If the earth is at one-fourth of its present | vedclass

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

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One-eighth the present year

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(B) According to Kepler's third law of planetary motion,the square of the time period $(T)$ of a planet is directly proportional to the cube of its semi-major axis $(r)$,i.e.,$T^2 \propto r^3$.Given that the new distance $r' = \frac{1}{4}r$.Using the ratio formula: $\left( \frac{T'}{T} \right)^2 = \left( \frac{r'}{r} \right)^3$.Substituting the value of $r'$: $\left( \frac{T'}{T} \right)^2 = \left( \frac{1}{4} \right)^3 = \frac{1}{64}$.Taking the square root on both sides: $\frac{T'}{T} = \sqrt{\frac{1}{64}} = \frac{1}{8}$.Therefore,the new duration of the year $T' = \frac{1}{8}T$.

If the earth is at one-fourth of its present distance from the sun,the duration of the year will be

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