If two particles each of mass `m` are placed at the two vertices of an equilateral triangle of side `a`, then the resultant gravitational force on mass `m` placed at the third vertex is

Options

A. `(sqrt3 Gm^2)/(2r^2)`
B. `(sqrt3 Gm^2)/(r^2)`
C. `(Gm^2)/(r^2)`
D. `(Gm^2)/(2r^2)`

Answer

B

Explanation

Gravitational force on particle at A C due to particle at A is

` F_1 = (Gmm)/(r^2) = (Gm^2)/(r^2) ` along CA
Gravitational force on particle at C due to particle at B is
`F_2 = (Gm^2)/(r^2) ` along CB
As `vecF_1` and `vecF_2` are inclined at an angle `60^@` , so the resultant gravitational force on particle at C is
` F = sqrt(F_1^2 + F_2^2 + 2F_1F_2 cos 60^@)`
` =sqrt(F_1^2 + F_1^2 + 2F_1 xx F_1 xx 1/2) = sqrt3F_1 = (sqrt3 Gm^2)/(r^2) " " ( because F_1 = F_2)`