A remote-sensing satellite of Earth revolves | vedclass

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(B) The orbital speed of a satellite is given by the formula $v_0 = \sqrt{\frac{GM}{R+h}}$.Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.Substituting

Vedclass · Accepted Answer

$7.76$

Vedclass · Answer

(B) The orbital speed of a satellite is given by the formula $v_0 = \sqrt{\frac{GM}{R+h}}$.Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.Substituting this into the orbital speed formula,we get $v_0 = \sqrt{\frac{gR^2}{R+h}} = R \sqrt{\frac{g}{R+h}}$.Given values: $R = 6.38 	imes 10^6 \, m$,$h = 0.25 	imes 10^6 \, m$,and $g = 9.8 \, m s^{-2}$.Total orbital radius $r = R + h = (6.38 + 0.25) 	imes 10^6 \, m = 6.63 	imes 10^6 \, m$.Now,$v_0 = \sqrt{\frac{9.8 	imes (6.38 	imes 10^6)^2}{6.63 	imes 10^6}} = \sqrt{\frac{9.8 	imes 40.7044 	imes 10^{12}}{6.63 	imes 10^6}} = \sqrt{60.16 	imes 10^6} \approx 7.756 	imes 10^3 \, m s^{-1}$.Thus,$v_0 \approx 7.76 \, km s^{-1}$.

$A$ remote-sensing satellite of Earth revolves in a circular orbit at a height of $0.25 \times 10^6 \, m$ above the surface of Earth. If Earth's radius is $6.38 \times 10^6 \, m$ and $g = 9.8 \, m s^{-2}$,then the orbital speed of the satellite is ...... $km s^{-1}$.

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