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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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Four similar particles of mass M are orbiting in a circle of radius r in the same angular direction because of their mutual gravitational attraction, velocity of particle is

$a)\; \bigg[\frac{GM}{r} \bigg(\frac{1+ 2 \sqrt 2}{4}\bigg)\bigg]^{\frac {1}{2}}\\ b)\;3 \sqrt {\frac{GM}{r}} \\ c)\;\sqrt {\frac{GM}{r} (1+ 2 \sqrt 2)} \\ d)\; \bigg[\frac{1}{2} \frac{GM}{r} \bigg(\frac{1+ 2 \sqrt 2}{2}\bigg)\bigg]^{\frac {1}{2}} $

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1 Answer

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The net gravitational force = centripetal force
$|F_1|$ force between A and B
$\qquad= \large\frac{GMM}{(\sqrt 2 r)^2}$
$|F_2|$ force between A and D
$\qquad= \large\frac{GMM}{(\sqrt 2 r)^2}$
$|F_3|$ force between A and C
$\qquad= \large\frac{GMM}{( 2 r)^2}$
The components of $F_1$ and $F_2$ along the radius
$|F_1| \cos 45$ and $|F_2| \cos 45 \qquad \bigg[|F_1|=|F_2|=F\bigg]$
Net force $= 2 f \cos 45 +F_3$
$\qquad= 2 \large\frac{GM^2}{(\sqrt 2 r)^2} \times \large\frac{1}{2} + \large\frac{GM^2}{4r^2}$
$\large\frac{Mv_0^2}{r}=\frac{GM^2}{4r^2}$$ [2 \sqrt 2+1]$
$v_0= \bigg[\large\frac{GM}{4r} $$(2 \sqrt 2+1)\bigg]^{1/2}$
Hence a is the orrect answer.

 

answered Aug 30, 2013 by meena.p
edited Feb 17, 2014 by meena.p
 

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