Net gravitational force at the center of a square is found to be F₁ when four particles having mass M, 2 M, 3 M and 4 M are placed at the four corners of the square as shown in figure and it is F₂ when the positions of 3 M and 4 M are interchanged. The ratio F₁ F₂ is 5 . The value of is _ _ _ _ .

Step-by-step solution

Let k = G r^2 2 where r = a/ 2 . Net field at center has components along x and y . Config 1 (original): g_x = k(-M+2M+3M-4M) = 0 , g_y = k(-M-2M+3M+4M) = 4kM . F₁ = 4kM . Config 2 (3M and 4M swapped): g_x = k(-M+2M+4M-3M) = 2kM , g_y = k(-M-2M+4M+3M) = 4kM . F₂ = kM 4+16 = 2 5 ,kM . F₁ F₂ = 4 2 5 = 2 5 = 2 .