Question 14 Marks
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 $=$ $6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$ )
(Mass of earth $=6 \times 10^{24} \mathrm{~kg}$, Radius of earth $=$ $6.4 \times 10^{6} \mathrm{~m}$, Gravitational constant $=$ $6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$ )
Answer
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$\mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{~m} \frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{r}}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2 \mathrm{r}}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)}$
$=\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 6.4 \times 10^{6}}{2\left(6.4 \times 10^{6}+2.7 \times 10^{5}\right)}=3 \times 10^{10} \mathrm{~J}$
$\mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{~m} \frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{r}}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2 \mathrm{r}}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)}$
$=\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 6.4 \times 10^{6}}{2\left(6.4 \times 10^{6}+2.7 \times 10^{5}\right)}=3 \times 10^{10} \mathrm{~J}$
