A geostationary satellite is orbiting around | vedclass

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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 $r$ $(T^2

Vedclass · Accepted Answer

$3$

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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 $r$ $(T^2 \propto r^3)$,which implies $T \propto r^{3/2}$.The orbital radius $r$ is the sum of the planet's radius $R$ and the height $h$ above the surface $(r = R + h)$.For the first satellite: $r_1 = R + 11R = 12R$ and $T_1 = 24\, 	ext{hours}$.For the second satellite: $r_2 = R + 2R = 3R$.Using the ratio: $\frac{T_1}{T_2} = \left(\frac{r_1}{r_2}ight)^{3/2}$.Substituting the values: $\frac{24}{T_2} = \left(\frac{12R}{3R}ight)^{3/2} = (4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$.Therefore,$T_2 = \frac{24}{8} = 3\, 	ext{hours}$.

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