Question 192: Gravitation

Question

Four identical particles of mass M are located
at the corners of a square of side 'a'. What
should be their speed if each of them revolves
under the influence of other's gravitational field
in a circular orbit circumscribing the square?

Accepted Answer

Net force on mass M at position B towards centre of circle is
$${F_{BO\,net}} = {F_{BD}} + {F_{BA}}\sin 45^\circ  + {F_{BC}}\sin 45^\circ $$
$$ = {{G{M^2}} \over {{{(\sqrt 2 a)}^2}}} + {{G{M^2}} \over {{a^2}}}\left( {{1 \over {\sqrt 2 }}} \right) + {{G{M^2}} \over {{a^2}}}\left( {{1 \over {\sqrt 2 }}} \right)$$
[where, diagonal length BD is $$\sqrt2$$a]
$$ = {{G{M^2}} \over {2{a^2}}} + {{G{M^2}} \over {{a^2}}}\left( {{2 \over {\sqrt 2 }}} \right) = {{G{M^2}} \over {{a^2}}}\left( {{1 \over 2} + \sqrt 2 } \right)$$
This force will act as centripetal force.
Distance of particle from centre of circle is $${a \over {\sqrt 2 }}$$.
Here, $${F_{centripetal}} = {{M{v^2}} \over r} = {{M{v^2}} \over {{a \over {\sqrt 2 }}}} = {{\sqrt 2 M{v^2}} \over a}$$ ($$\because$$ $$r = {a \over {\sqrt 2 }}$$)
So, for rotation about the centre,
$${F_{centripetal}} = {F_{BO(net)}}$$
$$ \Rightarrow \sqrt 2 {{M{v^2}} \over a} = {{G{M^2}} \over {{a^2}}}\left( {{1 \over 2} + \sqrt 2 } \right)$$
$$ \Rightarrow {v^2} = {{GM} \over a}\left( {1 + {1 \over {2\sqrt 2 }}} \right) = {{GM} \over a}(1.35)$$
$$ \Rightarrow v = 1.16\sqrt {{{GM} \over a}} $$

Answer

$$1.21\sqrt {{{GM} \over a}} $$

Answer

$$1.16\sqrt {{{GM} \over a}} $$

Answer

$$1.41\sqrt {{{GM} \over a}} $$

Answer

$$1.35\sqrt {{{GM} \over a}} $$

Gravitation

Four identical particles of mass M are located at the corners of a square of side 'a'. What should be their speed if each of them revolves under the influence of other's gravitational field in a circular orbit circumscribing the square?

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