Question

Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be :

• A. $\frac{\sqrt{(1+2\sqrt{2})G}}{2}$
• B. $\sqrt{\frac{G}{2}(1+2\sqrt{2})}$
• C. $\sqrt{G(1+2\sqrt{2})}$
• D. $\sqrt{\frac{G}{2}(2\sqrt{2}-1)}$

Hint:

For symmetrical particle systems, use vector addition of forces towards the center to find the required centripetal force.

Answer 1

The Correct Option is A

Step 1: Net force on one particle $F_{net} = \frac{mv^2}{R}$. Force from adjacent particles $F_1 = \frac{Gm^2}{(\sqrt{2}R)^2}$ at $45^\circ$. Force from opposite particle $F_2 = \frac{Gm^2}{(2R)^2}$.
Step 2: $F_{net} = 2F_1 \cos 45^\circ + F_2 = 2(\frac{Gm^2}{2R^2})\frac{1}{\sqrt{2}} + \frac{Gm^2}{4R^2} = \frac{Gm^2}{R^2}(\frac{1}{\sqrt{2}} + \frac{1}{4})$.
Step 3: $\frac{mv^2}{R} = \frac{Gm^2}{R^2}(\frac{2\sqrt{2}+1}{4}) \Rightarrow v^2 = \frac{Gm}{R}(\frac{2\sqrt{2}+1}{4})$.
Step 4: With $m=1, R=1$, $v = \frac{\sqrt{G(1+2\sqrt{2})}}{2}$.