A satellite is to be placed in an equatorial | vedclass

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(B) For a geostationary satellite, the orbital period $T$ is $24$ hours, which is $24 	imes 3600 = 86400 \,s$.The formula for the orbital radius $r$

Vedclass · Accepted Answer

$3.57 	imes 10^7 \,m$

Vedclass · Answer

(B) For a geostationary satellite, the orbital period $T$ is $24$ hours, which is $24 	imes 3600 = 86400 \,s$.The formula for the orbital radius $r$ is given by Kepler's Third Law: $T^2 = \frac{4\pi^2 r^3}{GM_{E}}$.Rearranging for $r$: $r = \left( \frac{GMT^2}{4\pi^2} ight)^{1/3}$.Substituting the values: $G = 6.67 	imes 10^{-11} \,Nm^2/kg^2$, $M_{E} = 6 	imes 10^{24} \,kg$, $T = 86400 \,s$.$r = \left( \frac{6.67 	imes 10^{-11} 	imes 6 	imes 10^{24} 	imes (86400)^2}{4 	imes (3.14)^2} ight)^{1/3} \approx 4.22 	imes 10^7 \,m$.The height $h$ of the satellite above the Earth's surface is $h = r - R_{E}$.$h = 4.22 	imes 10^7 \,m - 0.64 	imes 10^7 \,m = 3.58 	imes 10^7 \,m$.Rounding to the nearest given option, the height is approximately $3.57 	imes 10^7 \,m$.

$A$ satellite is to be placed in an equatorial geostationary orbit around the Earth for communication purposes. The height of such a satellite is $(M_{E} = 6 \times 10^{24} \,kg, R_{E} = 6400 \,km)$.

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