Take the mean distance of the moon and the sun from the earth to be $0.4 \times 10^6\ \text{km}$ and $150 \times 10^6\ \text{km}$, respectively. Their masses are $8 \times 10^{22}\ \text{kg}$ and $2 \times 10^{30}\ \text{kg}$, respectively. The radius of the earth is $6400\ \text{km}$. Let $\Delta F_1$ be the difference in the forces exerted by the moon at the nearest and farthest points on the earth, and $\Delta F_2$ be the difference in the forces exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to $\dfrac{\Delta F_1}{\Delta F_2}$ is,
Answer: (C) $2$
For a mass $m$ on the earth, a body of mass $M$ at distance $r$ exerts $F = \dfrac{GMm}{r^2}$. The nearest and farthest points differ in distance by $2R_e \ll r$, so
$$\Delta F \approx \left|\frac{dF}{dr}\right|(2R_e) = \frac{2GMm}{r^3}(2R_e) \;\propto\; \frac{M}{r^3}$$
Hence
$$\frac{\Delta F_1}{\Delta F_2} = \frac{M_\text{moon}}{M_\text{sun}}\left(\frac{r_\text{sun}}{r_\text{moon}}\right)^3 = \frac{8\times10^{22}}{2\times10^{30}}\left(\frac{150}{0.4}\right)^3$$
$$= 4\times10^{-8} \times (375)^3 = 4\times10^{-8}\times 5.27\times10^{7} \approx 2.1$$
The closest number is $2$.
Solution by Sreeraj P, M.Sc Physics