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

Options

A. `(M_(s))/(M_(m)) (r_(2))/(r_(1))xx100`
B. `(M_(s))/(M_(m))((r_(2))/(r_(1)))xx100`
C. `2((r_(1))/(r_(2)))^(2) (M_(m))/(M_(s))xx100`
D. `((r_(1))/(r_(2)))^(2)(M_(m))/(M_(s))xx100`

Answer

C

Explanation

During total eclipse total attraction due to the sun and the moon.
`F_(1)=(GM_(s)M_(e))/(r_(1)^(2))+(GM_(m)M_(e))/(R_(2)^(2)`
When the Moon goes on the opposite side to the earth, the effective force of attraction is
`F_(2)=(GM_(s)M_(e))/(r_(1)^(2))-(GM_(m)M_(e))/(r_(2)^(2))`
Change in force `/_\F=F_(1)-F_(2)=(2GM_(m)M_(e))/(r_(2)^(2)`
change in acceleration of the earth
`/_\a=(/_\F)/(M_(e))=(2Gmm)/(r_(2)^(2))`
Average force on the earth
`F_(av)=(F_(1)+F_(2))/2=(GM_(s)M_(e))/(r_(1)^(2))`
Average acceleration of the earth
`a_(av)+F_(av)/(M_(e))=(GM_(s))/(r_(1)^(2))`
Percentage change in acceleration is
`(/_\a)/a_(av)xx100=(2GM_(m))/(r_(2)^(2))xx(r_(1)^(2))/(GM_(s))xx100=2((r_(1))/(r_(2)))^(2)(M_(m))/(M_(s))xx100`