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. `((2GM)/(R ))^((1)/(2))`
D. `((GM)/(R ))^((1)/(2))`

Answer

To solve the problem of finding the speed of an object when it hits the surface of the moon after being released from a distance of \(3R\) from the center of the moon, we can use the principle of conservation of energy. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the System
We have:
- A stationary object of mass \(m\) released from a point \(P\) at a distance \(3R\) from the center of the moon.
- The moon has a radius \(R\) and mass \(M\).

### Step 2: Identify Potential Energy
The gravitational potential energy \(U\) at a distance \(r\) from the center of a mass \(M\) is given by the formula:
\[ U = -\frac{GMm}{r} \]
where \(G\) is the gravitational constant.

### Step 3: Calculate Initial Potential Energy
At point \(P\) (distance \(3R\) from the center of the moon):
\[ U_P = -\frac{GMm}{3R} \]

### Step 4: Calculate Final Potential Energy
When the object hits the surface of the moon (distance \(R\) from the center):
\[ U_Q = -\frac{GMm}{R} \]

### Step 5: Use Conservation of Energy
According to the conservation of energy, the change in potential energy will equal the kinetic energy gained by the object. Therefore:
\[ \Delta U = K \]
where \(K\) is the kinetic energy given by:
\[ K = \frac{1}{2}mv^2 \]

### Step 6: Calculate Change in Potential Energy
The change in potential energy as the object moves from point \(P\) to point \(Q\) is:
\[ \Delta U = U_Q - U_P = \left(-\frac{GMm}{R}\right) - \left(-\frac{GMm}{3R}\right) \]
\[ \Delta U = -\frac{GMm}{R} + \frac{GMm}{3R} \]
\[ \Delta U = -\frac{3GMm}{3R} + \frac{GMm}{3R} = -\frac{2GMm}{3R} \]

### Step 7: Set Change in Potential Energy Equal to Kinetic Energy
Now we set this equal to the kinetic energy:
\[ -\frac{2GMm}{3R} = \frac{1}{2}mv^2 \]

### Step 8: Solve for \(v^2\)
We can cancel \(m\) from both sides (assuming \(m \neq 0\)):
\[ -\frac{2GM}{3R} = \frac{1}{2}v^2 \]
Multiplying both sides by 2:
\[ -\frac{4GM}{3R} = v^2 \]

### Step 9: Take the Square Root
Taking the square root to find \(v\):
\[ v = \sqrt{\frac{4GM}{3R}} \]

### Final Expression
Thus, the speed of the object when it hits the moon is:
\[ v = \sqrt{\frac{4GM}{3R}} \]