Question

The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.

• A. \( 108 \text{ days} \)
• B. \( 115 \text{ days} \)
• C. \( 100 \text{ days} \)
• D. \( 54 \text{ days} \)

Hint:

Remember the conservation of angular momentum \(I_1 \omega_1 = I_2 \omega_2\). The moment of inertia of a uniform sphere is \(I = \frac{2}{5} MR^2\). If the radius changes and mass remains constant, the moment of inertia changes with the square of the radius.

Answer 1

The Correct Option is A

To determine the period of revolution if the Sun expands to twice its present radius, we will use the concept of conservation of angular momentum. The angular momentum \( L \) of a rotating object is given by:

\( L = I \omega \)

where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. For a sphere of uniform density, the moment of inertia \( I \) is:

\( I = \frac{2}{5}MR^2 \)

where \( M \) is the mass and \( R \) is the radius. Since there is no external torque, the angular momentum before and after expansion remains constant:

\( I_1 \omega_1 = I_2 \omega_2 \)

Initially, we have:

\( I_1 = \frac{2}{5}MR_1^2 \)

After expansion, the new radius is \( 2R_1 \), so:

\( I_2 = \frac{2}{5}M(2R_1)^2 = \frac{8}{5}MR_1^2 \)

Given that the initial period of rotation \( T_1 \) is 27 days, we know:

\( \omega_1 = \frac{2\pi}{T_1} \)

The new angular velocity \( \omega_2 \) is:

\( \omega_2 = \frac{2\pi}{T_2} \)

From conservation of angular momentum:

\( \frac{2}{5}MR_1^2 \cdot \frac{2\pi}{27} = \frac{8}{5}MR_1^2 \cdot \frac{2\pi}{T_2} \)

Canceling out the common terms \( \frac{2\pi}{5}MR_1^2 \), we get:

\( \frac{1}{27} = \frac{4}{T_2} \)

Solving for \( T_2 \):

\( T_2 = 4 \times 27 = 108 \text{ days} \)

Thus, if the Sun's radius doubles, the period of revolution will be \( 108 \text{ days} \).