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 $=6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}$ )

<p>The kinetic energy (KE) of a satellite revolving around the Earth can be expressed with the following formula:</p> <p>$ \mathrm{KE} = \frac{1}{2} m v^2 = \frac{1}{2} m \frac{GM_e}{r} = \frac{GM_e m}{2r} $</p> <p>For this satellite, the radius $ r $ is the sum of the Earth's radius $ R_E $ and the height $ h $ above the Earth's surface:</p> <p>$ r = R_E + h $</p> <p>Given:</p> <p><p>Mass of the satellite, $ m = 1000 \, \text{kg} $</p></p> <p><p>Mass of the Earth, $ M_e = 6 \times 10^{24} \, \text{kg} $</p></p> <p><p>Radius of the Earth, $ R_E = 6.4 \times 10^6 \, \text{m} $</p></p> <p><p>Height of orbit above the Earth's surface, $ h = 270 \, \text{km} = 2.7 \times 10^5 \, \text{m} $</p></p> <p><p>Gravitational constant, $ G = 6.67 \times 10^{-11} \, \text{Nm}^2 \, \text{kg}^{-2} $</p></p> <p>Substitute these values into the equation:</p> <p>$ \mathrm{KE} = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1000}{2(6.4 \times 10^6 + 2.7 \times 10^5)} $</p> <p>$ = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1000}{2 \times 6.67 \times 10^6} $</p> <p>$ = 3 \times 10^{10} \, \text{J} $</p> <p>Hence, the kinetic energy of the satellite in its orbit is $ 3 \times 10^{10} \, \text{J} $.</p>