A satellite of mass $1000\ \text{kg}$ is launched to revolve around the earth in an orbit at a height of $270\ \text{km}$ from the earth's surface. The kinetic energy of the satellite in this orbit is ______ $\times10^{10}\ \text{J}$.
(Mass of earth $= 6\times10^{24}\ \text{kg}$, radius of earth $= 6.4\times10^6\ \text{m}$, gravitational constant $= 6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}$)
Numerical value type. Enter your answer.
Answer: 3
For a circular orbit, $\dfrac{mv^2}{r} = \dfrac{GMm}{r^2}$, so $K = \tfrac12mv^2 = \dfrac{GMm}{2r}$.
$r = 6.4\times10^6 + 0.27\times10^6 = 6.67\times10^6\ \text{m}$:
$$K = \frac{6.67\times10^{-11}\times6\times10^{24}\times1000}{2\times6.67\times10^6} = 3\times10^{10}\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics