A geostationary satellite is orbiting the ear | vedclass

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(A) According to Kepler's Third Law,the time period $T$ of a satellite is related to its orbital radius $r$ by $T \propto r^{3/2}$.Here,the orbital ra

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$6 \sqrt{2}$

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(A) According to Kepler's Third Law,the time period $T$ of a satellite is related to its orbital radius $r$ by $T \propto r^{3/2}$.Here,the orbital radius $r$ is the distance from the center of the earth,so $r = R_{earth} + h$.For the geostationary satellite: $r_1 = R + 5R = 6R$. Its time period $T_1 = 24$ hours.For the second satellite: $r_2 = R + 2R = 3R$.Using the ratio: $\frac{T_1}{T_2} = \left( \frac{r_1}{r_2} \right)^{3/2}$.Substituting the values: $\frac{24}{T_2} = \left( \frac{6R}{3R} \right)^{3/2} = (2)^{3/2} = 2\sqrt{2}$.Therefore,$T_2 = \frac{24}{2\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2}$ hours.

$A$ geostationary satellite is orbiting the earth at a height of $5R$ above the surface of the earth,where $R$ is the radius of the earth. The time period of another satellite in hours at a height of $2R$ from the surface of the earth is:

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