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

`T \approx 586468.163\,days`

Step-by-step explanation:

The initial angular speed of the Sun is:

`\omega_(o) = (2\pi)/((24.47\,days)\cdot \left(86400\,(s)/(days) \right))`

`\omega_(o) \approx 2.972* 10^(-6)\,(rad)/(s)`

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

`I = (2)/(5)\cdot m \cdot r^(2)`

The initial moment of inertia is:

`I =(2)/(5)\cdot (1.989* 10^(30)\,kg)\cdot (6.983* 10^(8)\,m)^(2)`

`I = 3.879* 10^(47)\,kg\cdot m^(2)`

The angular momentum is:

`L = (3.879* 10^(47)\,kg\cdot m^(2))\cdot (2.972* 10^(-6)\,(rad)/(s) )`

`L = 1.153* 10^(42)\,kg\cdot (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 =(2)/(5)\cdot (1.989* 10^(30)\,kg)\cdot (1.081* 10^(11)\,m)^(2)`

`I = 9.297* 10^(51)\,kg\cdot m^(2)`

The final angular speed is:

`\omega_(f) = (1.153* 10^(42)\,kg\cdot (m^(2))/(s) )/(9.297* 10^(51)\,kg\cdot m^(2))`

`\omega_(f) = 1.240* 10^(-10)\,(rad)/(s)`

The period of rotation is:

`T = \left((2\pi)/(1.240* 10^(-10)\,(rad)/(s) ) \right)\cdot \left((1)/(86400)\,(d)/(s) \right)`

`T \approx 586468.163\,days`