To calculate the new period of revolution if the Sun's radius doubles, we will apply the principle of conservation of angular momentum. The angular momentum \( L \) of a rotating body is defined as:
\( L = I \omega \)
where \( I \) denotes the moment of inertia and \( \omega \) represents the angular velocity. For a sphere of uniform density, the moment of inertia \( I \) is given by:
\( I = \frac{2}{5}MR^2 \)
where \( M \) is the mass and \( R \) is the radius. Due to the absence of external torque, angular momentum remains constant throughout the expansion process:
\( I_1 \omega_1 = I_2 \omega_2 \)
The initial state is described by:
\( I_1 = \frac{2}{5}MR_1^2 \)
Following expansion, the radius becomes \( 2R_1 \), leading to:
\( I_2 = \frac{2}{5}M(2R_1)^2 = \frac{8}{5}MR_1^2 \)
Given that the initial rotational period \( T_1 \) is 27 days, the initial angular velocity is:
\( \omega_1 = \frac{2\pi}{T_1} \)
The new angular velocity \( \omega_2 \) is expressed as:
\( \omega_2 = \frac{2\pi}{T_2} \)
Applying conservation of angular momentum yields:
\( \frac{2}{5}MR_1^2 \cdot \frac{2\pi}{27} = \frac{8}{5}MR_1^2 \cdot \frac{2\pi}{T_2} \)
After canceling common factors \( \frac{2\pi}{5}MR_1^2 \), the equation simplifies to:
\( \frac{1}{27} = \frac{4}{T_2} \)
Solving for \( T_2 \):
\( T_2 = 4 \times 27 = 108 \text{ days} \)
Therefore, a doubling of the Sun's radius results in a revolution period of \( 108 \text{ days} \).