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(C) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is proportional to the cube of the semi-major axis $(r)$:

Vedclass · Accepted Answer

$\frac{1}{8}$

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(C) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is proportional to the cube of the semi-major axis $(r)$: $T^2 \propto r^3$ or $T \propto r^{3/2}$.Given that the new distance $r_2 = \frac{1}{4} r_1$.Substituting this into the ratio formula:$\frac{T_2}{T_1} = \left( \frac{r_2}{r_1} \right)^{3/2} = \left( \frac{1}{4} \right)^{3/2} = \frac{1}{8}$.Therefore,the length of the year (orbital period) is reduced to $\frac{1}{8}$ of its original value.

The Earth is revolving around the Sun. If the distance of the Earth from the Sun is reduced to $\frac{1}{4}$ of the present distance,then the length of the present day will be reduced by a factor of:

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