Four identical particles of mass $\mathrm{M}$ 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

  $1.35\sqrt{\frac{\mathrm{G}\mathrm{M}}{\mathrm{a}}}$
• B

  $1.21\sqrt{\frac{\mathrm{G}\mathrm{M}}{\mathrm{a}}}$
• C

  $1.41\sqrt{\frac{\mathrm{G}\mathrm{M}}{\mathrm{a}}}$
• Correct answer
  D

  $1.16\sqrt{\frac{\mathrm{G}\mathrm{M}}{\mathrm{a}}}$

Solution

Gravitational force on each mass 'M'
$=\sqrt{2}\frac{\mathrm{G}{\mathrm{M}}^2}{{\mathrm{a}}^2}+\frac{\mathrm{G}{\mathrm{M}}^2}{{\left(\sqrt{2}\mathrm{a}\right)}^2}=\left(\sqrt{2}+\frac{1}{2}\right)\frac{\mathrm{G}{\mathrm{M}}^2}{{\mathrm{a}}^2}$
This force will provide required centripetal force
$\left(\sqrt{2}+\frac{1}{2}\right)\frac{\mathrm{G}{\mathrm{M}}^2}{{\mathrm{a}}^2}=\frac{\mathrm{M}{\mathrm{v}}^2}{(\mathrm{a}/\sqrt{2})}$
$\mathrm{v}=\sqrt{\left(1+\frac{1}{2\sqrt{2}}\right)\frac{\mathrm{G}\mathrm{M}}{\mathrm{a}}}=1.16\sqrt{\frac{\mathrm{G}\mathrm{M}}{\mathrm{a}}}$

Hence, the correct answer is option: 'D'.