A) \[4\times 11.2km/s\]
B) \[\sqrt{15}\times 11.2km/s\]
C) zero
D) \[3\times 11.2\text{ }km/s\]
Correct Answer: B
Solution :
For a given gravitation field and a given position, the escape velocity is the minimum speed on object without propulsion, at that position, needs to have to move away indefinitely from the source field. From law of conservation of energy \[\frac{1}{2}m{{u}^{2}}=\frac{1}{2}mv_{e}^{2}+\frac{1}{2}m{{v}^{2}}\] \[\Rightarrow \] \[{{v}^{2}}={{u}^{2}}-v_{e}^{2}\] \[{{v}^{2}}={{(4{{v}_{e}})}^{2}}-v_{e}^{2}=12v_{e}^{2}\] \[\Rightarrow \] \[v=\sqrt{15}{{V}_{e}}\] Given, \[{{v}_{e}}=11.2km/s\] \[\therefore \] \[v=\sqrt{15}\times 11.2\,km/s\]
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