Question

Four identical particles of mass $M$ are located at the corners of a square of side ${}^{\prime }{a}^{\prime }$. What should be their speed if each of them revolves under the influence of the others' gravitational field in a circular orbit circumscribing the square?

• A. $1.21\sqrt{\frac{GM}{a}}$
• B. $1.41\sqrt{\frac{GM}{a}}$
• C. $1.16\sqrt{\frac{GM}{a}}$
• D. $1.35\sqrt{\frac{GM}{a}}$

Answer 1

The correct option is C $1.16\sqrt{\frac{GM}{a}}$

Net force on particle towards centre of circle is $F_C=\frac{G{M}^2}{2a^2}+\frac{G{M}^2}{a^2}\sqrt{2}$
$=\frac{G{M}^2}{a^2}(\frac{1}{2}+\sqrt{2})$
This force will act as centripetal force. Distance of particle from centre of circle is $\frac{a}{\sqrt{2}}$.
$r=\frac{a}{\sqrt{2}},F_C=\frac{M{v}^2}{r}$
$\frac{M{v}^2}{\frac{a}{\sqrt{2}}}=\frac{G{M}^2}{a^2}(\frac{1}{2}+\sqrt{2})$
$v^2=\frac{GM}{a}(\frac{1}{2\sqrt{2}}+1)$
$v^2=\frac{GM}{a}\left(1.35\right)$
$v=1.16\sqrt{\frac{GM}{a}}$.