The ratio of kinetic energy required to raise a satellite upto a height 'h' to the kinetic energy of the satellite at that height (R-radius of earth)

$(a)\;2h:R \quad (b)\; h:R\quad (c)\;R:2h \quad (d)\;R:h$

The energy needed to raise the satellite to a height 'h' is
$\Delta v=v_A -v_B$
$\qquad=\large\frac{-GMm}{R+h}-\bigg(\frac{-GMm}{R}\bigg)$
$\qquad=\large\frac{GMmh}{R(R+h)}$
$\qquad=\large\frac{GMmh}{R^2(1+\frac{h}{R})}\qquad \frac{GM}{R^2}$$=g \qquad=\large\frac{mgh}{(1+\frac{h}{R})}$$=E_1$
$E_2=$ energy of the satellite
$\qquad=\large\frac{1}{2}$$mv_0^2\qquad (v_0-orbital velocity ) r- distance of satellite from center of earth \qquad= \large\frac{1}{2}$$m \bigg(\large\frac{GM}{r}\bigg)$
$\qquad= \large\frac{1}{2}$$m \bigg(\large\frac{GM}{R+h}\bigg) \qquad= \large\frac{1}{2}$$ m \bigg(\large\frac{GM}{R^2}\bigg)\frac{R}{\bigg(1+\Large\frac{h}{R}\bigg)}$
$E_2=\large\frac{mgR}{2(1+\frac{h}{R})}$
$\therefore \large\frac{E_1}{E_2}$$=\large\frac{2h}{R}$
Hence a is the correct answer.

edited Feb 17, 2014 by meena.p