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Q. The minimum energy required to launch a satellite of mass m from the surface of Earth of mass M and radius R in a circular orbit at an altitude 2R from Earth’s surface is

Solution:

$-\frac {GMm}{R}+K.E.=\frac{-GMm}{3R}+\frac {1}{2}m$ $\left(\frac{GM}{3R}\right)$
K.E = $\frac {GMm}{R}-\frac {GMm}{3R}+\frac{GMm}{6R}$
$\therefore K.E. = \frac {5GMm}{6R}$