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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 orbital radius $(r)$,i.

Vedclass · Accepted Answer

$2^{-3/2}$ lunar month

Vedclass · Answer

(C) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is proportional to the cube of the orbital radius $(r)$,i.e.,$T^2 \propto r^3$.Let $T_s$ and $r_s$ be the time period and orbital radius of the satellite,and $T_m$ and $r_m$ be the time period and orbital radius of the moon.Given: $r_s = \frac{1}{2} r_m$ and $T_m = 1$ lunar month.Using the ratio: $\frac{T_s}{T_m} = \left( \frac{r_s}{r_m} \right)^{3/2}$.Substituting the values: $\frac{T_s}{1} = \left( \frac{1/2 r_m}{r_m} \right)^{3/2} = \left( \frac{1}{2} \right)^{3/2} = 2^{-3/2}$.Therefore,the satellite completes one revolution in $2^{-3/2}$ lunar month.

$A$ satellite moves in a circle around the earth. The radius of this circle is equal to one half of the radius of the moon's orbit. The satellite completes one revolution in

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