A geostationary satellite is orbiting the ear | vedclass

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Question

(B) 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

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

Vedclass · Answer

(B) 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 $\frac{T_1^2}{r_1^3} = \frac{T_2^2}{r_2^3}$.For a geostationary satellite,the time period $T_1 = 24 	ext{ hours}$ and the orbital radius $r_1 = R + 6R = 7R$.For the second satellite,the height is $2.5R$,so the orbital radius $r_2 = R + 2.5R = 3.5R$.Substituting these values into the formula:$\frac{24^2}{(7R)^3} = \frac{T_2^2}{(3.5R)^3}$$T_2^2 = 24^2 	imes \left(\frac{3.5R}{7R}ight)^3$$T_2^2 = 576 	imes \left(\frac{1}{2}ight)^3 = 576 	imes \frac{1}{8} = 72$$T_2 = \sqrt{72} = \sqrt{36 	imes 2} = 6\sqrt{2} 	ext{ hours}$.

$A$ geostationary satellite is orbiting the earth at a height of $6R$ above the surface of the earth ($R$ 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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