NEET · Physics · STD 11 - 7. gravitation
A particle of mass \('\mathrm{m}'\) is projected with a velocity \(v=\mathrm{kV}_{\mathrm{e}}(\mathrm{k}\,<\,1)\) from the surface of the earth. \(\left(\mathrm{V}_{\mathrm{e}}=\right.\text { escape velocity) }\) The maximum height above the surface reached by the particle is :
- A \(\mathrm{R}\left(\frac{\mathrm{k}}{1-\mathrm{k}}\right)^{2}\)
- B \(\mathrm{R}\left(\frac{\mathrm{k}}{1+\mathrm{k}}\right)^{2}\)
- C \(\frac{\mathrm{R}^{2} \mathrm{k}}{1+\mathrm{k}}\)
- D \(\frac{\mathrm{Rk}^{2}}{1-\mathrm{k}^{2}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{Rk}^{2}}{1-\mathrm{k}^{2}}\)
Step-by-step Solution
Detailed explanation
\(-\frac{G M m}{R}+\frac{1}{2} m k^{2} v_{e}^{2}=-\frac{G M m}{r}\) \(-\frac{G M m}{R}+\frac{1}{2} m k^{2} \frac{2 G M}{R}=-\frac{G M m}{r}\)
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