (h = R). Therefore, the initial height from the center of the Earth is R (the radius of the Earth), and the final height when the body reaches its maximum height is 2R (the radius of the Earth plus the height).

### Step 2: Write the Conservation of Mechanical Energy Equation
The total mechanical energy at the initial position (when the body is projected) must equal the total mechanical energy at the maximum height.

\[ \text{Initial Mechanical Energy} = \text{Final Mechanical Energy} \]

This can be expressed as:

\[ \text{Initial Kinetic Energy} + \text{Initial Potential Energy} = \text{Final Kinetic Energy} + \text{Final Potential Energy} \]

### Step 3: Define the Energies
1. **Initial Kinetic Energy (KE_initial)**:
\[ KE_{\text{initial}} = \frac{1}{2} mv^2 \]
where \( v \) is the initial velocity and \( m \) is the mass of the body.

2. **Initial Potential Energy (PE_initial)**:
The potential energy at the surface of the Earth is given by:
\[ PE_{\text{initial}} = -\frac{G M m}{R} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

3. **Final Kinetic Energy (KE_final)**:
At the maximum height, the velocity is 0, so:
\[ KE_{\text{final}} = 0 \]

4. **Final Potential Energy (PE_final)**:
At the maximum height (2R), the potential energy is:
\[ PE_{\text{final}} = -\frac{G M m}{2R} \]

### Step 4: Set Up the Equation
Substituting these expressions into the conservation of energy equation gives:

\[ \frac{1}{2} mv^2 - \frac{G M m}{R} = 0 - \frac{G M m}{2R} \]

### Step 5: Simplify the Equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):

\[ \frac{1}{2} v^2 - \frac{G M}{R} = -\frac{G M}{2R} \]

Rearranging gives:

\[ \frac{1}{2} v^2 = \frac{G M}{R} - \frac{G M}{2R} \]

Combining the terms on the right:

\[ \frac{1}{2} v^2 = \frac{G M}{2R} \]

### Step 6: Solve for \( v^2 \)
Multiplying both sides by 2:

\[ v^2 = \frac{G M}{R} \]

### Step 7: Take the Square Root
Taking the square root of both sides gives:

\[ v = \sqrt{\frac{G M}{R}} \]

### Step 8: Conclusion
Thus, the velocity with which the body was projected is:

\[ v = \sqrt{\frac{G M}{R}} \]