Question

A sky lab of mass 200 kg has to be lifted from one circular orbit of radius 2R to another circular orbit of radius 3R. Calculate the minimum energy required if $R_E=6.37\times {10}^{6}m$ and acceleration due to gravity on earth's surface$=9.8m{s}^{-2}$

Answer 1

Total energy of sky laboratory $E=-\frac{GMm}{2r}$
$E_1=-\frac{GMm}{2\left(2R\right)}$ and $E_2=-\frac{GMm}{2\left(3R\right)}$
Therefore, the energy required,
$\Delta E=E_2-E_1=GMm(\frac{1}{4R}-\frac{1}{6R})=\frac{GMm}{12R}$
$=\frac{\left(R^{2}g\right)m}{12R}=\frac{mgR}{12}=1.04\times {10}^{10}J$