JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2019
- Hard
- MCQ
A test particle is moving in a circular orbit in the gravitational field produced by a mass density \(\rho \left( r \right) = \frac{K}{{{r^2}}}\). Identify the correct relation between the radius \(R\) of the particle’s orbit and its period \(T\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(T/R\) is a constant
Step-by-step Solution
Detailed explanation
For circular motion of particle:
\(\frac{{m{V^2}}}{r} = mE\)
\( = m\left( {\frac{{GM}}{{{r^2}}}} \right)\)
where \(M = \int\limits_0^r {\left( {4\pi {x^2}dx} \right)} \left( {\frac{k}{{{x^2}}}} \right)\)
\( = 4\pi kr\)
\( \Rightarrow \frac{{m{V^2}}}{r} = m\left( {\frac{{G\left( {4\pi k} \right)}}{r}} \right)\)
\( \Rightarrow\,V =constant\)
\(T = \frac{{2\pi R}}{V}\)
\( \Rightarrow \) \(\frac{T}{R}=constant\)
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