A) 22.4 \[km{{s}^{-1}}\]
B) 31.7 \[km{{s}^{-1}}\]
C) 33.6 \[km{{s}^{-1}}\]
D) none of these
Correct Answer: B
Solution :
Key Idea: Applying conservation of energy. By law of conservation of energy \[{{(U+K)}_{surface}}={{(U+K)}_{\infty }}\] \[\Rightarrow \] \[-\frac{GMm}{R}+\frac{1}{2}m\,{{(3{{v}_{e}})}^{2}}=0+\frac{1}{2}\,m{{v}^{2}}\] \[\Rightarrow \] \[-\frac{GM}{R}+\frac{9\,v_{e}^{2}}{2}=\frac{1}{2}\,{{v}^{2}}\] Since, \[v_{e}^{2}=\frac{2GM}{2}\] \[\therefore \] \[-\frac{v_{e}^{2}}{2}+\frac{9v_{e}^{2}}{2}=\frac{1}{2}\,{{v}^{2}}\] \[\Rightarrow \] \[{{v}^{2}}=8\,v_{e}^{2}\] \[\therefore \] \[v=2\sqrt{2}\,{{v}_{e}}\] \[=2\sqrt{2}\times 11.2\] \[=317\,km{{s}^{-1}}\]
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