Answer
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 $F_{moon}$.
