JEE Mains · Physics · STD 11 - 7. gravitation
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 \(R_{\mathrm{e}}=6370 \mathrm{~km}\) )
- A \(11\)
- B \(9\)
- C \(12\)
- D \(10\)
Answer & Solution
Correct Answer
(A) \(11\)
Step-by-step Solution
Detailed explanation
\(\mathrm{h}=318.5 \approx\left(\frac{\mathrm{R}_{\mathrm{e}}}{20}\right)\) \(\mathrm{T} \cdot \mathrm{E}_{\mathrm{i}}=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}}\)…
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