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 the 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 _________.

Let the distance from the center of the square to each corner be $$r$$. The magnitude of the gravitational force from a mass $$m$$ on a test mass $$m_0$$ at the center is $$F = \frac{G m m_0}{r^2}$$. Let $$k = \frac{G M m_0}{r^2}$$.

So the force from a mass $$nM$$ has magnitude $$nk$$.

In the given figure:

Diagonal 1 (connecting $$M$$ and $$3M$$): The masses pull in opposite directions. The net force along this diagonal is $$3k - 1k = 2k$$ (acting towards the $$3M$$ mass).

Diagonal 2 (connecting $$2M$$ and $$4M$$): The net force along this diagonal is $$4k - 2k = 2k$$ (acting towards the $$4M$$ mass).

Since the diagonals of a square are perpendicular: $$F_1 = \sqrt{(2k)^2 + (2k)^2} = \sqrt{8}k = 2\sqrt{2}k$$

When the positions of $$3M$$ and $$4M$$ are interchanged:

The $$4M$$ mass is now at the Top-Right corner (paired with $$M$$).

The $$3M$$ mass is now at the Top-Left corner (paired with $$2M$$).

Diagonal 1 (connecting $$M$$ and $$4M$$): Net force $$= 4k - 1k = 3k$$.

Diagonal 2 (connecting $$2M$$ and $$3M$$): Net force $$= 3k - 2k = 1k$$.

$$F_2 = \sqrt{(3k)^2 + (1k)^2} = \sqrt{9k^2 + k^2} = \sqrt{10}k$$

$$\frac{F_1}{F_2} = \frac{\sqrt{8}k}{\sqrt{10}k} = \sqrt{\frac{8}{10}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$$

$$\frac{2}{\sqrt{5}} = \frac{\alpha}{\sqrt{5}} \implies \alpha = 2$$