A geostationary satellite is orbiting the ear | vedclass

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(B) According to Kepler's Third Law of planetary motion, the square of the orbital period $T$ is proportional to the cube of the semi-major axis $r$ o

Vedclass · Accepted Answer

$6\sqrt{2}$

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(B) According to Kepler's Third Law of planetary motion, the square of the orbital period $T$ is proportional to the cube of the semi-major axis $r$ of the orbit, i.e.,$T^2 \propto r^3$.For a satellite orbiting at a height $h$ above the Earth's surface, the orbital radius is $r = R + h$.For the geostationary satellite $(S_1)$: Height $h_1 = 4R$, so $r_1 = R + 4R = 5R$. The orbital period of a geostationary satellite is $T_1 = 24 	ext{ hours}$.For the second satellite $(S_2)$: Height $h_2 = 1.5R$, so $r_2 = R + 1.5R = 2.5R = 5R/2$. Let its period be $T_2 = T$.Using the ratio: $(T_2/T_1)^2 = (r_2/r_1)^3$.$(T/24)^2 = (2.5R / 5R)^3 = (1/2)^3 = 1/8$.$(T/24)^2 = 1/8 \implies T/24 = 1/\sqrt{8} = 1/(2\sqrt{2})$.$T = 24 / (2\sqrt{2}) = 12 / \sqrt{2} = 6 	imes 2 / \sqrt{2} = 6\sqrt{2} 	ext{ hours}$.

$A$ geostationary satellite is orbiting the earth at a height of $4R$ above the surface of the earth, where $R$ is the radius of the earth. Another satellite is orbiting the earth at a height $1.5R$ from the surface of the earth with periodic time $T$ in hours. The value of $T$ in hours is:

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