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Kepler's third law states that the square of period of revolution (T) of a planet around the sun, is proportional to third power of average distance of r between sun and planet i.e $T^{2} = {kr}^{3}$ where k is constant .If the mass of sun and planet are M, m respectively then as per Newton's law of gravitation force of attraction, $F = \frac{GMm}{r^{2}}$ , the relation between $G & K$ is describe as

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$GK = 4 π^{2}$

$GMK = 4 π^{2}$

$K = G$

$K = \frac{1}{G}$

answer is B.

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Detailed Solution

$\begin{array}{l} ∵ {mRω}^{2} = \frac{GMm}{R^{2}} \Rightarrow ω^{2} = \frac{GM}{R^{3}} \Rightarrow \frac{4 π^{2}}{T^{2}} = \frac{GM}{R^{3}} \\ ∴ T^{2} = \frac{4 π^{2}}{GM} R^{3} = {kR}^{3} \Rightarrow GMk = 4 π^{2} \end{array}$

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