Respuesta :
Answer:
[tex]r_f=137493m[/tex]
[tex]v=8638940m/s[/tex]
Explanation:
During this process the mass [tex]M=2\times10^{30}Kg[/tex] will be considered constant. We start from a radius [tex]r_i=7\times10^8m[/tex] and a period [tex]T_i=30\ days=(30)(24)(60)(60)s=2592000s[/tex]. The final period is [tex]T_f=0.1s[/tex].
Angular momentum L is conserved in this process. We can use the formula [tex]L=I\omega[/tex], where I is the momentum of inertia (which for a solid sphere is [tex]I=\frac{2mr^2}{5}[/tex]) and [tex]\omega=\frac{2\pi }{T}[/tex] is the angular velocity, so we can write the star's angular momentum as:
[tex]L=I\omega=\frac{2mr^2}{5}\frac{2\pi }{T}=\frac{4\pi mr^2 }{5T}[/tex]
Since [tex]L_f=L_i[/tex] we have:
[tex]\frac{4\pi mr_f^2 }{5T_f}=\frac{4\pi mr_i^2 }{5T_i}[/tex]
Which can be simplified as:
[tex]\frac{r_f^2 }{T_f}=\frac{r_i^2 }{T_i}[/tex]
Which means:
[tex]r_f=\sqrt{\frac{r_i^2 T_f}{T_i}}=r_i \sqrt{\frac{T_f}{T_i}}[/tex]
Which for our values is:
[tex]r_f=r_i \sqrt{\frac{T_f}{T_i}}=(7\times10^8m) \sqrt{\frac{0.1s}{2592000s}}=137493m[/tex]
And we calculate the speed of a point on the equator by dividing the final circumference over the final period:
[tex]v=\frac{C_f}{T_f}=\frac{2\pi r_f}{T_f}=\frac{2\pi (137493m)}{(0.1s)}=8638940m/s[/tex]