Gravitation Part 2

Concept:The gravitational force between two masses is given by Newton's law: $F = \frac{G m_1 m_2}{r^2}$, and forces are vectors that add vectorially.Explanation:Let the square side be $a$; distance from each corner to centre is $r = \frac{a}{\sqrt{2}}$.Place a test mass $m_0$ at the centre. The force due to mass $M$ is $F_0 = \frac{GM m_0}{r^2}$.Forces from other corners: $2M$ gives $2F_0$, $3M$ gives $3F_0$, $4M$ gives $4F_0$.Case 1 (initial):Corner masses (in order around square): $4M$, $3M$, $2M$, $M$.Forces on centre: $4F_0$ (top left), $3F_0$ (top right), $2F_0$ (bottom right), $F_0$ (bottom left).Along one diagonal: $3F_0$ and $F_0$ oppose → net $2F_0$ toward $3F_0$.Along other diagonal: $4F_0$ and $2F_0$ oppose → net $2F_0$ toward $4F_0$.These two net forces are perpendicular. Resultant $F_1 = \sqrt{(2F_0)^2 + (2F_0)^2} = 2\sqrt{2}\,F_0$.Case 2 (interchange $3M$ and $4M$):New order: $3M$, $4M$, $2M$, $M$.Forces: $3F_0$, $4F_0$, $2F_0$, $F_0$ in same positions.Along one diagonal: $4F_0$ and $2F_0$ oppose → net $2F_0$ toward $4F_0$. (Wait careful: Actually need to re-evaluate positions. Let's set coordinates: top-left 4M → 3M now, top-right 3M → 4M, bottom-right 2M unchanged, bottom-left M unchanged. Then diagonal top-left to bottom-right: forces from top-left (3F0) and bottom-right (2F0) oppose → net 1F0 toward top-left. Other diagonal top-right (4F0) and bottom-left (1F0) oppose → net 3F0 toward top-right. Then resultant $F_2 = \sqrt{(1F_0)^2 + (3F_0)^2} = \sqrt{10}\,F_0$.)Thus $\frac{F_1}{F_2} = \frac{2\sqrt{2}F_0}{\sqrt{10}F_0} = \frac{2}{\sqrt{5}}$.Given $\frac{F_1}{F_2} = \frac{\alpha}{\sqrt{5}}$, so $\alpha = 2$.Answer:$\alpha = 2$, option (A).