A*R*C*S 14.7.1. The Merry-Go-Round I.
The Sun has mass \(>2 \times 10^{30} \mathrm{~kg}\text{,}\) radius \(7\times 10^{5} \mathrm{~km}\text{,}\) and rotational period of approximately 28 days. If the Sun should collapse into a white dwarf of radius \(3.5 \times 10^3 \mathrm{~km}\text{,}\) what would its period be if no mass were ejected and a sphere of uniform density can model the Sun both before and after?
Solution.
For this system, angular momentum is conserved since there will be no torque that acts on the system. This means the initial angular momentum equals the final angular momentum: \(L_i = L_f\text{.}\) You can use the moment of inertia of a solid uniform sphere, and relate the angular speed to the period of the rotation:
\begin{equation*}
\frac{2}{5} M_s R_s ^2 (\frac{2\pi}{T_s}) = \frac{2}{5} M_s R_n ^2 (\frac{2\pi}{T_n})
\end{equation*}
Solving for the new time:
\begin{equation*}
T_n = T_s \frac{R_n^2}{R_s^2} = (28 \ \mathrm{~days})\frac{(3.5\times10^3)^2}{(7\times10^5)^2} = 0.0007 \ \mathrm{~days} = 60.5 \mathrm{~s}
\end{equation*}
