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JEE MAIN 2025
02-04-2025 SHIFT-2
Question
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 $\_\_\_\_$ $\times 10^{10} J$.
(Mass of earth $=6 \times 10^{24} \mathrm{~kg}$, Radius of earth $=6.4 \times 10^{6} \mathrm{~m}$, Gravitational constant $=6.67 \times 10^{-11} N m^{2} k g^{-2}$ )
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Solution
$K E=\frac{1}{2} m v^{2}=\frac{1}{2} m \frac{G M_{e}}{r}=\frac{G M_{e} m}{2 r}=\frac{G M_{e} m}{2\left(R_{E}+h\right)}$
$=\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 6.4 \times 10^{6}}{2\left(6.4 \times 10^{6}+2.7 \times 10^{5}\right)}=3 \times 10^{10} J$
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