A geostationary satellite is orbiting the ear | vedclass

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(D) The orbital radius $r$ is the distance from the center of the earth,given by $r = R + h$.For the geostationary satellite,$h_1 = 6R$,so $r_1 = R +

Vedclass · Accepted Answer

$6\sqrt{2} 	ext{ hr}$

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(D) The orbital radius $r$ is the distance from the center of the earth,given by $r = R + h$.For the geostationary satellite,$h_1 = 6R$,so $r_1 = R + 6R = 7R$. The time period $T_1 = 24 	ext{ hr}$.For the second satellite,$h_2 = 2.5R$,so $r_2 = R + 2.5R = 3.5R$.According to Kepler's Third Law,$T^2 \propto r^3$,which implies $\frac{T_2}{T_1} = \left( \frac{r_2}{r_1} ight)^{3/2}$.Substituting the values: $\frac{T_2}{24} = \left( \frac{3.5R}{7R} ight)^{3/2} = \left( \frac{1}{2} ight)^{3/2}$.$T_2 = 24 	imes \frac{1}{2\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2} 	ext{ hr}$.

$A$ geostationary satellite is orbiting the earth at a height of $6R$ above the surface of the earth,where $R$ is the radius of the earth. The time period of another satellite at a height of $2.5R$ from the surface of the earth is

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