A planet revolves around the sun whose mean d | vedclass

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(D) According to Kepler's third law of planetary motion, the square of the time period of revolution $(T)$ is proportional to the cube of the semi-maj

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$2$

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(D) According to Kepler's third law of planetary motion, the square of the time period of revolution $(T)$ is proportional to the cube of the semi-major axis $(r)$ of the orbit: $T^2 \propto r^3$.Given that the mean distance of the planet $(r_p)$ is $1.588$ times the mean distance of the earth $(r_e)$, we have $r_p = 1.588 	imes r_e$.The ratio of the time periods is given by: $\frac{T_p}{T_e} = \left( \frac{r_p}{r_e} ight)^{3/2}$.Substituting the given values: $\frac{T_p}{T_e} = (1.588)^{3/2}$.Since $1.588 \approx (2)^{2/3}$, we have $(1.588)^{3/2} \approx ((2)^{2/3})^{3/2} = 2^1 = 2$.Therefore, $T_p = 2 	imes T_e$. Since $T_e = 1 	ext{ year}$, the revolution time of the planet is $2 	ext{ years}$.

$A$ planet revolves around the sun whose mean distance is $1.588$ times the mean distance between the earth and the sun. The revolution time of the planet will be ........... $years$.

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