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(B) According to Kepler's third law,the square of the orbital period $(T)$ is proportional to the cube of the semi-major axis $(r)$,i.e.,$T^2 \propto

Vedclass · Accepted Answer

$129$

Vedclass · Answer

(B) According to Kepler's third law,the square of the orbital period $(T)$ is proportional to the cube of the semi-major axis $(r)$,i.e.,$T^2 \propto r^3$.Let $T_1 = 365$ days and $r_1$ be the present distance.Given $r_2 = \frac{1}{2} r_1$.Then,$\left( \frac{T_2}{T_1} ight)^2 = \left( \frac{r_2}{r_1} ight)^3 = \left( \frac{1}{2} ight)^3 = \frac{1}{8}$.Taking the square root on both sides,$\frac{T_2}{T_1} = \sqrt{\frac{1}{8}} = \frac{1}{2\sqrt{2}}$.Therefore,$T_2 = \frac{T_1}{2\sqrt{2}} = \frac{365}{2 	imes 1.414} \approx \frac{365}{2.828} \approx 129$ days.

If the distance between the earth and the sun becomes half its present value,the number of days in a year would have been

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