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?

• 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

Answer: (C)

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

Answer 2

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