A geostationary satellite orbits the Earth at | vedclass

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Question

(B) Mass of the Earth,$M = 6.0 	imes 10^{24} \; kg$.Radius of the Earth,$R = 6400 \; km = 6.4 	imes 10^{6} \; m$.Height of the geostationary satelli

Vedclass · Accepted Answer

$-9.4 	imes 10^{6} \; J/kg$

Vedclass · Answer

(B) Mass of the Earth,$M = 6.0 	imes 10^{24} \; kg$.Radius of the Earth,$R = 6400 \; km = 6.4 	imes 10^{6} \; m$.Height of the geostationary satellite,$h = 36000 \; km = 3.6 	imes 10^{7} \; m$.The distance of the satellite from the center of the Earth is $r = R + h = 6.4 	imes 10^{6} \; m + 36.0 	imes 10^{6} \; m = 42.4 	imes 10^{6} \; m = 4.24 	imes 10^{7} \; m$.Gravitational potential $V$ at a distance $r$ is given by $V = -\frac{GM}{r}$.Substituting the values: $V = -\frac{6.67 	imes 10^{-11} 	imes 6.0 	imes 10^{24}}{4.24 	imes 10^{7}}$.$V = -\frac{40.02 	imes 10^{13}}{4.24 	imes 10^{7}}$.$V \approx -9.44 	imes 10^{6} \; J/kg$.

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