A) \[{{E}_{k}}\]
B) \[2{{E}_{k}}\]
C) \[\frac{1}{2}{{E}_{k}}\]
D) \[3{{E}_{k}}\]
Correct Answer: A
Solution :
If \[{{\upsilon }_{0}}\] be the orbital speed of the satellite of mass m, then\[{{E}_{k}}=\frac{1}{2}m\upsilon _{0}^{2}\] If \[{{\upsilon }_{c}}\] is the escape velocity, then we have \[{{\upsilon }_{e}}=\sqrt{2}{{\upsilon }_{0}}\] Thus kinetic energy required to escape \[E{{}_{K}}=\frac{1}{2}m\upsilon _{e}^{2}=\frac{1}{2}m{{(\sqrt{2}{{\upsilon }_{0}})}^{2}}\] \[=2\left( \frac{1}{2}m\upsilon _{0}^{2} \right)=2{{E}_{K}}\] Therefore, additional kinetic energy required \[=2EK-EK=EK\]
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