A) \[2.8km/s\]
B) \[15.6km/s\]
C) \[22.4km/s\]
D) \[44.8km/s\]
Correct Answer: C
Solution :
At a certain velocity of projection the body will go out of the gravitational field of the earth and will never return to the earth, this velocity is known as escape velocity. \[{{v}_{e}}=\sqrt{\frac{2G{{M}_{e}}}{{{R}_{e}}}}\] Given, \[{{M}_{e}}={{M}_{p}},\,\,{{R}_{p}}=\frac{{{R}_{e}}}{4}\] \[\therefore \] \[\frac{{{v}_{p}}}{{{v}_{e}}}=\sqrt{\frac{{{M}_{e}}}{{{M}_{e}}}\times \frac{{{R}_{e}}}{{{R}_{e}}/4}}=\sqrt{4}=2\] \[\Rightarrow \,\,\,\,\,{{v}_{p}}\,=2{{v}_{e}}\,=2\times 11.2\] \[=22.4km/s\]
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