A) circle
B) hyperbola
C) falls on the earth
D) ellipse
Correct Answer: D
Solution :
\[{{v}_{0}}=R\sqrt{\frac{g}{(R+h)}}=\sqrt{\frac{g{{R}^{2}}}{(R+h)}}=\]orbit velocity \[{{v}_{e}}=\]Escape velocity \[=\sqrt{2gR}=\sqrt{\frac{2(R+gh)}{R}}=\sqrt{\frac{g{{R}^{2}}}{(R+h)}}\] The coefficients of \[\sqrt{\frac{g{{R}^{2}}}{(R+h)}}\] is \[1\] for orbital velocity and \[\sqrt{\frac{2(R+h)}{R}}\] for escape velocity. The coefficients \[\sqrt{\frac{2(R+h)}{R}}=\sqrt{22}\] for \[h=10\,\,R\]given velocity lies between \[{{v}_{0}}\] and\[{{v}_{e}}\]. Thus path is elliptical If \[v<{{v}_{0}}\] satellite falls on earth in parabolic path or move on elliptical path with the earth at near focus. \[v={{v}_{0}}\] satellite orbits in circular path \[{{v}_{e}}>v>{{v}_{0}}\]satellite orbits in elliptical path with the earth at far focus. \[v>{{v}_{e}},\] satellite escapes in hyperbolic path
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