Question

The sun currently rotates once every 24.47 days. The solar radius is 6.963x10^8 m. As the Sun ages it will evolve into a type of star called a red giant. Astrophysicists estimate that when the Sun reaches this phase of its life, its radius will be at least as large as the distance from Venus to the Sun, engulfing the planet. The distance between Venus and the Sun is 1.081x10^11 m. How many days will it take for the Sun to rotate with its new radius?

Answer 1

Answer:

$T \approx 586468.163\,days$

Explanation:

The initial angular speed of the Sun is:

$\omega_{o} = \frac{2\pi}{(24.47\,days)\cdot \left(86400\,\frac{s}{days} \right)}$

$\omega_{o} \approx 2.972\times 10^{-6}\,\frac{rad}{s}$

Let suppose that Sun can be modelled as an uniform sphere, the moment of inertia is:

$I = \frac{2}{5}\cdot m \cdot r^{2}$

The initial moment of inertia is:

$I =\frac{2}{5}\cdot (1.989\times 10^{30}\,kg)\cdot (6.983\times 10^{8}\,m)^{2}$

$I = 3.879\times 10^{47}\,kg\cdot m^{2}$

The angular momentum is:

$L = (3.879\times 10^{47}\,kg\cdot m^{2})\cdot (2.972\times 10^{-6}\,\frac{rad}{s} )$

$L = 1.153\times 10^{42}\,kg\cdot \frac{m^{2}}{s}$

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:

$I =\frac{2}{5}\cdot (1.989\times 10^{30}\,kg)\cdot (1.081\times 10^{11}\,m)^{2}$

$I = 9.297\times 10^{51}\,kg\cdot m^{2}$

The final angular speed is:

$\omega_{f} = \frac{1.153\times 10^{42}\,kg\cdot \frac{m^{2}}{s} }{9.297\times 10^{51}\,kg\cdot m^{2}}$

$\omega_{f} = 1.240\times 10^{-10}\,\frac{rad}{s}$

The period of rotation is:

$T = \left(\frac{2\pi}{1.240\times 10^{-10}\,\frac{rad}{s} } \right)\cdot \left(\frac{1}{86400}\,\frac{d}{s} \right)$

$T \approx 586468.163\,days$