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Question

(C) 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

$\frac{1}{8}$ times

Vedclass · Answer

(C) 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 the initial radius be $r_1$ and the initial time period be $T_1$. Let the new radius be $r_2 = \frac{1}{4} r_1$ and the new time period be $T_2$.Using the proportionality: $\frac{T_2^2}{T_1^2} = \left( \frac{r_2}{r_1} \right)^3$.Substituting the values: $\frac{T_2^2}{T_1^2} = \left( \frac{1}{4} \right)^3 = \frac{1}{64}$.Taking the square root on both sides: $\frac{T_2}{T_1} = \sqrt{\frac{1}{64}} = \frac{1}{8}$.Therefore,the new duration of the year will be $\frac{1}{8}$ times the original duration.

If the radius of the Earth's orbit is made $\frac{1}{4}$ of its original value,the duration of a year will become:

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