## Physics (10th Edition)

Published by Wiley

# Chapter 8 - Rotational Kinematics - Problems - Page 213: 18

#### Answer

a) $\theta_{moon}=9.04\times10^{-3}rad$ and $\theta_{sun}=9.27\times10^{-3}rad$ b) A total eclispe is not actually total. c) The ratio is $95.1\%$

#### Work Step by Step

a) The moon has a diameter $d=3.48\times10^6m$ and a distance from earth $r=3.85\times10^8m$. The angle subtended by the moon is $$\theta_{moon}=\frac{d}{r}=9.04\times10^{-3}rad$$ The sun has a diameter $d=1.39\times10^9m$ and a distance from earth $r=1.5\times10^{11}m$. The angle subtended by the sun is $$\theta_{sun}=\frac{d}{r}=9.27\times10^{-3}rad$$ b) A total eclipse of the sun is actually not entirely total, because the angle subtended by the moon is slightly less than that by the sun, so the moon cannot cover all of the sun during the eclipse, because the sun still appears slightly bigger to people on Earth. c) From Earth, the moon and the sun appear to us as having the same distance away from the Earth. We call this apparent distance $r_{a}$ From here, we can calculate the apparent diameter of the moon and the sun: $$d_{a, moon}=r_a\theta_{moon}$$ $$d_{a, sun}=r_a\theta_{sun}$$ The apparent area of the moon and the sun is $$A_{a, moon}=2\pi\Big(\frac{d_{a, moon}}{2}\Big)^2=\frac{\pi d^2_{a, moon}}{2}=\frac{\pi r_a^2}{2}\theta^2_{moon}$$ $$A_{a, sun}=2\pi\Big(\frac{d_{a, sun}}{2}\Big)^2=\frac{\pi d^2_{a, sun}}{2}=\frac{\pi r_a^2}{2}\theta^2_{sun}$$ The ratio between the two is $$\frac{A_{a, moon}}{A_{a, sun}}\times100\%=\Big(\frac{\theta_{moon}}{\theta_{sun}}\Big)^2=95.1\%$$

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.