Question

Four identical particles of equal masses $1 kg$ made to move along the circumference of a circle of radius $1 m$ 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

The Correct Option is D

$F _{1}=\frac{ Gmm }{(2 R )^{2}}=\frac{ Gm ^{2}}{4 R ^{2}}$
$F _{2}=\frac{ Gmm }{(\sqrt{2} R )^{2}}=\frac{ Gm ^{2}}{2 R ^{2}}$
$F _{3}=\frac{ Gmm }{(\sqrt{2} R )^{2}}=\frac{ Gm ^{2}}{2 R ^{2}}$
$\Rightarrow F _{ net }= F _{1}+ F _{2} \cos 45^{\circ}+ F _{3} \cos 45^{\circ}$
$=\frac{ Gm ^{2}}{4 R ^{2}}+\frac{ Gm ^{2}}{2 R ^{2}} \frac{1}{\sqrt{2}}+\frac{ Gm ^{2}}{2 R ^{2}} \frac{1}{\sqrt{2}}$
$=\frac{ Gm ^{2}}{ R ^{2}}\left(\frac{1}{4}+\frac{1}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}\right)$
$=\frac{ Gm ^{2}}{ R ^{2}}\left(\frac{1}{4}+\frac{1}{\sqrt{2}}\right)=\frac{ Gm ^{2}}{4 R ^{2}}(1+2 \sqrt{2})$
$F _{ net }=\frac{ Gm ^{2}}{4 R ^{2}}(1+2 \sqrt{2})=\frac{ mv ^{2}}{ R }$
$\Rightarrow v =\frac{\sqrt{ G (1+2 \sqrt{2})}}{2}$