Practicing Success

Target Exam

IIT-JEE

Subject

-- Physics - Section A

Chapter

Gravitation

Question:

Three identical particles, each of mass m, are located in space at the vertices of an equilateral triangle of side length a. They are revolving in a circular orbital under mutual gravitational attraction. What is the acceleration of the centre of mass of a system comprising of any two particles.

Options:

$\frac{\sqrt{3}}{2}\frac{Gm}{a^2}$

$\frac{3}{2}\frac{Gm}{a^2}$

$\frac{\sqrt{2}}{3}\frac{Gm}{a}$

$\frac{2}{3}\frac{Gm}{a}$

Correct Answer:

$\frac{\sqrt{3}}{2}\frac{Gm}{a^2}$

Explanation:

Consider our system to be made up of B and C. External force on this system is due to A. Net external force = 2F sin 60°.

$=\sqrt{3}F=\frac{\sqrt{3}Gm^2}{a^2}$

 

$∴a_{cm}=\left(\frac{\sqrt{3}Gm^2}{a^2}\right)×\frac{1}{2m}$

$=\frac{\sqrt{3}}{2}\frac{Gm}{a^2}$ (towards the centre of the triangle)