To solve this problem it is necessary to apply the kinematic equations of linear and angular motion, as well as the given definitions of the period.
Centripetal acceleration can be found through the relationship
[tex]a_c = \frac{v^2}{R}[/tex]
Where
v = Tangential Velocity
R = Radius
At the same time linear velocity can be expressed in terms of angular velocity as
[tex]v = R\omega[/tex]
Where,
R = Radius
[tex]\omega =[/tex] Angular Velocity
PART A) From this point on, we can use the values used for the period given in the exercise because the angular velocity by definition is described as
[tex]\omega = \frac{2\pi}{T}[/tex]
T = Period
So replacing we have to
[tex]\omega = \frac{2\pi}{2.6*10^8years}\\\omega = 2.4166*10^{-8}rad/years\\\omega = 2.4166*10^{-8}rad/years(\frac{1years}{365days})(\frac{1day}{86400s})\\\omega = 7.663*10^{-16}rad/s[/tex]
Since [tex]1 Light year = 9.48*10^{15}m[/tex]
Then the radius in meters would be
[tex]R = (3*10^4ly)(\frac{9.48*10^{15}m}{1ly})[/tex]
[tex]R = 2.844*10^{20}m[/tex]
Then the centripetal acceleration would be
[tex]a_c = \frac{v^2}{R}\\a_c = \frac{(R\omega)^2}{R}\\a_c = R\omega^2 \\a_c = 2.844*10^{20}(7.663*10^{-16})^2\\a_c = 1.67*10^{-10}m/s^2[/tex]
From the result obtained, considering that it is an unimaginably low value of an order of less than [tex]10^{-10}[/tex] it is possible to conclude that it supports the assertion on the inertial reference frame.
