Need homework help? Click here.

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.

Ask Question

Tag:MathPhyChemBioOther

Take Test

...