Net gravitational force at the center of a square is found to be $F_1$ when four particles having ma

Question

Net gravitational force at the center of a square is found to be $F_1$ 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 $F_2$ when the positions of $3 M$ and $4 M$ are interchanged. The ratio $\frac{F_1}{F_2}$ is $\frac{\alpha}{\sqrt{5}}$. The value of $\alpha$ is $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

Let $r$ be the distance from the corners to the center. Initially, opposite pairs are (M, 3M) and (2M, 4M). The net fields along the diagonals are $\frac{2GM}{r^2}$ and $\frac{2GM}{r^2}$. Because they are perpendicular, $F_1 = \sqrt{2^2 + 2^2} \frac{GM}{r^2} = 2\sqrt{2}\frac{GM}{r^2}$. After swapping 3M and 4M, the opposite pairs are (M, 4M) and (2M, 3M). The net fields along the diagonals are $\frac{3GM}{r^2}$ and $\frac{1GM}{r^2}$. The new net field is $F_2 = \sqrt{3^2 + 1^2}\frac{GM}{r^2} = \sqrt{10}\frac{GM}{r^2}$. The ratio is $\frac{F_1}{F_2} = \frac{2\sqrt{2}}{\sqrt{10}} = \frac{2}{\sqrt{5}}$, so $\alpha = 2$.