Answer
a) $\omega=8.3\times 10^{-16}\frac{rad}{s}$
b) $r=1.7\times 10^{17}mi$
Work Step by Step
(a) We can find the angular speed as follows:
$\omega=\frac{\Delta \theta}{\Delta t}$
We plug in the known values to obtain:
$\omega=\frac{2\pi rad}{240\times 10^6yr(3.16\times 10^7\frac{s}{yr})}=8.3\times 10^{-16}\frac{rad}{s}$
(b) We can find the radius as below
$v=r\omega$
This can be rearranged as
$r=\frac{v}{\omega}$
$r=\frac{137\frac{mi}{s}}{8.3\times 10^{-16}\frac{rad}{s}}=1.7\times 10^{17}mi$
