First slide

Gravitational force and laws

Question

Three point masses each of mass ‘m’ rotate in a circle of radius r with constant angular velocity $ω$ due to their mutual gravitational attraction. If at any instant, the masses are on the vertex of an equilateral triangle of side ‘a’, then the value of $ω$ is

Moderate

A

$\sqrt{\frac{G m}{a^{3}}}$
B

$\sqrt{\frac{3 G m}{a^{3}}}$
C

$\sqrt{\frac{G m}{3 a^{3}}}$
D

Zero

Solution

$\begin{array}{l} F = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} \cos θ} \Rightarrow I f F_{1} = F_{2} t h e n F_{n e t} = 2 F \cos \frac{θ}{2} \\ θ = 60 ∴ F_{n e t} = 2 F \cos \frac{60}{2} = 2 F \cos 30 = \frac{2 G m m}{a^{2}} (\frac{\sqrt{3}}{2}) = \frac{\sqrt{3} G m m}{a^{2}} \\ F_{n e t} p r o v i d e s c e n t r i p e t a l f o r c e \\ \frac{\sqrt{3} G M m}{a^{2}} = m r ω^{2} \\ \frac{\sqrt{3} G M}{a^{2}} = (\frac{a}{\sqrt{3}}) ω^{2} \\ ω^{2} = \frac{3 G M}{a^{3}} \\ ω = \sqrt{\frac{3 G M}{a^{3}}} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App