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Imagine a light planet revolving around a very massive star in a circular orbit of radius r with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional $r^{\frac{- 5}{2}}$ , then the square of the time period will be proportional to

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r^2.5

r^3.5

r³

r²

answer is D.

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Detailed Solution

$\frac{{mv}^{2}}{r} = \frac{GMm}{r^{5 / 2}} [∵ Fα \frac{1}{r^{5 / 2}}] or {rω}^{2} = \frac{GM}{r^{5 / 2}} or r \frac{4 π^{2}}{T^{2}} = \frac{GM}{r^{5 / 2}} or T^{2} {αr}^{7 / 2} or T^{2} {αr}^{3 .5}$

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