Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 4 - Forces and Newton's Laws of Motion - Problems - Page 115: 35


The sun exerts a greater gravitational force than the moon. $\frac{F_{sun}}{F_{moon}}=177.5$

Work Step by Step

- The mass of the person $m$ - The mass of the sun $M_s=1.99\times10^{30}kg$ - The mass of the moon $M_m=7.35\times10^{22}kg$ - The distance between the sun and the person on earth $r_{se}=1.5\times10^{11}m$ - The distance between the moon and the person on earth $r_{me}=3.84\times10^{8}m$ - Gravitational constant $G=6.67\times10^{-11}Nm^2/kg^2$ We have $F_{sun}=G\frac{mM_s}{r_{se}^2}$ and $F_{moon}=G\frac{mM_m}{r_{me}^2}$ Then, $$\frac{F_{sun}}{F_{moon}}=\frac{G\frac{mM_s}{r_{se}^2}}{G\frac{mM_m}{r_{me}^2}}=\frac{r_{me}^2M_s}{r_{se}^2M_m}$$ We have $\frac{M_s}{M_m}=2.71\times10^7$ and $\frac{r_{me}^2}{r_{se}^2}=6.55\times10^{-6}$ Therefore, $$\frac{F_{sun}}{F_{moon}}=(2.71\times10^7)\times(6.55\times10^{-6})=177.5$$ So $F_{sun}$ is greater than $F_{moon}$.
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