Net gravitational force at the center of a square is found to be F₁… | JEE Main 2026 (22 Jan, Shift 1)

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Newton's Law of Gravitation

Net gravitational force at the center of a square is found to be $F_1$ when four particles having mass $M, 2M, 3M$ and $4M$ are placed at the four corners of the square as shown in figure and it is $F_2$ when the positions of $3M$ and $4M$ are interchanged. The ratio $\frac{F_1}{F_2}$ is $\frac{\alpha}{\sqrt{5}}$ . The value of $\alpha$ is ______.

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Answer: (a)

Each corner is at the same distance $r$ from the centre, so each pull is $\frac{GM_im}{r^2}$ along the unit vector to that corner, $\frac{(\pm1, \pm1)}{\sqrt{2}}$ .
Take $M(-1,-1)$ , $2M(1,-1)$ , $3M(1,1)$ , $4M(-1,1)$ : $\sum M_i(x,y) = M(0, 4)$ , so $F_1 \propto \frac{4}{\sqrt{2}}$ .
Swap $3M$ and $4M$ : $\sum M_i(x,y) = M(2, 4)$ , so $F_2 \propto \frac{\sqrt{20}}{\sqrt{2}}$ .
$\frac{F_1}{F_2} = \frac{4}{\sqrt{20}} = \frac{2}{\sqrt{5}} \Rightarrow \alpha = 2$
So the answer is (a) 2.

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