Question

Three particles of equal mass M are situated at the vertices of an equilateral triangle of side L. What should be the velocity of each particle so that they move on a circular path without changing L

• A. $\sqrt{\frac{GM}{2L}}$
• B. $\sqrt{\frac{GM}{L}}$
• C. $\sqrt{\frac{2GM}{L}}$
• D. $\sqrt{\frac{GM}{3L}}$

Answer 1

Answer: (B)

$\sqrt{\frac{GM}{L}}$

The particle to move in a circular path without changing distance ${}^{\prime }{L}^{\prime }$ will happen if the resultant gravitational force on each particle provides the centripetal force necessary for the motion.

$\frac{{mv}^2}{r}=\sqrt{F^2+F^2+{2F}^2\cos 60°}\frac{{mv}^2}{r}=\sqrt{3}F$

Radius of the path is:

$r=\frac{\sqrt{3}}{2}L\times \frac{2}{3}=\frac{L}{\sqrt{3}}\therefore v=\sqrt{\frac{GM}{L}}$