Answer:
[tex]T \approx 586468.163\,days[/tex]
Explanation:
The initial angular speed of the Sun is:
[tex]\omega_{o} = \frac{2\pi}{(24.47\,days)\cdot \left(86400\,\frac{s}{days} \right)}[/tex]
[tex]\omega_{o} \approx 2.972\times 10^{-6}\,\frac{rad}{s}[/tex]
Let suppose that Sun can be modelled as an uniform sphere, the moment of inertia is:
[tex]I = \frac{2}{5}\cdot m \cdot r^{2}[/tex]
The initial moment of inertia is:
[tex]I =\frac{2}{5}\cdot (1.989\times 10^{30}\,kg)\cdot (6.983\times 10^{8}\,m)^{2}[/tex]
[tex]I = 3.879\times 10^{47}\,kg\cdot m^{2}[/tex]
The angular momentum is:
[tex]L = (3.879\times 10^{47}\,kg\cdot m^{2})\cdot (2.972\times 10^{-6}\,\frac{rad}{s} )[/tex]
[tex]L = 1.153\times 10^{42}\,kg\cdot \frac{m^{2}}{s}[/tex]
Given the absence of external forces exerted on the Sun, the final angular speed can found by the Principle of Angular Momentum Conservation. The final moment of inertia is:
[tex]I =\frac{2}{5}\cdot (1.989\times 10^{30}\,kg)\cdot (1.081\times 10^{11}\,m)^{2}[/tex]
[tex]I = 9.297\times 10^{51}\,kg\cdot m^{2}[/tex]
The final angular speed is:
[tex]\omega_{f} = \frac{1.153\times 10^{42}\,kg\cdot \frac{m^{2}}{s} }{9.297\times 10^{51}\,kg\cdot m^{2}}[/tex]
[tex]\omega_{f} = 1.240\times 10^{-10}\,\frac{rad}{s}[/tex]
The period of rotation is:
[tex]T = \left(\frac{2\pi}{1.240\times 10^{-10}\,\frac{rad}{s} } \right)\cdot \left(\frac{1}{86400}\,\frac{d}{s} \right)[/tex]
[tex]T \approx 586468.163\,days[/tex]
