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Variation of G Due to Height and Depth

A satellite i...

Question

A satellite is revolving around the earth in a circular orbit of radius $a$ with velocity $v_{0}$ . A particle is projected from the satellite in forward direction with relative velocity $V = [(\sqrt{5} / 4) - 1] v_{0}$ . During the subsequent motion of the particle, Match the following

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Solution

Angular momentum of particle= $m (v_{0} + v) a$ where $v_{0} = \sqrt{\frac{G M_{e}}{a}}$ .
Total energy of particle= $\frac{1}{2} m (v_{0} + v^{2}) - \frac{G M_{e} m}{a}$
$= \frac{5}{8} \frac{G M_{e} m}{a} - \frac{G M_{e} m}{a} = \frac{- 3}{8} \frac{G M_{e} m}{a}$
At any distance 'r', total energy= $\frac{1}{2} m u^{2} - \frac{G M_{e} m}{r}$
But angular momentum conservation gives,
$m u r = m \sqrt{\frac{5 G M_{e}}{4 a}} a$
$⟹ u = \sqrt{\frac{5}{4} \frac{G M_{e} a}{r^{2}}}$
Therefore total energy= $\frac{1}{2} m \frac{5}{4} \frac{G M_{e} a}{r^{2}} - \frac{G M_{e} m}{r}$ .
According to conservation of energy this is equal to the initial enegy.
Hence, $\frac{1}{2} m \frac{5}{4} \frac{G M_{e} a}{r^{2}} - \frac{G M_{e} m}{r}$ $= - \frac{3 G M_{e} m}{8 a}$
Soving this gives $r = a, \frac{5}{3} a$ .

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