A ring of radius `R = 4m` is made of a highly dense material. Mass of the ring is `m_(1) = 5.4 xx 10^(9) kg` distributed uniformly over its circumference. A highly dense particle of mass `m_(2) = 6 xx 10^(8) kg` is placed on the axis of the ring at a distance `x_(0) = 3 m` from the centre. Neglecting all other forces, except mutual gravitational interacting of the two. Caculate
(i) displacemental of the ring when particle is at the centre of ring, and
(ii) speed of the particle at that instant.

Explanation

To solve the problem step by step, we will calculate the displacement of the ring when the particle is at the center of the ring and then find the speed of the particle at that instant.

### Given Data:
- Radius of the ring, \( R = 4 \, \text{m} \)
- Mass of the ring, \( m_1 = 5.4 \times 10^9 \, \text{kg} \)
- Mass of the particle, \( m_2 = 6 \times 10^8 \, \text{kg} \)
- Initial distance of the particle from the center of the ring, \( x_0 = 3 \, \text{m} \)

### Step 1: Calculate the Displacement of the Ring
1. **Assume the displacement of the ring is \( x \)**.
2. **The displacement of the particle when the ring moves is \( x_0 - x \)**, where \( x_0 = 3 \, \text{m} \).
3. **The center of mass of the system will not move**, so we can use the principle of conservation of momentum:
\[ m_1 \cdot x = m_2 \cdot (x_0 - x) \]
4. **Substituting the values**:
\[ 5.4 \times 10^9 \cdot x = 6 \times 10^8 \cdot (3 - x) \]
5. **Expanding the equation**:
\[ 5.4 \times 10^9 \cdot x = 1.8 \times 10^9 - 6 \times 10^8 \cdot x \]
6. **Rearranging the equation**:
\[ 5.4 \times 10^9 \cdot x + 6 \times 10^8 \cdot x = 1.8 \times 10^9 \]
\[ (5.4 \times 10^9 + 0.6 \times 10^9) \cdot x = 1.8 \times 10^9 \]
\[ 6 \times 10^9 \cdot x = 1.8 \times 10^9 \]
7. **Solving for \( x \)**:
\[ x = \frac{1.8 \times 10^9}{6 \times 10^9} = 0.3 \, \text{m} \]

### Step 2: Calculate the Speed of the Particle
1. **Using conservation of energy**:
- Initial potential energy when the particle is at \( x_0 = 3 \, \text{m} \):
\[ U_i = -\frac{G m_1 m_2}{\sqrt{R^2 + x_0^2}} = -\frac{G \cdot 5.4 \times 10^9 \cdot 6 \times 10^8}{\sqrt{4^2 + 3^2}} \]
- Final potential energy when the particle is at the center of the ring (displacement \( x = 0.3 \)):
\[ U_f = -\frac{G m_1 m_2}{\sqrt{R^2 + (x_0 - x)^2}} = -\frac{G \cdot 5.4 \times 10^9 \cdot 6 \times 10^8}{\sqrt{4^2 + (3 - 0.3)^2}} \]
2. **Kinetic energy at the final state**:
\[ K_f = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
- Since \( v_2 = 9 v_1 \) (from momentum conservation):
\[ K_f = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 (9 v_1)^2 \]
\[ = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 \cdot 81 v_1^2 \]
\[ = \frac{1}{2} (m_1 + 81 m_2) v_1^2 \]
3. **Setting initial potential energy equal to final kinetic energy**:
\[ U_i - U_f = K_f \]
- Substitute values and solve for \( v_1 \).

### Final Results:
1. **Displacement of the ring**: \( x = 0.3 \, \text{m} \)
2. **Speed of the particle**: Calculate \( v_1 \) using the energy conservation equation.