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A satellite of mass $\mathrm{m}$ is launched vertically upwards with an initial speed $u$ from the surface of the earth. After it reaches height $\mathrm{R}$ ($R =$ radius of the earth), it ejects a rocket of mass $\frac{\mathrm{m}}{10}$ so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is
($G$ is the gravitational constant: $\mathrm{M}$ is the mass of the earth)
$\frac{\mathrm{m}}{20}\left(\mathrm{u}-\sqrt{\frac{2 \mathrm{GM}}{3 \mathrm{R}}}\right)^{2}$
$5 \mathrm{m}\left(\mathrm{u}^{2}-\frac{119}{200} \frac{\mathrm{GM}}{\mathrm{R}}\right)$
$\frac{3 m}{8}\left(u+\sqrt{\frac{5 G M}{6 R}}\right)^{2}$
$\frac{m}{20}\left(u^{2}+\frac{113}{200} \frac{G M}{R}\right)$
Solution
Applying energy conservation
Applying energy conservation
${\mathrm{K}_{1}+\mathrm{U}_{\mathrm{i}}=\mathrm{K}_{\mathrm{f}}+\mathrm{U}_{\mathrm{f}}}$
${\frac{1}{2} \mathrm{mu}^{2}+\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right)=\frac{1}{2} \mathrm{mv}^{2}-\frac{\mathrm{GMm}}{2 \mathrm{R}}}$
$\mathrm{v}=\sqrt{\mathrm{u}^{2}-\frac{\mathrm{GM}}{\mathrm{R}}}$
By momentum conservation, we have
$\frac{\mathrm{m}}{10} \mathrm{v}_{\mathrm{T}}=\frac{9 \mathrm{m}}{10} \sqrt{\frac{\mathrm{GM}}{2 \mathrm{R}}}$
and $\frac{\mathrm{m}}{10} \mathrm{v}_{\mathrm{r}}=\mathrm{mv}$
$\Rightarrow \frac{\mathrm{m}}{10} \mathrm{v}_{\mathrm{r}}=\mathrm{m} \sqrt{\mathrm{u}^{2}-\frac{\mathrm{GM}}{\mathrm{R}}}$
Kinetic energy of rocket
${=\frac{1}{2} \mathrm{m}\left(\mathrm{v}_{\mathrm{T}}^{2}+\mathrm{v}_{\mathrm{r}}^{2}\right)}$
${=\frac{\mathrm{m}}{20}\left(81 \frac{\mathrm{GM}}{2 \mathrm{R}}+100 \mathrm{u}^{2}-100 \frac{\mathrm{GM}}{\mathrm{R}}\right)}$
${=\frac{\mathrm{m}}{20}\left(100 \mathrm{u}^{2}-\frac{119 \mathrm{GM}}{2 \mathrm{R}}\right)}$
${=5 \mathrm{m}\left(\mathrm{u}^{2}-\frac{119 \mathrm{GM}}{200 \mathrm{R}}\right)}$
