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