Question

Four particles each of mass $M$, move along a circle of radius $R$ under the action of their mutual gravitational attraction as shown in figure. The speed of each particle is:

• A. $\frac{1}{2}\sqrt{\frac{GM}{R(2\sqrt{2}+1)}}$
• B. $\frac{1}{2}\sqrt{\frac{GM}{R}(2\sqrt{2}+1)}$
• C. $\frac{1}{2}\sqrt{\frac{GM}{R}(2\sqrt{2}-1)}$
• D. $\sqrt{\frac{GM}{R}}$

Answer 1

Answer: (B)

$F_{\text{net }}=\frac{M{V}^2}{R}$
$\sqrt{2}F+F_1=\frac{M{V}^2}{R}$
$\sqrt{2}\frac{GMM}{(\sqrt{2}R{)}^2}+\frac{GMM}{(2R{)}^2}=\frac{M{V}^2}{R}$
$\frac{GM}{R}\left(\frac{1}{\sqrt{2}}+\frac{1}{4}\right)=V^2$
$\frac{GM}{R}\left(\frac{4+\sqrt{2}}{4\sqrt{2}}\right)=V^2$
$V=\sqrt{\frac{GM(4+\sqrt{2})}{R4\sqrt{2}}}$
$V=\frac{1}{2}\sqrt{\frac{GM(2\sqrt{2}+1)}{R}}$