## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Let $M_s$ be mass of the sun. Let $R_s$ be the distance from the moon to the sun. Let $M_m$ be the mass of the moon. We can write an expression for the force of gravity $F_s$ that the sun exerts on the moon. $F_s = \frac{G~M_s~M_m}{R_s^2}$ Let $M_e$ be the mass of the earth. Let $R_e$ be the distance from the earth to the moon. We can write an expression for the force of gravity $F_e$ that the earth exerts on the moon. $F_e = \frac{G~M_e~M_m}{R_e^2}$ We can find the ratio of $\frac{F_s}{F_e}$. $\frac{F_s}{F_e} = \frac{(\frac{G~M_s~M_m}{R_s^2})}{(\frac{G~M_e~M_m}{R_e^2})}$ $\frac{F_s}{F_e} = \frac{M_s~R_e^2}{M_e~R_s^2}$ $\frac{F_s}{F_e} = \frac{(1.989\times 10^{30}~kg)(3.84\times 10^8~m)^2}{(5.98\times 10^{24}~kg)(1.50\times 10^{11}~m)^2}$ $\frac{F_s}{F_e} = 2.18$ The ratio of the sun's gravitational force on the moon and the earth's gravitational force on the moon is 2.18