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(D) According to Kepler's third law of planetary motion,the square of the orbital period $T$ is directly proportional to the cube of the semi-major ax

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$3\sqrt{3} 	ext{ years}$

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(D) According to Kepler's third law of planetary motion,the square of the orbital period $T$ is directly proportional to the cube of the semi-major axis $R$ of its orbit: $T^2 \propto R^3$.Given the initial distance is $R$ and the initial period $T = 1 	ext{ year}$.When the new distance $R' = 3R$,the new period $T'$ is given by the ratio:$\frac{T'}{T} = \left(\frac{R'}{R}ight)^{3/2}$.Substituting the values:$\frac{T'}{1} = \left(\frac{3R}{R}ight)^{3/2} = (3)^{3/2}$.$T' = 3^{1} \cdot 3^{1/2} = 3\sqrt{3} 	ext{ years}$.

The distance between the Sun and the Earth is $R$. What will be the duration of the year if the distance between the Sun and the Earth becomes $3R$?

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