Take the mean distance of the moon and the sun from the earth to be $0.4 \times 10^6 \mathrm{~km}$ and $150…
Take the mean distance of the moon and the sun from the earth to be $0.4 \times 10^6 \mathrm{~km}$ and $150 \times 10^6 \mathrm{~km}$ respectively. Their masses are $8 \times 10^{22} \mathrm{~kg}$ and $2 \times$ $10^{30} \mathrm{~kg}$ respectively. The radius of the earth is $6400 \mathrm{~km}$. Let $\Delta F_1$ be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and $\Delta \mathrm{F}_2$ be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to $\frac{\Delta F_1}{\Delta F_2}$ is:
$2$
$6$
$10^{-2}$
$0.6$
Solution
As we know, Gravitational force of attraction,
$\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{R}^2}$
$\mathrm{~F}_1=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{r}_1^2}$ and $\mathrm{F}_2=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{M}_{\mathrm{s}}}{\mathrm{r}_2^2}$
$\Delta \mathrm{F}_1=\frac{2 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{r}_1^3} \Delta \mathrm{r}_1$ and $\Delta \mathrm{F}_2=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{M}_{\mathrm{s}}}{\mathrm{r}_2^3} \Delta \mathrm{r}_2$
Using $\Delta \mathrm{r}_1=\Delta \mathrm{r}_2=2 \mathrm{R}_{\text {earth }} ; \mathrm{m}=8 \times 10^{22} \mathrm{~kg}$;
$\mathrm{M}_{\mathrm{s}}=2 \times 10^{30} \mathrm{~kg}$
$\mathrm{r}_1=0.4 \times 10^6 \mathrm{~km}^2$ and $\mathrm{r}_2=150 \times 10^6 \mathrm{~km}$
$\frac{\Delta \mathrm{F}_1}{\Delta \mathrm{F}_2}=\left(\frac{8 \times 10^{22}}{2 \times 10^{30}}\right)\left(\frac{150 \times 10^6}{0.4 \times 10^6}\right)^3 \times 1 \cong 2$