Four particles of mass m, 2m, 3m, and 4, are kept in sequence at the corners of a square of side a. The magnitude of gravitational force acting on a particle of mass m placed at the centre of the square will be

Options

A. `(24 m^(2)G)/(a^(2))`
B. `(6 m^(2) G)/(a^(2))`
C. `(4 sqrt(2) Gm^(2))/(a^(2))`
D. Zero

Answer

C

Explanation

If two particle of mass m are placed x distance apart then force of attraction `(G m m )/( x^(2)) = F ` (Let )
Now according to problem particle of mass m is placed at the centre (P) of square . Then it will experience four forces .
`F_(PA) `= force at point P due to particle `A = (G m m )/( x^(2)) = F `
Similarly `F_(PB) = (G2 m m )/( x^(2)) = 2 F , F_(PC) = (G 3 m m )/(x^(2)) = 3F and F_(PD) = (G 4 m m )/(x^(2)) = 4 F `
Hence the net force on `P vec(F) _(" net ") = vec(F) _(PA) = vec(F) _(PB) + vec(F)_(PC) + vec(F)_(PD) = 2 sqrt(2) F `
`: vec(F)_("net") = 2 sqrt(2) (G m m )/( x^(2)) = 2 sqrt(2) (Gm^(2)) /((a // sqrt(2))^(2)) ` [`x = (1)/( sqrt(2))` = half of the diagonal of the square ]
`= (4 sqrt(2) G m^(2))/( a^(2))`