Question

The minimum energy required to launch a $m\,kg $ satellite from earths surface in a circular orbit at an altitude of $ 2R, 'R' $ is the radius of earth, will be:

• A. $ 3\,mgR $
• B. $ \frac{5}{6}\,mgR $
• C. $ 2\,mgR $
• D. $ \frac{1}{5}\,mgR $

Answer 1

Answer: (B)

The kinetic energy at altitude $2 R$ is $=\frac{ GMm }{6 R }$
The gravitational potential energy at altitude $2 R$ is $=\frac{- GMm }{3 R }$
Total energy $=$ Kinetic energy $+$ potential energy
$=\frac{ GMm }{6 R }+\frac{- GMm }{3 R }$
Potential energy at the surfce is $=\frac{ GMm }{ R }$
Therefore, required kinetic energy $=-\frac{ GMm }{6 R }+\frac{ GMm }{ R }=\frac{5 GMm }{6 R }$