Answer
a) $v_T=2.41\times10^5m/s$
b) $\sum F=5.27\times10^{20}N$
Work Step by Step
a) We have $r=2.3\times10^4$ light-years $\times\frac{9.5\times10^{15}m}{1\text{ light-year}}=2.19\times10^{20}m$ and $\omega=1.1\times10^{-15}rad/s$
$$v_T=\omega r=2.41\times10^5m/s$$
b) The centripetal acceleration of the sun is $$a_c=r\omega^2=2.65\times10^{-10}m/s^2$$
We assume the sun has a uniform circular motion, which means $v_T$ is constant and $a_T=0$. Therefore, the total acceleration of the sun is $a=a_c$
The sun's mass $m=1.99\times10^{30}kg$. The magnitude of the net force acting on the sun is $$\sum F=ma=5.27\times10^{20}N$$
