Get 2 FREE Instant-Explanations on Filo with code FILOAPP

Home

Class 11

Physics

Gravitation

Gravitational force

The masses m each are placed at the corners of an equilateral

Question

Hard

Solving time: 4 mins

The masses $m$ each are placed at the corners of an equilateral triangle of side a and a mass $\frac{m}{3}$ is placed at the centre of the triangle. The 3 masses are in uniform circular motion with velocity $v$ on the circumcircle of the triangle, the centripetal force being provided by their mutual attraction. Then

a) $v = \frac{GM}{a} \frac{( 3 + 1 )}{3}$
(b) $v = \frac{GM}{a} \frac{( 3 - 1 )}{3}$
(c) The mass $\frac{m}{3}$ experiences no force
(d) If $\frac{m}{3}$ is absent, $v = \frac{GM}{a}$

Views: 5,980 students

Book:

IIT JEE Super Course in Physics Mechanics II (Pearson)

Found 6 tutors discussing this question

Charlotte Discussed

The masses

m

each are placed at the corners of an equilateral triangle of side a and a mass

\frac{m}{3}

is placed at the centre of the triangle. The 3 masses are in uniform circular motion with velocity

v

on the circumcircle of the triangle, the centripetal force being provided by their mutual attraction. Then

v = \frac{GM}{a} \frac{( 3 + 1 )}{3}

(b)

v = \frac{GM}{a} \frac{( 3 - 1 )}{3}

\frac{m}{3}

experiences no force
(d) If

\frac{m}{3}

is absent,

v = \frac{GM}{a}

9 mins ago

Discuss this question LIVE

9 mins ago

Text solutionVerified

F_{r} = F^{'} + 2 F cos 3 0^{\circ} = F^{'} + 3 F = \frac{Gm \frac{m}{3}}{a ^{2} /3} + \frac{3 Gmm}{a ^{2}} = \frac{Gm ^{2}}{a ^{2}} [1 + 3]

Since this provides the centripetal force:

\frac{Gm ^{2}}{a ^{2}} (1 + 3) = \frac{mv ^{2}}{( \frac{a}{3} )}

∴ v = \frac{Gm}{a} (\frac{1 + 3}{3}) \Rightarrow If \frac{m}{3} is absent, then F_{r} = 3 F \Rightarrow ∴ 3 \frac{Gmm}{a ^{2}} = \frac{mv ^{2}}{\frac{a}{3}} \Rightarrow v = \frac{Gm}{a}

Resultant force on

\frac{m}{3} \Rightarrow= 0

Was this solution helpful?

Report

One destination for complete JEE/NEET preparation

One destination to cover all your homework and assignment needs

Learn Practice Revision Succeed

Instant 1:1 help, 24x7

60, 000+ Expert tutors

Textbook solutions

Big idea maths, McGraw-Hill Education etc

Essay review

Get expert feedback on your essay

Schedule classes

High dosage tutoring from Dedicated 3 experts

Trusted by 4 million+ students

Stuck on the question or explanation?

Connect with our Physics tutors online and get step by step solution of this question.

231 students are taking LIVE classes

Question Text	The masses $m$ each are placed at the corners of an equilateral triangle of side a and a mass $\frac{m}{3}$ is placed at the centre of the triangle. The 3 masses are in uniform circular motion with velocity $v$ on the circumcircle of the triangle, the centripetal force being provided by their mutual attraction. Then a) $v = \frac{GM}{a} \frac{( 3 + 1 )}{3}$ (b) $v = \frac{GM}{a} \frac{( 3 - 1 )}{3}$ (c) The mass $\frac{m}{3}$ experiences no force (d) If $\frac{m}{3}$ is absent, $v = \frac{GM}{a}$
Topic	Gravitation
Subject	Physics
Class	Class 11
Answer Type	Text solution:1
Upvotes	80

Text solutionVerified

Practice questions from IIT JEE Super Course in Physics Mechanics II (Pearson)

Practice more questions from Gravitation

Practice questions on similar concepts asked by Filo students

Text solutionVerified

Practice questions from IIT JEE Super Course in Physics Mechanics II (Pearson)

Practice more questions from Gravitation

Practice questions on similar concepts asked by Filo students

Related books for questions on Gravitation