Question 226: Gravitation

Question

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 :

Accepted Answer

All those particles are moving due to their mutual gravitational attraction.
The force between each masses are repulsive force.
On mass M at C, due to mass at D the repulsive force is F in the vertical direction.
On mass M at C, due to mass at B the repulsive force is F in the horizontal direction.
On mass M at C, due to mass at A the repulsive force is F'
Net force acting on particle at C,
= $$2F\,\cos \,{45^ \circ } + F'$$
Where $$F = {{G{M^2}} \over {{{\left( {\sqrt 2 R} ight)}^2}}}$$ and $$F' = {{G{M^2}} \over {4{R^2}}}$$
$$ \Rightarrow {{2 	imes G{M^2}} \over {\sqrt 2 {{\left( {R\sqrt 2 } ight)}^2}}} + {{G{M^2}} \over {4{R^2}}}$$
$$ \Rightarrow {{G{M^2}} \over R^2}\left[ {{1 \over 4} + {1 \over {\sqrt 2 }}} ight] $$
This net force will balance by the centripetal force F
cp
= $${{M{v^2}} \over R}$$
$$	herefore$$ $${{M{v^2}} \over R} = $$ $${{G{M^2}} \over R^2}\left[ {{1 \over 4} + {1 \over {\sqrt 2 }}} ight] $$
$$ \Rightarrow $$ $$v = \sqrt {{{Gm} \over R}\left( {{{\sqrt 2  + 4} \over {4\sqrt 2 }}} ight)} $$
$$ = {1 \over 2}\sqrt {{{Gm} \over R}\left( {1 + 2\sqrt 2 } ight)} $$

Answer

$$\sqrt {{{GM} \over R}} $$

Answer

$$\sqrt {2\sqrt 2 {{GM} \over R}} $$

Answer

$$\sqrt {{{GM} \over R}\left( {1 + 2\sqrt 2 } \right)} $$

Answer

$${1 \over 2}\sqrt {{{GM} \over R}\left( {1 + 2\sqrt 2 } \right)} $$

Gravitation

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 :

Choose the correct answer: