Four particles under Gravitation

Classical Mechanics Level 3

Four particles, each of mass $m$ , are moving along a circle of radius $r$ under the influence of mutual gravitational attraction. The speed of each particle will be $\text{\_\_\_\_\_\_\_\_\_\_}.$

\sqrt{ \frac {GM}r \cdot \frac{2\sqrt2+1}4 }

\sqrt{ \frac {GM}r \cdot{(2\sqrt2+1)} }

\sqrt{ \frac {GM}r}

\sqrt{ \frac {2\sqrt2 GM}r}

1 solution

Swapnil Das
Jan 16, 2017

Concept : Gravitational force from the other three particles provides centripetal force to the fourth particle.

Adding the Gravitational force vectors and equating with centripetal force, we get:

$\displaystyle \frac { G{ m }^{ 2 }\left( 2\sqrt { 2 } +1 \right) }{ 4{ r }^{ 2 } } =\frac { m{ v }^{ 2 } }{ r }$

$\displaystyle v= \sqrt{ \frac {GM}r \cdot \frac{2\sqrt2+1}4 }$

Four particles under Gravitation

1 solution

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