A) 1.4147 Rg
B) 1.414 \[\sqrt{Rg}\]
C) 0.414 Rg
D) 0.414 \[\sqrt{Rg}\]
Correct Answer: D
Solution :
Let a spaceship is launched in a circular orbit of orbital velocity \[{{v}_{o}}.\]That spaceship should have escape velocity \[{{v}_{es}}\]to overcome the earths gravitational pull. Now suppose v is the additional velocity to be imparted to the spaceship. Then according to above statement \[{{v}_{0}}+v={{v}_{es}}\] \[\left[ \begin{align} & \because \,{{v}_{0}}=\sqrt{Rg} \\ & {{v}_{es}}=\sqrt{2}Rg \\ \end{align} \right]\] or \[v={{v}_{es}}-{{v}_{o}}\] \[v=\sqrt{2}{{v}_{o}}-{{v}_{o}}={{v}_{o}}(\sqrt{2}-1)={{v}_{o}}(1.414-1)\] \[=0.414\sqrt{Rg}\]
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