A) 11.2 km/s
B) 44.8 km/s
C) 22.4 km/s
D) 5.6 km/s
Correct Answer: C
Solution :
Key Idea: The escape velocity from earth's surface is \[\sqrt{\frac{2G{{M}_{e}}}{{{R}_{e}}}}\] Escape velocity is given by \[{{v}_{es}}=\sqrt{\frac{2G{{M}_{e}}}{{{R}_{e}}}}\] From a planet, \[v_{es}^{'}=\sqrt{\frac{2G{{M}_{p}}}{{{R}_{p}}}}\] Therefore, \[\frac{v_{es}^{'}}{{{v}_{es}}}=\sqrt{\frac{2G{{M}_{p}}}{{{R}_{p}}}}\times \sqrt{\frac{{{R}_{e}}}{2G{{M}_{e}}}}\] It is given that, mass of planet = Mass of earth i.e., \[{{M}_{n}}={{M}_{e}}\] \[So,\frac{v_{es}^{'}}{{{v}_{es}}}=\sqrt{\frac{{{R}_{e}}}{{{R}_{p}}}}....(i)\] \[Given{{R}_{P}}=\frac{{{R}_{e}}}{4}\Rightarrow \,\frac{{{R}_{P}}}{{{R}_{e}}}=\frac{1}{4}\] \[and{{v}_{es}}=11.2\,km/s\] Substituting in Eq. (i) we have \[\frac{v_{es}^{'}}{11.2}=\sqrt{\frac{4}{1}}=2\] \[v_{es}^{'}=11.22=22.4\,km/s\]
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