A) \[\frac{R}{1-{{k}^{2}}}\]
B) \[\frac{R}{{{k}^{2}}}\]
C) \[\frac{1-{{k}^{2}}}{R}\]
D) \[\frac{{{k}^{2}}}{R}\]
Correct Answer: A
Solution :
[a] : Let the projectile be fired vertically upward from the earths surface with velocity v and it reaches a maximum height h. By the law of conservation of mechanical energy, \[\frac{1}{2}m{{v}^{2}}-\frac{GMm}{R}=-\frac{GMm}{R+h}\] or\[\frac{1}{2}m{{v}^{2}}=GMm\left[ \frac{1}{R}-\frac{1}{R+h} \right]=\frac{GMmh}{(R)(R+h)}\] or\[\frac{1}{2}m{{v}^{2}}=\frac{mghR}{(R+h)}\left( \because g=\frac{GM}{{{R}^{2}}} \right)\] As per question\[v=k{{v}_{e}}=k\sqrt{2gR}\]and\[h=r-R\] \[\left( \because {{v}_{e}}=\sqrt{2gR} \right)\]where r is the distance from the centre of the earth. \[\therefore \]\[\frac{1}{2}m{{k}^{2}}2gR=\frac{mg(r-R)R}{r}\] or\[{{k}^{2}}=\frac{r-R}{r}or(1-{{k}^{2}})r=R\,or\,r=\frac{R}{1-{{k}^{2}}}\]
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