A body mass 'm' is dropped from heinght `R/2`, from earth's surface, where 'R' is the radius of earth. Its speed when it will hit the earth's surface is `(v_(e)="escape velocity from earth's surface")`.

Options

A. `sqrt3 v_(e)`
B. `(v_(e))/(sqrt3)`
C. `(v_(e))/(sqrt2)`
D. `sqrt2 v_(e)`

Answer

To solve the problem of finding the speed of a body dropped from a height of \( R/2 \) above the Earth's surface when it hits the Earth's surface, we can use the principle of conservation of energy. Here's a step-by-step solution:

### Step 1: Understand the Initial and Final Potential Energy
1. **Initial Height**: The body is dropped from a height of \( R/2 \) above the Earth's surface. Therefore, the distance from the center of the Earth is:
\[ R + \frac{R}{2} = \frac{3R}{2} \]
2. **Final Height**: When the body hits the Earth's surface, the distance from the center of the Earth is \( R \).

### Step 2: Calculate the Potential Energy at Initial and Final Points
1. **Initial Potential Energy (PE_initial)** at height \( \frac{3R}{2} \):
\[ PE_{\text{initial}} = -\frac{GMm}{\frac{3R}{2}} = -\frac{2GMm}{3R} \]
2. **Final Potential Energy (PE_final)** at the Earth's surface:
\[ PE_{\text{final}} = -\frac{GMm}{R} \]

### Step 3: Calculate the Change in Potential Energy
The change in potential energy as the body falls from height \( \frac{3R}{2} \) to the surface is:
\[ \Delta PE = PE_{\text{final}} - PE_{\text{initial}} = \left(-\frac{GMm}{R}\right) - \left(-\frac{2GMm}{3R}\right) \]
\[ \Delta PE = -\frac{GMm}{R} + \frac{2GMm}{3R} = -\frac{3GMm}{3R} + \frac{2GMm}{3R} = -\frac{GMm}{3R} \]

### Step 4: Relate Change in Potential Energy to Kinetic Energy
According to the conservation of energy, the change in potential energy is equal to the kinetic energy gained:
\[ \Delta PE = KE = \frac{1}{2} mv^2 \]
Thus, we can write:
\[ -\frac{GMm}{3R} = \frac{1}{2} mv^2 \]

### Step 5: Solve for Velocity \( v \)
1. Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ -\frac{GM}{3R} = \frac{1}{2} v^2 \]
2. Rearranging gives:
\[ v^2 = -\frac{2GM}{3R} \]
3. Taking the square root:
\[ v = \sqrt{\frac{2GM}{3R}} \]

### Step 6: Relate to Escape Velocity
The escape velocity \( v_e \) from the Earth's surface is given by:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
To express \( v \) in terms of \( v_e \):
\[ v = \sqrt{\frac{2GM}{3R}} = \sqrt{\frac{2GM}{R} \cdot \frac{1}{3}} = \frac{1}{\sqrt{3}} v_e \]

### Final Result
Thus, the speed of the body when it hits the Earth's surface is:
\[ v = \frac{1}{\sqrt{3}} v_e \]