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(B) According to Kepler's Third Law of Planetary Motion, the square of the orbital period $T$ is directly proportional to the cube of the semi-major a

Vedclass · Accepted Answer

$4$

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(B) According to Kepler's Third Law of Planetary Motion, the square of the orbital period $T$ is directly proportional to the cube of the semi-major axis (orbital radius) $r$ of the orbit.$T^2 \propto r^3$This implies $T \propto r^{3/2}$ or $r \propto T^{2/3}$.Given that the new orbital period $T' = 8T$.Substituting this into the proportionality:$r' / r = (T' / T)^{2/3}$$r' / r = (8T / T)^{2/3} = (8)^{2/3}$$r' / r = (2^3)^{2/3} = 2^2 = 4$.Therefore, the orbital radius becomes $4$ times the original radius.

If the orbital period of a satellite becomes $8$ times, then by how many times does its orbital radius increase (in $\times$)?

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