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 o 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

The correct option is B ${10}^{-2}$

$\Delta F=F_1-F_2$

${\Delta F}_2=F_1^1-F_2^1$

${\Delta F}_1=\frac{{Gm}_m{m}_e}{(0.4\times {10}^6-R_e)}\frac{-{Gm}_m{m}_e}{(0.4\times {10}^6+R_e)}$

${\Delta F}_2=\frac{{Gm}_m{m}_5}{(150\times {10}^6-R_e)}\frac{-{Gm}_m{m}_e}{(150\times {10}^6+R_e)}$

$\frac{\Delta F_2}{\Delta F_1}=\frac{{Gm}_m{m}_e(\frac{1}{0.4\times {10}^6-R_e}-\frac{1}{0.4\times {10}^6+R_e})}{{Gm}_e{m}_s(\frac{1}{150\times {10}^6-R_e}-\frac{1}{150\times {10}^6+R_e})}$

$=\frac{\frac{m_m}{m_s}\left\{\frac{0.4\times {10}^6+R_e-0.4\times {10}^6+R_e}{(0.4\times {10}^6-R_e)(0.4\times {10}^6+R_e)}\right\}}{\left\{\left(\frac{150\times {10}^6+R_e-150\times {10}^6+R_e}{(150\times {10}^6-R_e)(150\times {10}^6+R_e)}\right)\right\}}$

$=\frac{m_m}{m_s}\times \frac{(150\times {10}^6-R_e)(150\times {10}^6+R_e)}{(0.4\times {10}^6-R_e)(0.4\times {10}^6+R_e)}$

$=\frac{(150-.0064)(150+.0064)}{(0.4-.0064)(0.4+0.0064)}$

$=\frac{m_m}{m_s}\frac{149.99\times 150.0064}{0.3936\times .4064}$

$=140657.633\times \frac{8\times {10}^{22}}{2\times {10}^{30}}$

$=0.0056$

$=0.01$

$={10}^{-2}$