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, $\Delta F=F_1-F_2=\frac{2 G _m M_e}{r_2^2}$
Change in acceleration of earth
$\Delta a=\frac{\Delta F}{M_e}=\frac{2 G M_m}{r_2^2}$
Average force on earth, $F_{a v}=\frac{F_{a v}}{M_e}=\frac{ G M_s}{\eta^2} \%$ age change in acceleration
$=\frac{\Delta a}{a_{a v}} \times 100=\frac{2 G M_m}{r_2^2} \times \frac{r_1^2}{ G M_s} \times 100=2\left(\frac{r_1}{r_2}\right)^2 \frac{M_m}{M_s} \times 100$