All point on the circumference of the ring are at distance \\sqrt{R^2 + x^2} from the center O of the ring.Force on the mass m at P : F = \\dfrac{ GMm}{R^2 +x^2} \ imes (2 \\cos \ heta ) where \\cos \ heta = \\dfrac{x}{\\sqrt{R^2 + x^2} } Due to symmetry, vertical components of the forces from two symmetrical elements cancel out, the horizontal components add up, hence we get a factor of 2. \ herefore F = \\dfrac{ GMm}{R^2 +x^2} \ imes 2\\dfrac{x}{\\sqrt{R^2 + x^2} } When F is maximum, \\frac{dF}{dx} =0 \\Rightarrow (R^2 + x^2) ^{-\\dfrac{3}{2}} +x ( -\\dfrac{3}{2} (R^2 + x^2) ^{-\\dfrac{3}{2} -1} \ imes 2x =0 \\Rightarrow (R^2 + x^2) = 3x^2 \\Rightarrow x = \\dfrac{R}{\\sqrt{2}}