JEE MAIN - Physics (2021 - 1st September Evening Shift - No. 17)
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 :
_1st_September_Evening_Shift_en_17_1.png)
$${1 \over 2}\sqrt {{{GM} \over {R(2\sqrt 2 + 1)}}} $$
$${1 \over 2}\sqrt {{{GM} \over R}(2\sqrt 2 + 1)} $$
$${1 \over 2}\sqrt {{{GM} \over R}(2\sqrt 2 - 1)} $$
$$\sqrt {{{GM} \over R}} $$
Explanation
Let us consider the gravitational force acting on each mass M by adjacent particles be F.
and the gravitational force acting on each mass M diagonally be F1.
_1st_September_Evening_Shift_en_17_2.png)
$${F_{net}} = {{M{V^2}} \over R}$$
Along the centre of circle,
$$\sqrt 2 F + {F_1} = {{M{V^2}} \over R}$$
$$\sqrt 2 {{GMM} \over {{{\left( {\sqrt 2 R} \right)}^2}}} + {{GMM} \over {{{(2R)}^2}}} = {{M{V^2}} \over R}$$
$${{GM} \over R}\left( {{1 \over {\sqrt 2 }} + {1 \over 4}} \right) = {V^2}$$
$${{GM} \over R}\left( {{{4 + \sqrt 2 } \over {4\sqrt 2 }}} \right) = {V^2}$$
$$V = \sqrt {{{GM\left( {4 + \sqrt 2 } \right)} \over {R4\sqrt 2 }}} $$
$$V = {1 \over 2}\sqrt {{{GM\left( {2\sqrt 2 + 1} \right)} \over R}} $$
Option (b)
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