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$$\textbf{Step1:Draw diagram and find radius of circular orbit}$$

$$(Hypotaneous)^2=a^2+a^2$$

$$Hypotaneous=\sqrt 2$$

$$Diameter=\sqrt 2$$

$$ \text{Radius} =\dfrac{\text{Diameter}}{\sqrt 2}$$

$$\textbf{Step2:Net force on particle towarda centre of circle}$$

Both $$F=\dfrac{GM^2}{a^2}$$ are at $$90^0$$ so their resultant is $$\dfrac{\sqrt 2GM^2}{a^2}$$ and act at angle bisector.

Net force on particle towards centre of circle is $$F_{C} = \dfrac {GM^{2}}{2a^{2}} + \dfrac {GM^{2}}{a^{2}} \sqrt {2}$$$$= \dfrac {GM^{2}}{a^{2}} \left (\dfrac {1}{2} + \sqrt {2}\right )$$

$$\textbf{Step3:Calculation of speed}$$

This resultant force will act as centripetal force. Distance of particle from centre of circle is $$\dfrac {a}{\sqrt {2}}$$.

$$r = \dfrac {a}{\sqrt {2}}, F_{C} = \dfrac {mv^{2}}{r}$$

$$\dfrac {mv^{2}}{\dfrac {a}{\sqrt {2}}} = \dfrac {GM^{2}}{a^{2}} \left (\dfrac {1}{2} + \sqrt {2}\right )$$

$$v^{2} = \dfrac {GM}{a} \left (\dfrac {1}{2\sqrt {2}} + 1\right )$$

$$v^{2} = \dfrac {GM}{a} (1.35)$$

$$v = 1.16\sqrt {\dfrac {GM}{a}}$$.

This resultant force will act as centripetal force. Distance of particle from centre of circle is $$\dfrac {a}{\sqrt {2}}$$.

$$r = \dfrac {a}{\sqrt {2}}, F_{C} = \dfrac {mv^{2}}{r}$$

$$\dfrac {mv^{2}}{\dfrac {a}{\sqrt {2}}} = \dfrac {GM^{2}}{a^{2}} \left (\dfrac {1}{2} + \sqrt {2}\right )$$

$$v^{2} = \dfrac {GM}{a} \left (\dfrac {1}{2\sqrt {2}} + 1\right )$$

$$v^{2} = \dfrac {GM}{a} (1.35)$$

$$v = 1.16\sqrt {\dfrac {GM}{a}}$$.

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