A) 2.8km/s
B) 15.6 km/s
C) 22.4 km/s
D) 44.8 km/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 of the earth, this velocity is known as escape velocity \[{{v}_{e}}=\sqrt{\frac{2G{{M}_{e}}}{{{R}_{e}}}}\] Given, \[{{M}_{e}}={{M}_{p}},\,{{R}_{p}}=\frac{{{R}_{e}}}{4}\] \[\therefore \] \[\frac{{{v}_{p}}}{{{v}_{e}}}=\sqrt{\frac{{{M}_{e}}}{{{M}_{e}}}\times \frac{{{R}_{e}}}{{{R}_{e/4}}}}=\sqrt{4}=2\] \[\Rightarrow \] \[{{v}_{p}}=2{{v}_{e}}=2\times 11.2\] \[=22.4\,\,km/s\]You need to login to perform this action.
You will be redirected in
3 sec