JEE Mains · Physics · STD 11 - 7. gravitation
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)
- A \(\frac{\mathrm{m}}{20}\left(\mathrm{u}-\sqrt{\frac{2 \mathrm{GM}}{3 \mathrm{R}}}\right)^{2}\)
- B \(5 \mathrm{m}\left(\mathrm{u}^{2}-\frac{119}{200} \frac{\mathrm{GM}}{\mathrm{R}}\right)\)
- C \(\frac{3 m}{8}\left(u+\sqrt{\frac{5 G M}{6 R}}\right)^{2}\)
- D \(\frac{m}{20}\left(u^{2}+\frac{113}{200} \frac{G M}{R}\right)\)
Answer & Solution
Correct Answer
(B) \(5 \mathrm{m}\left(\mathrm{u}^{2}-\frac{119}{200} \frac{\mathrm{GM}}{\mathrm{R}}\right)\)
Step-by-step Solution
Detailed explanation
Applying energy conservation Applying energy conservation \({\mathrm{K}_{1}+\mathrm{U}_{\mathrm{i}}=\mathrm{K}_{\mathrm{f}}+\mathrm{U}_{\mathrm{f}}}\)…
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