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

Vedclass · Accepted Answer

$1.38^{3/2}$

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(B) According to Kepler's third law of planetary motion,the square of the time period of revolution $(T)$ is directly proportional to the cube of the semi-major axis $(a)$ of the orbit: $T^2 \propto a^3$.Given the ratio of distances (semi-major axes) of two planets from the sun is $\frac{a_1}{a_2} = 1.38$.We need to find the ratio of their periods of revolution $\frac{T_1}{T_2}$.From Kepler's third law: $\left(\frac{T_1}{T_2}\right)^2 = \left(\frac{a_1}{a_2}\right)^3$.Taking the square root on both sides: $\frac{T_1}{T_2} = \left(\frac{a_1}{a_2}\right)^{3/2}$.Substituting the given value: $\frac{T_1}{T_2} = (1.38)^{3/2}$.

The ratio of the distances of two planets from the sun is $1.38$. The ratio of their period of revolution around the sun is

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