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Four massive particles, each of mass \mathrm{m, } are kept at the vertices of square of side \mathrm{\ell.} With what speed should each particle be projected so as to stabilise the system?
 

Option: 1

\mathrm{\sqrt{\frac{\mathrm{Gm}}{\ell}\left(1+\frac{1}{4 \sqrt{2}}\right)}}
 


Option: 2

\mathrm{\sqrt{\frac{\mathrm{Gm}}{\ell}\left(1+\frac{1}{2 \sqrt{2}}\right)}}


Option: 3

\mathrm{\sqrt{\frac{\mathrm{Gm}}{2 \boldsymbol{\ell}}\left(1+\frac{1}{2 \sqrt{2}}\right)}}

 


Option: 4

\mathrm{\sqrt{\frac{\mathrm{Gm}}{\ell}\left(1+\frac{1}{\sqrt{2}}\right)}}


Answers (1)

In order to stabilise the system, each mass should move in a circular path of the same radius about the centre of the square. How is it possible? It is possible when each particle moves perpendicular to the corresponding diagonal with a speed, let v. As a result the net gravitational force on the particle due to the others will provide necessary centripetal acceleration \mathrm{v^2 / r, where \: r=\ell / \sqrt{2}}

\mathrm{ \Rightarrow F_g=m \frac{v^2}{r}=\frac{m v^2}{\left(\frac{\ell}{\sqrt{2}}\right)} }          ...........(1)

where \mathrm{ F_g= } Net gravitational force on each particle.

Resolving the individual gravitational force along the diameter, we obtain,

\mathrm{ F_g=\frac{F}{\sqrt{2}}+\frac{F}{\sqrt{2}}+F^{\prime}=F^{\prime}+\sqrt{2} F }

Putting \mathrm{ \mathrm{F}=\frac{\mathrm{Gm}^2}{\ell^2} \: and \: \mathrm{F}^{\prime}=\frac{\mathrm{Gm}^2}{2 \ell^2} }

\mathrm{ \Rightarrow F_g=\frac{G m^2}{\ell^2}\left(\sqrt{2}+\frac{1}{2}\right) }    .......(2)

Using (1) and (2), we obtain,

\mathrm{ v=\sqrt{\frac{G m}{\ell}\left(1+\frac{1}{2 \sqrt{2}}\right)} . }

Hence option 2 is correct.
 

 

Posted by

Ramraj Saini

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