Question 184: Gravitation

Question

Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be :

Accepted Answer

Given, m = 1 kg, R = 1 m
We know that,
$$F = {{G{m_1}{m_2}} \over {{r^2}}}$$
$$\because$$ $${F_1} = {{Gmm} \over {{{(2R)}^2}}} = {{G{m^2}} \over {4{R^2}}}$$
and $${F_2} = {{Gmm} \over {{{(\sqrt 2 R)}^2}}} = {{G{m^2}} \over {2{R^2}}}$$
Net force on one particle,
$${F_{net}} = {F_1} + {F_2}\cos 45^\circ  + {F_2}\cos 45^\circ $$
$$ = {F_1} + 2{F_2}\cos 45^\circ $$
$$ = {{G{m^2}} \over {4{R^2}}} + 2\left( {{{G{m^2}} \over {2{R^2}}}} \right).{1 \over {\sqrt 2 }}$$
$$ = {{G{m^2}} \over {4{R^2}}} + {{G{m^2}} \over {\sqrt 2 {R^2}}}$$
$$ = {{G{m^2}} \over {{R^2}}}\left[ {{1 \over 4} + {1 \over {\sqrt 2 }}} \right]$$
As the gravitational force provides the necessary centripetal force, so
$${F_{net}} = {F_C} = {{m{v^2}} \over R}$$
Here, F
C
= centripetal force.
$$ \Rightarrow {{G{m^2}} \over {{R^2}}}\left[ {{1 \over 4} + {1 \over {\sqrt 2 }}} \right] = {{m{v^2}} \over R}$$
$$ \Rightarrow v = {1 \over 2}\sqrt {{{Gm} \over R}(1 + 2\sqrt 2 )} $$
$$ \Rightarrow v = {1 \over 2}\sqrt {G(1 + 2\sqrt 2 )} $$

Answer

$$\sqrt {{G \over 2}(1 + 2\sqrt 2 )} $$

Answer

$$\sqrt {{G \over 2}(2\sqrt 2  - 1)} $$

Answer

$$\sqrt {G(1 + 2\sqrt 2 )} $$

Answer

$${1\over2}\sqrt {G(1 + 2\sqrt 2 )} $$

Gravitation

Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be :

Choose the correct answer: