$a)\; \bigg[\frac{GM}{r} \bigg(\frac{1+ 2 \sqrt 2}{4}\bigg)\bigg]^{\frac {1}{2}}\\ b)\;3 \sqrt {\frac{GM}{r}} \\ c)\;\sqrt {\frac{GM}{r} (1+ 2 \sqrt 2)} \\ d)\; \bigg[\frac{1}{2} \frac{GM}{r} \bigg(\frac{1+ 2 \sqrt 2}{2}\bigg)\bigg]^{\frac {1}{2}} $

The net gravitational force = centripetal force

$|F_1|$ force between A and B

$\qquad= \large\frac{GMM}{(\sqrt 2 r)^2}$

$|F_2|$ force between A and D

$\qquad= \large\frac{GMM}{(\sqrt 2 r)^2}$

$|F_3|$ force between A and C

$\qquad= \large\frac{GMM}{( 2 r)^2}$

The components of $F_1$ and $F_2$ along the radius

$|F_1| \cos 45$ and $|F_2| \cos 45 \qquad \bigg[|F_1|=|F_2|=F\bigg]$

Net force $= 2 f \cos 45 +F_3$

$\qquad= 2 \large\frac{GM^2}{(\sqrt 2 r)^2} \times \large\frac{1}{2} + \large\frac{GM^2}{4r^2}$

$\large\frac{Mv_0^2}{r}=\frac{GM^2}{4r^2}$$ [2 \sqrt 2+1]$

$v_0= \bigg[\large\frac{GM}{4r} $$(2 \sqrt 2+1)\bigg]^{1/2}$

Hence a is the orrect answer.

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Why Here force is taken out from

Center of circle