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An astronaut takes a ball of mass $m$ from earth to space. He throws the ball into a circular orbit about earth at an altitude of $318.5 \mathrm{~km}$. From earth's surface to the orbit, the change in total mechanical energy of the ball is $x \frac{\mathrm{GM}_e \mathrm{~m}}{21 R_e}$. The value of $x$ is (take $R_{\mathrm{e}}=6370 \mathrm{~km}$ ):
$11$
$9$
$12$
$10$
Solution
$\mathrm{h}=318.5 \approx\left(\frac{\mathrm{R}_{\mathrm{e}}}{20}\right)$
$\mathrm{T} \cdot \mathrm{E}_{\mathrm{i}}=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}}$
$\mathrm{T} \cdot \mathrm{E}_{\mathrm{r}}=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2\left(\mathrm{R}_{\mathrm{e}}+\mathrm{h}\right)}=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2\left(\mathrm{R}_{\mathrm{e}}+\frac{\mathrm{R}_{\mathrm{e}}}{20}\right)}$
$\Rightarrow \mathrm{T} \cdot \mathrm{E}_{\mathrm{f}}=\frac{-10 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{R}_{\mathrm{e}}}$
Change in total mechanical energy
$=\mathrm{TE}_{\mathrm{f}}-\mathrm{TE}$
$=\frac{\mathrm{GM}_{\mathrm{i}} \mathrm{m}}{\mathrm{Re}}\left[1-\frac{10}{21}\right]=\frac{11 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{Re}}$
