A satellite of mass 1000 kg is launched to revolve around the earth in an orbit at a height of 270 km from the earth's surface. Kinetic energy of the satellite in this orbit is ______ $\times 10^{10} \mathrm{~J}$.
(Mass of earth $=6 \times 10^{24} \mathrm{~kg}$, Radius of earth $=$ $6.4 \times 10^6 \mathrm{~m}$, Gravitational constant $=$
$\left.6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}\right)$

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{KE}=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \mathrm{~m} \frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{r}}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2 \mathrm{r}}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}ight)} \ & =\frac{6.67 	imes 10^{-11} 	imes 6 	imes 10^{24} 	imes 6.4 	imes 10^6}{2\left(6.4 	imes 10^6+2.7 	imes 10^5ight)}=3 	imes 10^{10} \mathrm{~J}\end{aligned}$

A satellite of mass 1000 kg is launched to revolve around the earth in an orbit at a height of 270 km from…

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