Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A test particle is moving in a circular orbit in a gravitational field produced by a mass density \rho (r)=\frac{K}{r^{2}} . Identity the correct relation between the radius R of the particles orbit and its period T :

  • Option 1)

    T/R\:\:is\:a \:constant

  • Option 2)

    T^{2}/R^{3}\:\:is\:a \:constant

  • Option 3)

    T/R^{2}\:\:is\:a \:constant

  • Option 4)

    TR\:\:is\:a \:constant

 

Answers (1)

best_answer

Given

 

\rho =\frac{K}{r^{2}}

E=\frac{G}{R^2}\int_{0}^{R}\frac{K}{r^{2}}.4\pi r^{2}dr=\frac{GK4\pi}{R}=\frac{C}{R}

Now 

\frac{C}{R}m=mw^{2}R=m\left ( \frac{2\pi}{T} \right )^{2}R

 

\Rightarrow\frac{T}{R}=constant


Option 1)

T/R\:\:is\:a \:constant

Option 2)

T^{2}/R^{3}\:\:is\:a \:constant

Option 3)

T/R^{2}\:\:is\:a \:constant

Option 4)

TR\:\:is\:a \:constant

Posted by

solutionqc

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

Parliamentary panel to review mushrooming of coaching centres, PM SHRI schools, HECI plans
anam-khan

JEE Main 2026 Registration LIVE: NTA session 1 application ends tomorrow; apply at jeemain.nta.nic.in
anam-khan

Kota: JEE aspirant dies after falling from 9th floor; police say likely suicide

Latest Question