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  • Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is :<div cl

Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is :

Option: 1

\sqrt{\frac{GM}{R}}


Option: 2

\sqrt{2\sqrt{2}\frac{GM}{R}}


Option: 3

\sqrt{\frac{GM}{R}(1+2\sqrt{2})}


Option: 4

\frac{1}{2}\sqrt{\frac{GM}{R}(1+2\sqrt{2})}


Answers (1)

best_answer

  

2FCos45^{\circ}+F^{1}=\frac{M\mathrm{u}^{2}}{R}

F=\frac{GM^{2}}{(\sqrt{2}R)^{2}}      and F^{1}=\frac{GM^{2}}{4R^{2}}

=> \frac{2GM^{2}}{\sqrt{2}(\sqrt{2}R)^{2}}+\frac{GM^{2}}{4R^{2}}=\frac{M\mathrm{u}^{2}}{R}

=> \frac{GM^{2}}{R}\:[\frac{1}{4}+\frac{1}{\sqrt{2}}]=M\mathrm{u}^{2}=> \mathrm{u}=\sqrt{\frac{GM}{R}(\frac{\sqrt{2}+4}{4\sqrt{2}}})\\\mathrm{u}=\frac{1}{2}\sqrt{\frac{GM}{R}(1+2\sqrt{2}})

 

Posted by

Shailly goel

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