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Q.

If the gravitational force were to vary inversely as $m^{t h}$ power of the distance, then the time period of a planet in circular orbit of radius $r$ around the Sun will be proportional to :

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a

$r^{- 3 m / 2}$

b

$r^{3 m / 2}$

c

$r^{m + \frac{1}{2}}$

d

$r^{(m + 1) / 2}$

answer is D.

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Detailed Solution

$F \propto \frac{1}{r^{m}}; F = \frac{C}{r^{m}}$

This force will provide the required centripetal force

Therefore,

$m ω^{2} r = \frac{C}{r^{m}}; ω^{2} = \frac{C}{m r^{m + 1}}$

$T = \frac{2 π}{ω} \Rightarrow T \propto r^{(m + 1) / 2}$

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