Q. Suppose the gravitational force varies inversely as the $nth$ 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) $R^{\left( \frac{n - 2}{2} \right)}$
Explanation

(a) The necessary centripetal force required for a planet to move round the sun = gravitational force exerted on it
$ie,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{mv^{2}}{R} = \frac{GM_{e}m}{R^{n}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = \left( \frac{GM_{e}}{R^{n - 1}} \right)^{1/2}$
$Now,\ \ T = \frac{2\pi R}{v} = 2\pi R \times \left( \frac{R^{n - 1}}{GM_{e}} \right)^{1/2}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2\pi\left( \frac{{R^{2} \times R}^{n - 1}}{GM_{e}} \right)^{1/2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2\pi\left( \frac{R^{(n + 1)/2}}{{(GM_{e})}^{1/2}} \right)$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ T \propto R^{(n + 1)/2}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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