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The period of a planet around sun is 27 times that of earth. The ratio of radius of planet's orbit to the radius of the earth's orbit is

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answer is 9.

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Detailed Solution

$Given : T_{P} = 27 T_{e}$

According to Kapler's law,

$T^{2} \propto R^{3}$

$\Rightarrow \frac{T_{e}^{2}}{T_{p}^{2}} = \frac{R_{e}^{3}}{R_{p}^{3}}$

$\Rightarrow \frac{T_{e}^{2}}{{(27 T_{e})}^{2}} = \frac{R_{e}^{3}}{R_{p}^{3}}$

$\Rightarrow \frac{R_{p}}{R_{e}} = {(27)}^{\frac{2}{3}} = 3^{2} = 9$

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