Question

Four identical particles of equal masses $1kg$ made to move along the circumference of a circle of radius $1m$ under the action of their own mutual gravitational attraction. The speed of each particle will be :

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

Answer 1

Answer: (D)

$F_1=\frac{Gmm}{(2R{)}^2}=\frac{G{m}^2}{4R^2}$
$F_2=\frac{Gmm}{(\sqrt{2}R{)}^2}=\frac{G{m}^2}{2R^2}$
$F_3=\frac{Gmm}{(\sqrt{2}R{)}^2}=\frac{G{m}^2}{2R^2}$
$\Rightarrow F_{net}=F_1+F_2\cos 45^{\circ }+F_3\cos 45^{\circ }$
$=\frac{G{m}^2}{4R^2}+\frac{G{m}^2}{2R^2}\frac{1}{\sqrt{2}}+\frac{G{m}^2}{2R^2}\frac{1}{\sqrt{2}}$
$=\frac{G{m}^2}{R^2}\left(\frac{1}{4}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}\right)$
$=\frac{G{m}^2}{R^2}\left(\frac{1}{4}+\frac{1}{\sqrt{2}}\right)=\frac{G{m}^2}{4R^2}(1+2\sqrt{2})$
$F_{net}=\frac{G{m}^2}{4R^2}(1+2\sqrt{2})=\frac{m{v}^2}{R}$
$\Rightarrow v=\frac{\sqrt{G(1+2\sqrt{2})}}{2}$