Two particles of equal mass m move in a circl | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The centripetal force required for circular motion is provided by the gravitational force of attraction between the two particles.The distance bet

Vedclass · Accepted Answer

$v = \frac{1}{2}\sqrt {\frac{{Gm}}{R}} $

Vedclass · Answer

(C) The centripetal force required for circular motion is provided by the gravitational force of attraction between the two particles.The distance between the two particles is $2R$.The gravitational force between them is $F_g = \frac{G \cdot m \cdot m}{(2R)^2} = \frac{Gm^2}{4R^2}$.The centripetal force required for a particle of mass $m$ moving with speed $v$ in a circle of radius $R$ is $F_c = \frac{mv^2}{R}$.Equating the two forces:$\frac{mv^2}{R} = \frac{Gm^2}{4R^2}$Solving for $v$:$v^2 = \frac{Gm^2}{4R^2} \cdot \frac{R}{m}$$v^2 = \frac{Gm}{4R}$$v = \sqrt{\frac{Gm}{4R}} = \frac{1}{2}\sqrt{\frac{Gm}{R}}$

Two particles of equal mass $m$ move in a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is

Similar Questions

The force of gravitation is:

Gravitational forces operate among which of the following?

The $SI$ unit of $\frac{G}{g}$ is $(g = \text{acceleration due to gravity}, G = \text{gravitational constant})$

State Newton's universal law of gravitation and represent it in a mathematical expression.

Explain the gravitational force exerted by an extended object.

Vedclass Test Series

Exam Paper Generator

Online Exam Module