A satellite of mass 1000 kg is launched to revolve around the earth...

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)

Correct answer

Step-by-step solution

\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}\right)} \\ & =\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 6.4 \times 10^6}{2\left(6.4 \times 10^6+2.7 \times 10^5\right)}=3 \times 10^{10} \mathrm{~J}\end{aligned}

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