An object is released from a height twice the radius of the earth on the surface of earth. Find the speed with which it will collide with group by neglecting effect of air. (Take, R radius of earth and mass of earth as M)

Options

A. `2sqrt((GM)/(3R))`
B. `3sqrt((GM)/(2R))`
C. `2 sqrt((GM)/(R))`
D. `3 sqrt((GM)/(R))`

Explanation

To find the speed with which an object will collide with the ground after being released from a height twice the radius of the Earth, we can use the principle of conservation of energy.

### Step-by-Step Solution:

1. **Identify the Initial and Final Positions**:
- The object is released from a height \( h = 2R \) (where \( R \) is the radius of the Earth).
- The final position is the surface of the Earth.

2. **Calculate the Initial Potential Energy (U_initial)**:
- The gravitational potential energy at a distance \( r \) from the center of the Earth is given by:
\[ U = -\frac{GMm}{r} \]
- At the height \( h = 2R \), the distance from the center of the Earth is \( 3R \):
\[ U_{\text{initial}} = -\frac{GMm}{3R} \]

3. **Calculate the Final Potential Energy (U_final)**:
- At the surface of the Earth (distance \( R \) from the center):
\[ U_{\text{final}} = -\frac{GMm}{R} \]

4. **Apply the Conservation of Energy**:
- The change in kinetic energy (\( \Delta KE \)) is equal to the negative change in potential energy (\( \Delta U \)):
\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = 0 - 0 = 0 \]
- The change in potential energy is:
\[ \Delta U = U_{\text{final}} - U_{\text{initial}} = -\frac{GMm}{R} - \left(-\frac{GMm}{3R}\right) \]
- Simplifying this gives:
\[ \Delta U = -\frac{GMm}{R} + \frac{GMm}{3R} = -\frac{GMm}{R} + \frac{GMm}{3R} = -\frac{2GMm}{3R} \]

5. **Relate Change in Kinetic Energy to Change in Potential Energy**:
- From conservation of energy:
\[ \Delta KE = -\Delta U \]
- Thus:
\[ \frac{1}{2}mv^2 = \frac{2GMm}{3R} \]

6. **Solve for the Speed (v)**:
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2}v^2 = \frac{2GM}{3R} \]
- Multiply both sides by 2:
\[ v^2 = \frac{4GM}{3R} \]
- Taking the square root gives:
\[ v = \sqrt{\frac{4GM}{3R}} = 2\sqrt{\frac{GM}{3R}} \]

### Final Answer:
The speed with which the object will collide with the ground is:
\[ v = 2\sqrt{\frac{GM}{3R}} \]