## Physics (10th Edition)

Published by Wiley

# Chapter 8 - Rotational Kinematics - Problems - Page 215: 51

#### 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 $$\sum F=ma=5.27\times10^{20}N$$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.