Physics - Gravitation Question with Solution | TestHub

$\text{F}={\text{F}}_1={\text{F}}_2\text{=}\frac{{\text{GM}}^2}{2{\text{R}}^2}$

${\text{F}}_3=\frac{{\text{GM}}^2}{4{\text{R}}^2}$

$\therefore$ F_Total towards the centre.

$=\sqrt{2}\text{ F}+{\text{F}}_3$

$=\sqrt{2}\frac{{\text{GM}}^2}{{\text{2R}}^2}+\frac{{\text{GM}}^2}{4{\text{R}}^2}$

$=\frac{{\text{GM}}^2}{4{\text{R}}^2}\left[2\sqrt{2}+1\right]$

${\text{F}}_{\text{T}}{=\text{F}}_{\text{CP}}$

$\frac{{\text{GM}}^2}{4{\text{R}}^2}\left[2\sqrt{2}+1\right]=\frac{{\text{Mv}}^2}{\text{R}}$

$\therefore \text{v}=\sqrt{\frac{\text{GM}}{4\text{R}}\left[2\sqrt{2}+1\right]}$.