Question 14 of 223

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$\dfrac{1}{2}\sqrt{\dfrac{GM}{R}\left(2\sqrt{2}+1\right)}$B$\dfrac{1}{2}\sqrt{\dfrac{GM}{R(2\sqrt{2}+1)}}$C$\dfrac{1}{2}\sqrt{\dfrac{GM}{R}\left(2\sqrt{2}-1\right)}$D$\sqrt{\dfrac{GM}{R}}$

Show answer and solution▾

Answer: (A) $\dfrac{1}{2}\sqrt{\dfrac{GM}{R}\left(2\sqrt{2}+1\right)}$

.
${F}_{\text{net}}={F}_{1}+2{F}_{2}\cos 45^{\circ}$
$=\dfrac{G{M}^{2}}{(2R{)}^{2}}+2\times \dfrac{G{M}^{2}}{(\sqrt{2}R{)}^{2}}\times \dfrac{1}{\sqrt{2}}=\dfrac{G{M}^{2}}{4{R}^{2}}+\dfrac{G{M}^{2}}{\sqrt{2}{R}^{2}}=\dfrac{(4+\sqrt{2})G{M}^{2}}{4\sqrt{2}{R}^{2}}$
Now ${F}_{C}={F}_{\text{net}}$
$\dfrac{M{V}^{2}}{R}=\dfrac{(4+\sqrt{2})G{M}^{2}}{4\sqrt{2}{R}^{2}}$
$V=\sqrt{\dfrac{(4+\sqrt{2})}{4\sqrt{2}}\dfrac{GM}{R}}$

JEE Main Physics practice question on Gravitation (Newton's Law of Gravitation), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.