The sun rotates round itself once in 27 days. What will be period of revolution if the sun were to expand to twice its present radius? Assume the sun to be a sphere of uniform density.

Answer

To solve the problem of determining the period of revolution of the sun if it expands to twice its present radius, we will use the principle of conservation of angular momentum. Here’s a step-by-step solution:

### Step 1: Understand the Initial Conditions
The sun currently rotates once in 27 days. We denote:
- Initial period of rotation, \( T = 27 \) days.
- Initial angular velocity, \( \omega \).

### Step 2: Calculate the Initial Angular Velocity
The angular velocity \( \omega \) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \omega = \frac{2\pi}{27} \text{ radians per day} \]

### Step 3: Determine the Moment of Inertia
Assuming the sun is a solid sphere, the moment of inertia \( I \) is given by:
\[ I = \frac{2}{5} m r^2 \]
where \( m \) is the mass of the sun and \( r \) is its radius.

### Step 4: Calculate the New Moment of Inertia After Expansion
If the sun expands to twice its present radius, the new radius \( r' \) is:
\[ r' = 2r \]
The new moment of inertia \( I' \) will be:
\[ I' = \frac{2}{5} m (r')^2 = \frac{2}{5} m (2r)^2 = \frac{2}{5} m \cdot 4r^2 = \frac{8}{5} m r^2 \]

### Step 5: Apply Conservation of Angular Momentum
Since no external torques are acting on the sun, angular momentum is conserved:
\[ I \omega = I' \omega' \]
Substituting the expressions for \( I \) and \( I' \):
\[ \frac{2}{5} m r^2 \cdot \frac{2\pi}{27} = \frac{8}{5} m r^2 \cdot \omega' \]
The mass \( m \) and \( r^2 \) cancel out:
\[ \frac{2}{5} \cdot \frac{2\pi}{27} = \frac{8}{5} \cdot \omega' \]

### Step 6: Solve for the New Angular Velocity
Rearranging the equation gives:
\[ \omega' = \frac{2}{8} \cdot \frac{2\pi}{27} = \frac{1}{4} \cdot \frac{2\pi}{27} = \frac{2\pi}{108} \text{ radians per day} \]

### Step 7: Calculate the New Period of Revolution
The new period \( T' \) can be found using:
\[ T' = \frac{2\pi}{\omega'} \]
Substituting the value of \( \omega' \):
\[ T' = \frac{2\pi}{\frac{2\pi}{108}} = 108 \text{ days} \]

### Final Answer
The period of revolution if the sun were to expand to twice its present radius would be **108 days**.

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