A geostationary satellite is orbiting the ear | vedclass

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Question

(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.,$

Vedclass · Accepted Answer

$6\sqrt{2} \, hr$

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.,$T^2 \propto r^3$ or $T \propto r^{3/2}$.The orbital radius $r$ is the distance from the center of the Earth,given by $r = R + h$,where $h$ is the height above the surface.For the geostationary satellite,$h_1 = 5R$,so $r_1 = R + 5R = 6R$. The time period $T_1 = 24 \, hr$.For the second satellite,$h_2 = 2R$,so $r_2 = R + 2R = 3R$.Using the ratio formula: $\frac{T_2}{T_1} = \left( \frac{r_2}{r_1} \right)^{3/2}$.Substituting the values: $\frac{T_2}{24} = \left( \frac{3R}{6R} \right)^{3/2} = \left( \frac{1}{2} \right)^{3/2} = \frac{1}{2\sqrt{2}}$.Therefore,$T_2 = \frac{24}{2\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2} \, hr$.

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

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