Question

Three identical bodies (each mass $M$) are placed at vertices of an equilateral triangle of arm $L,$ keeping the triangle as such, by what angular speed the bodies should be rotated in their gravitational fields,so that the triangle moves along circumference of circular orbit.

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

Answer 1

The correct option is A $\sqrt{\frac{3GM}{L^3}}$

As shown in figure, $F_c=$ Centripetal force and $F_g=$ Gravitational force. We take upper body for calculation.

Let angular speed of bodies are $\omega$.

Given side of equilateral triangle is $L$ so radius of orbit is $\frac{L}{\sqrt{3}}$.

$\Rightarrow F_c=M\frac{L}{\sqrt{3}}{\omega }^2$

$\Rightarrow F_g=\frac{G{M}^2}{L^2}$

$\therefore$ Net gravitational force acting towards the center of the triangle $F_{Net}=2F_{g}cos\left({30}^o\right)$ $\Rightarrow F_{Net}=\frac{\sqrt{3}G{M}^2}{L^2}$,

$\Rightarrow F_{Net}=F_c$ $\Rightarrow \frac{ML{\omega }^2}{\sqrt{3}}=\frac{\sqrt{3}G{M}^2}{L^2}$

$\Rightarrow \omega =\sqrt{\frac{3GM}{L^3}}$