The time period of a geostationary satellite | vedclass

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Question

(A) 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 $(r)$: $

Vedclass · Accepted Answer

$6 \sqrt{2} \; h$

Vedclass · Answer

(A) 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 $(r)$: $T^2 \propto r^3$ or $T \propto r^{3/2}$.The orbital radius $r$ is the distance from the center of the Earth,given by $r = R_{E} + h$,where $h$ is the height above the surface.For the first satellite: $r_1 = R_{E} + 6 R_{E} = 7 R_{E}$ and $T_1 = 24 \; h$.For the second satellite: $r_2 = R_{E} + 2.5 R_{E} = 3.5 R_{E}$.Using the ratio: $\frac{T_2}{T_1} = \left( \frac{r_2}{r_1} ight)^{3/2} = \left( \frac{3.5 R_{E}}{7 R_{E}} ight)^{3/2} = \left( \frac{1}{2} ight)^{3/2} = \frac{1}{2 \sqrt{2}}$.Therefore,$T_2 = T_1 	imes \frac{1}{2 \sqrt{2}} = \frac{24}{2 \sqrt{2}} = \frac{12}{\sqrt{2}} = 6 \sqrt{2} \; h$.

The time period of a geostationary satellite is $24 \; h$,at a height $6 R_{E}$ ($R_{E}$ is the radius of the Earth) from the surface of the Earth. The time period of another satellite whose height is $2.5 R_{E}$ from the surface will be:

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