Suppose gravitational force varies inversely as n^{th} power of distance. Then time period of a plan

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7.Gravitation

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Suppose the gravitational force varies inversely as the $n^{th}$ power of distance. Then the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to

A

${R^{\left( {\frac{{n + 1}}{2}} \right)}}$

B

${R^{\left( {\frac{{n - 1}}{2}} \right)}}$

C

$R^n$

D

${R^{\left( {\frac{{n - 2}}{2}} \right)}}$

Solution

(c) $F=K R^{-n}=M R \omega^{2} \Rightarrow \omega^{2}=K R^{-n+1}$

or $\omega=K \frac{R^{(n+1)}}{2}$

$\frac{2 \pi}{T} \propto R^{-\frac{n+1}{2}} \therefore T \propto R^{+\frac{n+1}{2}}$

Standard 11

Physics

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