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JEE Mains · Physics · STD 11 - Gravitation
  • JEE Mains 2020
  • Hard
  • MCQ

The mass density of a spherical galaxy varies as \(\frac{ K }{ r }\) over a large distance \('r'\) from its centre. In that region, a small star is in a circular orbit of radius \(R\). Then the period of revolution, \(T\) depends on \(R\) as

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Answer & Solution

Correct Answer

(C) \(T ^{2} \propto R\)

Step-by-step Solution

Detailed explanation

\(dm =\rho dv\)

\(dm =\left(\frac{ k }{ r }\right)\left(4 \pi r ^{2} dr \right)\)

\(dm =4 \pi krdr\)

\(M =\int_{0}^{ R } d m =\int_{0}^{ R } 4 \pi krdr\)

\(M=\left.4 \pi k \frac{r^{2}}{2}\right|_{0} ^{R}\)

\(M =2 \pi k \left( R ^{2}-0\right)\)

\(M =2 \pi kR ^{2}\)

for circular motion gravitational force will provide required centripital force or

\(\frac{ GMm }{ R ^{2}}=\frac{ mv ^{2}}{ R }\)

\(\frac{G\left(2 \pi kR ^{2}\right) m }{ R ^{2}}=\frac{ mv ^{2}}{ R } \Rightarrow v =\sqrt{2 \pi GkR }\)

Time period \(\quad T =\frac{2 \pi R }{ v }\)

\(T =\frac{2 \pi R }{\sqrt{2 \pi GkR }} \propto \sqrt{ R }\)

or \(T^{2} \propto R\)

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