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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

$2$

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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 $r$,i.e.,$T^2 \propto r^3$.For a geostationary satellite,$T_1 = 24\, h$ and $r_1 = 36000\, km$.For a satellite orbiting near the Earth's surface,$r_2 \approx R_{	ext{Earth}} = 6400\, km$.Using the ratio formula: $\frac{T_2}{T_1} = \left( \frac{r_2}{r_1} ight)^{3/2}$.Substituting the values: $T_2 = 24 	imes \left( \frac{6400}{36000} ight)^{3/2}$.$T_2 = 24 	imes \left( \frac{64}{360} ight)^{3/2} = 24 	imes \left( \frac{8}{18.97} ight)^3 \approx 24 	imes (0.177)^{3/2} \approx 24 	imes 0.075 \approx 1.8\, h$.Rounding to the nearest integer,the time period is approximately $2\, hours$.

$A$ geostationary satellite orbits around the earth in a circular orbit of radius $36000\, km$. Then,the time period of a satellite orbiting a few hundred kilometres above the earth's surface $(R_{\text{Earth}} = 6400\, km)$ will approximately be ....... $hours$.

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