Question

The percentage change in the acceleration of the earth towards the sun from a total eclipse of the sun to the point where the moon is on a side of earth directly opposite to the sun is

• A. $\frac{M_s}{M_m}\frac{r_2}{\eta }\times 100$
• B. $\frac{M_s}{M_m}{\left(\frac{r_2}{r_1}\right)}^2\times 100$
• C. $2{\left(\frac{r_1}{r_2}\right)}^2\frac{M_m}{M_s}\times 100$
• D. ${\left(\frac{r_1}{r_2}\right)}^2\frac{M_s}{M_m}\times 100$

Answer 1

Answer: (C)

During total eclipse:
Total attraction due to sun and moon,
$F_1=\frac{G{M}_s{M}_e}{r_1^2}+\frac{G{M}_m{M}_e}{r_2^2}$
When moon goes on the opposite side of earth. Effective force of attraction,
$F_2=\frac{G{M}_s{M}_e}{r_1^2}-\frac{G{M}_m{M}_e}{r_2^2}$
Change in force, $\Delta F=F_1-F_2=\frac{2G_m{M}_e}{r_2^2}$
Change in acceleration of earth
$\Delta a=\frac{\Delta F}{M_e}=\frac{2G{M}_m}{r_2^2}$
Average force on earth, $F_{av}=\frac{F_{av}}{M_e}=\frac{G{M}_s}{{\eta }^2}\%$ age change in acceleration
$=\frac{\Delta a}{a_{av}}\times 100=\frac{2G{M}_m}{r_2^2}\times \frac{r_1^2}{G{M}_s}\times 100=2{\left(\frac{r_1}{r_2}\right)}^2\frac{M_m}{M_s}\times 100$