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

Vedclass · Accepted Answer

$2.8$

Vedclass · Answer

(B) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is directly proportional to the cube of the semi-major axis $(R)$ of the orbit: $T^2 \propto R^3$.Let $T_1$ and $R_1$ be the time period and orbital radius of the Earth,and $T_2$ and $R_2$ be the time period and orbital radius of the planet.Given: $T_1 = 1 	ext{ year}$,$R_1 = R$,and $R_2 = 2R$.Using the ratio formula: $\frac{T_2}{T_1} = \left( \frac{R_2}{R_1} ight)^{3/2}$.Substituting the values: $T_2 = 1 	imes \left( \frac{2R}{R} ight)^{3/2} = 2^{3/2}$.Calculating the value: $2^{3/2} = \sqrt{2^3} = \sqrt{8} \approx 2.828 	ext{ years}$.Rounding to the nearest option,the time period is $2.8 	ext{ years}$.

$A$ planet has an orbital radius twice that of the Earth's orbital radius. The time period of the planet is .......... $years$.

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