The time period of rotation of the sun about its axis is 27 days. If the sun expands to thrice of its present diameter, what will be its new period of rotation?

Answer

To solve the problem, we need to understand how the period of rotation changes when the radius of the sun changes. The key concept here is the conservation of angular momentum.

1. **Understanding the Initial Conditions**:
- The initial diameter of the sun is \( D \), which gives an initial radius \( R = \frac{D}{2} \).
- The initial period of rotation \( T = 27 \) days.

2. **Understanding the Final Conditions**:
- The sun expands to a diameter of \( 3D \), which gives a new radius \( R' = \frac{3D}{2} = 3R \).

3. **Using the Conservation of Angular Momentum**:
- Angular momentum \( L \) is given by the formula:
\[ L = I \omega \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
- The moment of inertia for a solid sphere is:
\[ I = \frac{2}{5} m R^2 \]
- The angular velocity \( \omega \) is related to the period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]

4. **Setting Up the Conservation Equation**:
- Initially, the angular momentum \( L_i \) is:
\[ L_i = I_i \omega_i = \frac{2}{5} m R^2 \cdot \frac{2\pi}{T} \]
- After expansion, the new moment of inertia \( I_f \) is:
\[ I_f = \frac{2}{5} m (3R)^2 = \frac{2}{5} m \cdot 9R^2 = \frac{18}{5} m R^2 \]
- The final angular momentum \( L_f \) is:
\[ L_f = I_f \omega_f = \frac{18}{5} m R^2 \cdot \frac{2\pi}{T'} \]

5. **Equating Initial and Final Angular Momentum**:
\[ L_i = L_f \]
\[ \frac{2}{5} m R^2 \cdot \frac{2\pi}{T} = \frac{18}{5} m R^2 \cdot \frac{2\pi}{T'} \]
- We can cancel \( \frac{2\pi}{5} m R^2 \) from both sides:
\[ \frac{1}{T} = \frac{18}{T'} \]

6. **Solving for the New Period of Rotation \( T' \)**:
\[ T' = 18T \]
- Substituting \( T = 27 \) days:
\[ T' = 18 \times 27 = 486 \text{ days} \]

Thus, the new period of rotation of the sun after it expands to thrice its diameter will be **486 days**.