A stationary object is released from a point P a distance 3R from the centre of the moon which has radius R and mass M. which one of the following expressions gives the speed of the object on hitting the moon?

Options

A. `((2GM)/(3R))^(1//2)`
B. `((4GM)/(3R))^(1//2)`
C. `((GM)/(3R))^(1//2)`
D. `((GM)/R)^(1//2)`

Answer

To find the speed of the object when it hits the moon, we can use the principle of conservation of energy. The total mechanical energy (potential energy + kinetic energy) at the point of release (point P) must equal the total mechanical energy just before the object hits the moon.

### Step-by-Step Solution:

1. **Identify the Initial and Final States**:
- The object is released from a distance of \(3R\) from the center of the moon.
- The potential energy at this point and the kinetic energy is zero since the object is stationary.
- The final state is just before the object hits the moon's surface.

2. **Write the Conservation of Energy Equation**:
\[ \text{Potential Energy at P} + \text{Kinetic Energy at P} = \text{Potential Energy at surface} + \text{Kinetic Energy at surface} \]

3. **Calculate Potential Energy at Point P**:
- The gravitational potential energy \(U\) at a distance \(r\) from the center of a mass \(M\) is given by:
\[ U = -\frac{GMm}{r} \]
- At point P (distance \(3R\)):
\[ U_P = -\frac{GMm}{3R} \]
- Kinetic energy at point P is zero since the object is stationary:
\[ K.E._P = 0 \]

4. **Calculate Potential Energy at the Surface of the Moon**:
- At the surface of the moon (distance \(R\)):
\[ U_{surface} = -\frac{GMm}{R} \]
- Let \(v\) be the speed of the object just before hitting the moon. The kinetic energy at the surface is:
\[ K.E._{surface} = \frac{1}{2}mv^2 \]

5. **Set Up the Energy Conservation Equation**:
\[ -\frac{GMm}{3R} + 0 = -\frac{GMm}{R} + \frac{1}{2}mv^2 \]

6. **Simplify the Equation**:
- Rearranging gives:
\[ -\frac{GMm}{3R} + \frac{GMm}{R} = \frac{1}{2}mv^2 \]
- Combine the left side:
\[ \frac{GMm}{R} - \frac{GMm}{3R} = \frac{GMm}{R} \left(1 - \frac{1}{3}\right) = \frac{GMm}{R} \cdot \frac{2}{3} = \frac{2GMm}{3R} \]
- Therefore, we have:
\[ \frac{2GMm}{3R} = \frac{1}{2}mv^2 \]

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

### Final Expression:
The speed of the object on hitting the moon is:
\[ v = \sqrt{\frac{4GM}{3R}} \]