Question

Take the mean distance of the moon and the sun from the earth to be $0.4\times 10^{6}km$ and $150\times 10^{6}km$ respectively. Their masses are $8\times 10^{22}kg$ and $2\times 10^{30}kg$ respectively. The radius of the earth is $6400km.$ 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 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 :

• A. 2
• B. $10^{-2}$
• C. 0.6
• D. 6

Answer 1

Answer: (A)

We have $\Delta F_1=$ difference in the forces exerted by the moon at the nearest and farthest points on the Earth. Therefore,
$F_1=\frac{G{m}_m{m}_e}{r_1^2}$
$\Rightarrow \Delta F_1=-\frac{G{m}_m{m}_e}{r_1^3}\Delta r_1$
where $m_m$ is mass of moon, $m_e$ is mass of Earth, $r_1$ is mean distance of the moon from the Earth. Now, $\Delta F_2=$ difference in the force exerted by the Sun at the nearest and farthest points on the Earth. Therefore,
$F_2=\frac{G{m}_c{m}_s}{r_2^2}$
$\Rightarrow \Delta F_2=-2\frac{G{m}_c{m}_s}{r_2^3}\Delta r_2$
where $m_s$ is mass of Sun, $m_e$ is mass of Earth, $r_2$ is mean distance of the Sun from the Earth. Thus,
$\frac{\Delta F_1}{\Delta F_2}=\frac{G{m}_m{m}_e}{G{m}_m{m}_s}\times \frac{r_2^3}{r_1^3}\times \frac{{\Delta }_1}{{\Delta }_2}$
Since, $\Delta r_1=\Delta r_2=2R_e$, therefore,
$\frac{\Delta F_1}{\Delta F_2}=\frac{8\times 10^{22}}{2\times 10^{30}}\times \frac{{\left(150\times 10^6\right)}^3}{{\left(0.4\times 10^6\right)}^3}\times \frac{2R_e}{2R_e}=2$