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  • 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: <img alt="" src="https://cdn.entrance360.com/media/uploads/2022/06/20/143932_1.j

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:
Option: 1 \begin{aligned} &\frac{1}{2} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(2 \sqrt{2}+1)} \\ \end{aligned}
Option: 2 \frac{1}{2} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}(2 \sqrt{2}+1)}} \\
Option: 3 \frac{1}{2} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(2 \sqrt{2}-1)} \\
Option: 4 \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}

Answers (1)

best_answer


r= \sqrt{2}R

Let Fnet be the force acting on particle 1


F_{net}= F_{4}+\left ( F_{2}+F_{3} \right )\cos 45
         = \frac{GM^{2}}{4R^{2}}+\frac{2GM^{2}}{2R^{2}}\times \frac{1}{\sqrt{2}}
F_{net}= \frac{GM^{2}}{R^{2}}\left ( \frac{1}{4}+\frac{1}{\sqrt{2}} \right )
F_{net}= \frac{MV^{2}}{R}
\frac{GM^{2}}{4R^{2}}\left ( 1+2\sqrt{2} \right )= \frac{MV^{2}}{R}
\Rightarrow V= \frac{1}{2}\sqrt{\frac{GM^{2}}{R}\left ( 2\sqrt{2}+1 \right )}
The correct option is (1)

       
 

Posted by

vishal kumar

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