A) 11.2 km/s
B) 22.4 km/s
C) 44.8 km/s
D) None of these
Correct Answer: B
Solution :
Escape velocity of earth \[{{v}_{e}}=\sqrt{\frac{2GMe}{{{\operatorname{R}}_{e}}}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,{{v}_{e}}\propto \sqrt{\frac{Me}{{{\operatorname{R}}_{e}}}}\] Let new escape velocity is \[{{v}_{e}}'\] According to question, \[\frac{{{v}_{e}}}{{{v}_{e}}'}\,=\,\,\sqrt{\frac{Me}{\operatorname{Re}}\times \frac{0.5\,\operatorname{Re}}{2Me}}\] \[=\,\,\,\sqrt{\frac{1}{4}}=\frac{1}{2}\] \[\Rightarrow \,\,\,\,\,\operatorname{V}_{e}^{'}\,\,=\,\,2{{v}_{e}}=2\times 11.2\,\,=\,\,22.4\,\,km/s\]You need to login to perform this action.
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