Take the mean distance of the moon and the sun from the earth to be`0.4 xx 10^(6) km` and `150 xx 10^(6) km` respectively. Their masses are `8 xx 10^(22) kg` and `2 xx 10^(30) kg` respectively. The radius of the earth of is `6400 km`. Let `DeltaF_(1)`be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and `DeltaF_(2)` be the difference in the force exerted b the sun at the nearest and farthest points on the earth. Then, the number closest to `(DeltaF_(1))/(DeltaF_(2))` is :

Options

A. 6
B. `10^(2)`
C. 2
D. `0.6`

Answer

C

Explanation

Gravitional force between the earth and the moon
`F_(1) = (GM_(e)M_(m))/(r_(1)^(2))`
Gravitational force between the earth and the sun
`F_(2) = (GM_(e)M_(s))/(r_(2)^(2))`
Differentiate w.r.t. distance `r_(1)` and `r_(2)` respectively
`Delta F_(1) = (2GM_(6)M_(m))/(r_(1)^(3)) Delta r_(1) , Delta F_(2) = (2GM_(e) M_(s))/(r_(2)^(3)) Delta r_(2)`
`(Delta F_(1))/(Delta F_(2)) = (M_(m) Delra r_(1))/(r_(1)^(3))(r_(2)^(3))/(M_(s) Delta r_(2)) = ((M_(m))/(M_(s)))((r_(2))/(r_(1)))^(3)((Delta r_(1))/(Delta r_(2)))`
Using `Delta r_(1) = Delta r_(2) = 2 R_(e)`
`M_(m) = 8 xx 10^(22) kg, M_(s) = 2 xx 10^(30) kg`
`r_(1) = 0.4 xx 10^(6) km, r_(2) = 150 xx 10^(6) km`
we get `(Delta F_(1))/(Delta F_(2)) ~~ 2`
hence option (c ) is correct.