An astronaut takes a ball of mass $m$ from earth to space. He throws the ball into a circular orbit about…

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 $x \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{R}_{\mathrm{e}}}$. The value of $x$ is (take $\left.\mathrm{R}_{\mathrm{e}}=6370 \mathrm{~km}\right)$ :
  1. 10
  2. 12
  3. 9
  4. 11

Solution

$\begin{aligned} & \mathrm{h}=318.5 \approx\left(\frac{\mathrm{R}_{\mathrm{e}}}{20}\right) \\ & \mathrm{T} \cdot \mathrm{E}_{\mathrm{i}}=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}} \\ & \mathrm{T} \cdot \mathrm{E}_{\mathrm{f}}=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2\left(\mathrm{R}_{\mathrm{e}}+\mathrm{h}\right)}=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2\left(\mathrm{R}_{\mathrm{e}}+\frac{\mathrm{R}_{\mathrm{e}}}{20}\right)} \\ & \Rightarrow \mathrm{T} \cdot \mathrm{E}_{\mathrm{f}}=\frac{-10 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{R}_{\mathrm{e}}}\end{aligned}$ Change in total mechanical energy $\begin{aligned} & =\mathrm{TE}_{\mathrm{f}}-\mathrm{TE}_{\mathrm{i}} \\ & =\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\operatorname{Re}}\left[1-\frac{10}{21}\right]=\frac{11 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \operatorname{Re}}\end{aligned}$

Asked in: JEE Main 2024 (09 Apr Shift 1)

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