GravitationMedium
Question
Net gravitational force at the centre 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 $\_\_\_\_$ .
Options
A.2
B.3
C.1
D.$2\sqrt{5}$
Solution
Initial configuration
$$F = 2\sqrt{2}\frac{{Gmm}_{0}}{r^{2}} $$New configuration
$${F' = \sqrt{10}\frac{{Gmm}_{0}}{r^{2}} \Rightarrow \frac{\text{ }F}{{\text{ }F}'} = 2\sqrt{2} \cdot \frac{1}{\sqrt{10}} = \frac{2}{\sqrt{5}} }{\therefore\alpha = 2}$$
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