JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2024
- Hard
- MCQ
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 \(R_{\mathrm{e}}=6370 \mathrm{~km}\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(11\)
Step-by-step Solution
Detailed explanation
\(\mathrm{h}=318.5 \approx\left(\frac{\mathrm{R}_{\mathrm{e}}}{20}\right)\)
\(\mathrm{T} \cdot \mathrm{E}_{\mathrm{i}}=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}}\)
\(\mathrm{T} \cdot \mathrm{E}_{\mathrm{r}}=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2\left(\mathrm{R}_{\mathrm{e}}+\mathrm{h}\right)}=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2\left(\mathrm{R}_{\mathrm{e}}+\frac{\mathrm{R}_{\mathrm{e}}}{20}\right)}\)
\(\Rightarrow \mathrm{T} \cdot \mathrm{E}_{\mathrm{f}}=\frac{-10 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{R}_{\mathrm{e}}}\)
Change in total mechanical energy
\(=\mathrm{TE}_{\mathrm{f}}-\mathrm{TE}\)
\(=\frac{\mathrm{GM}_{\mathrm{i}} \mathrm{m}}{\mathrm{Re}}\left[1-\frac{10}{21}\right]=\frac{11 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{Re}}\)
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