Question

Four similar particles of mass m are orbiting in a circle of radius r in the same angular direction because of their mutual gravitational attractive force. Velocity of a particle is given by.

• A. ${\left[\frac{Gm}{r}\left(\frac{1+2\sqrt{2}}{4}\right)\right]}^{\frac{1}{2}}$
• B. $\sqrt{\frac{Gm}{r}}$
• C. $\sqrt{\frac{Gm}{2}(1+2\sqrt{2})}$
• D. $\frac{1}{2}{\left[\frac{Gm}{r}\left(\frac{1+2\sqrt{2}}{2}\right)\right]}^{\frac{1}{2}}$

Answer 1

The correct option is A ${\left[\frac{Gm}{r}\left(\frac{1+2\sqrt{2}}{4}\right)\right]}^{\frac{1}{2}}$

$F_1=F_2=\frac{G{m}^2}{2r^2}{F}_s=\frac{G{m}^2}{4r^2}$

Centripetal force

$\frac{m{v}^2}{r}=\frac{2G{m}^2}{2r^2}\times \sin 45°+\frac{G{m}^2}{4r^2}\Rightarrow \frac{m{v}^2}{r}=\frac{G{m}^2}{r^2}\times \frac{1}{\sqrt{2}}+\frac{G{m}^2}{4r^2}\Rightarrow v^2=\frac{Gm}{r}[\frac{1}{\sqrt{2}}+\frac{1}{4}]v^2=\frac{GM}{r}\left[\frac{2\sqrt{2}+1}{4}\right]v={\left(\frac{GM}{r}\left[\frac{2\sqrt{2}+1}{4}\right]\right)}^{\frac{1}{2}}$