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. $6$
• C. ${10}^{-2}$
• D. $0.6$

Answer 1

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

$\Delta F_1=F_1-F_2$

$\Delta F_2=F_1^{\prime }-F_2^{\prime }$

$\Delta F_1=\frac{G{m}_m{m}_e}{(0.4\times {10}^6-Re)}-\frac{G{m}_m{m}_e}{(0.4\times {10}^6+Re)}$

$\Delta F_2=\frac{G{m}_m{m}_s}{(150\times {10}^6-Re)}-\frac{Gm{e}_m{m}_e}{(150\times {10}^6+Re)}$

$\frac{\Delta F_1}{F_2}=\frac{G{M}_{m}Me\{\frac{1}{0.4\times {10}^6-Re}-\frac{1}{0.4\times {10}^6+Re}\}}{G{M}_{e}Ms\{\frac{1}{150\times {10}^6-Re}-\frac{1}{150\times {10}^6+Re}\}}$

$=\frac{\frac{Mm}{Ms}\left\{\frac{0.4\times {10}^6+Re-0.4\times {10}^6+Re}{(0.4\times {10}^6-Re)(0.4\times {10}^6+Re)}\right\}}{\left\{\frac{150\times {10}^6+Re-150\times {10}^6+Re}{(150\times {10}^6-Re)(150\times {10}^6+Re)}\right\}}$

$\frac{Mm}{Ms}\times \frac{(150\times {10}^6-Re)(150\times {10}^6+Re)}{(0.4\times {10}^6-Re)(0.4\times {10}^6+Re)}=\frac{(150-.0064)(150+.0064)}{(0.4-.0064)(0.4+.0064)}$

$=\frac{Mm}{Ms}\frac{149.99\times 150.0064}{0.3936\times .4064}$

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

$\approx 0.0056$

$\approx 0.01$

$\approx {10}^{-2}$