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:
$r_{1}$ is the distance of earth from sun, $r_{2}$ is the distance of earth from moon.

\frac{M_{s}}{M_{m}} \frac{r_{2}}{r_{1}} \times 100

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\frac{M_{s}}{M_{m}} {(\frac{r_{2}}{r_{1}})}^{2} \times 100

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2 {(\frac{r_{1}}{r_{2}})}^{2} \frac{M_{m}}{M_{s}} \times 100

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{(\frac{r_{1}}{r_{2}})}^{2} \frac{M_{s}}{M_{m}} \times 100

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Solution

The correct option is B $2 {(\frac{r_{1}}{r_{2}})}^{2} \frac{M_{m}}{M_{s}} \times 100$
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,
$△ F = F_{1} - F_{2} = \frac{2 G M_{m} M_{e}}{r_{2}^{2}}$
Change in acceleration of earth
$△ a = \frac{△ F}{M_{e}} = \frac{2 G M_{m}}{r_{2}^{2}}$
Average force on earth, $F_{a v} = \frac{F_{a v}}{M_{e}} = \frac{G M_{s}}{r_{1}^{2}}$
Percentage change in acceleration
= $\frac{△ a}{a_{a v}} \times 100 = \frac{2 G M_{m}}{r_{2}^{2}} \times \frac{r_{1}^{2}}{G M_{s}} \times 100 = 2 {(\frac{r_{1}}{r_{2}})}^{2} \frac{M_{m}}{M_{s}} \times 100$

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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:r1 is the distance of earth from sun, r2 is the distance of earth from moon.

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:
$r_{1}$ is the distance of earth from sun, $r_{2}$ is the distance of earth from moon.