Four particles having masses, m, 2m, 3m, and 4m are placed at the four corners of a square of edge a. Find the gravitational force acting on a particle of mass m placed at the centre.

Options

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

Answer

C

Explanation

If two particles of mass m are pleced x distance apart then force of attraction `(Gmm)/(x^(2))=F(let)`

Now according to problem particle of a 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=(Gmm)/(x^(2))=F`
Simillarly, `F_(PB)=(G2mm)/(x^(2))=2F, F_(PC)=(G3mm)/(x^(2))=3F`
`F_(PD)=(G4mm)/(x^(2))=4F`
Hence the net force on P
`vecF_(n et)=vecF_(PA)+vecF_(PB)+vecF_(PC)+vecF_(PD)=2sqrt(2)F`
`:. vecF_(n et)=2sqrt(2) (Gmm)/(x^(2))=2sqrt(2)(Gm^(2))/((a//sqrt(2))^(2))`
[`x=a/(sqrt(2))`=half of the diagonal of the square `=(4sqrt(2)Gm^(2))/(a^(2))`