A small body of mass is projected with a velocity just sufficient to make it reach from the surface of a planet ( or radius `2R` and mass `3M`) to the surface of another planet (or radius `R` and mass `M`). The distance between the centres of the two spherical planets is `6R`. the distance of the body from the centre of bigger planet is 'x' at any moment. During the journey, find the distance 'x' where the speed of the body is (a) maximum (b) minimum. Assume motion of body along the line joining centres of planets.

Explanation

To solve the problem, we need to analyze the motion of the small body as it travels from the surface of the larger planet to the surface of the smaller planet. We will use the concepts of gravitational potential energy and kinetic energy to find the points where the speed of the body is maximum and minimum.

### Given Data:
- Mass of larger planet, \( M_1 = 3M \)
- Radius of larger planet, \( R_1 = 2R \)
- Mass of smaller planet, \( M_2 = M \)
- Radius of smaller planet, \( R_2 = R \)
- Distance between the centers of the two planets, \( d = 6R \)

### Step 1: Determine the gravitational potential energy (U) at different positions
The gravitational potential energy \( U \) at a distance \( x \) from the center of the larger planet is given by:

\[ U_1 = -\frac{G \cdot M_1 \cdot m}{x} \quad \text{(for } x \geq 2R\text{)} \]

For the smaller planet, at a distance \( d - x \) from the center of the smaller planet, the potential energy is:

\[ U_2 = -\frac{G \cdot M_2 \cdot m}{d - x} \quad \text{(for } d - x \geq R\text{)} \]

### Step 2: Total mechanical energy conservation
The total mechanical energy \( E \) is conserved during the motion. At the surface of the larger planet, the total energy is:

\[ E = K + U_1 = \frac{1}{2}mv^2 - \frac{G \cdot M_1 \cdot m}{2R} \]

At the surface of the smaller planet, the total energy is:

\[ E = K + U_2 = 0 - \frac{G \cdot M_2 \cdot m}{R} \]

Setting these equal gives us:

\[ \frac{1}{2}mv^2 - \frac{G \cdot 3M \cdot m}{2R} = -\frac{G \cdot M \cdot m}{R} \]

### Step 3: Solve for initial velocity
From the above equation, we can solve for the initial velocity \( v \):

\[ \frac{1}{2}mv^2 = -\frac{G \cdot M \cdot m}{R} + \frac{G \cdot 3M \cdot m}{2R} \]

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

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

### Step 4: Analyze speed at different points
The speed of the body will be maximum when the gravitational potential energy is minimum (i.e., when it is closest to the center of the larger planet) and minimum when it is closest to the center of the smaller planet.

#### (a) Maximum speed
The maximum speed occurs when \( x = 2R \) (the surface of the larger planet):

\[ v_{\text{max}} = \sqrt{\frac{G \cdot M}{R}} \]

#### (b) Minimum speed
The minimum speed occurs when \( x = 4R \) (the distance from the center of the larger planet to the center of the smaller planet minus the radius of the smaller planet):

\[ v_{\text{min}} = 0 \quad \text{(at the surface of the smaller planet)} \]

### Final Answer:
- (a) The distance \( x \) where the speed is maximum is \( 2R \).
- (b) The distance \( x \) where the speed is minimum is \( 4R \).