Looking for classes? Ksquare Career Institute, Bengaluru →
The Sun rotates around its centre once in $27\text{ days}$. 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.
Explanation
Conservation of angular momentum gives $I_1 \omega_1 = I_2 \omega_2$. Since $I \propto R^2$, $T_2 = T_1 (R_2/R_1)^2 = 27 \times 4 = 108\text{ days}$.
Detailed Solution
In the absence of external torque, angular momentum is conserved: $L = I_1 \omega_1 = I_2 \omega_2$. For a uniform sphere, $I = \frac{2}{5} M R^2$. When the radius doubles ($R_2 = 2 R_1$), the moment of inertia becomes $I_2 = \frac{2}{5} M (2R_1)^2 = 4 I_1$. Since $\omega = \frac{2\pi}{T}$, we have $I_1 \left(\frac{2\pi}{T_1}\right) = (4 I_1) \left(\frac{2\pi}{T_2}\right) \implies T_2 = 4 T_1 = 4 \times 27\text{ days} = 108\text{ days}$.