A) 11.2 km/s
B) 22.4 km/s
C) 15.00 km/s
D) 5.8 km/s
Correct Answer: B
Solution :
At a certain velocity of projection, the body will go out of the gravitational field to the earth and will never return, this initial velocity is called escape velocity \[{{\nu }_{e}}=\sqrt{\frac{2G{{M}_{e}}}{{{\operatorname{R}}_{e}}}}\] where G is gravitational constant, \[{{M}_{e}}\] is mass of earth and \[{{R}_{e}}\] is radius. For planet \[{{\operatorname{R}}_{p}} = 2{{\operatorname{R}}_{e}},\, {{d}_{p}} = {{d}_{p}}\] Also since earth is assumed spherical in shape, its mass is given by \[M= \frac{4}{3}\pi {{R}^{3}}\times d\] \[\therefore \,\,\,\,\,\,\,{{\nu }_{e}}=\sqrt{\frac{2G}{{{\operatorname{R}}_{e}}}\times \frac{4}{3}\pi {{R}^{3}}d}\] \[{{\nu }_{p}}=\sqrt{\frac{2G}{2{{\operatorname{R}}_{e}}}\times \frac{4}{3}\pi {{(2{{R}_{e}})}^{3}}d}\] [b] Dividing Eq. [a] by Eq. [b], we get \[\frac{{{\nu }_{e}}}{{{\nu }_{p}}}=\frac{1}{2}\] \[\Rightarrow \,\,\,{{\nu }_{p}} =2{{\nu }_{e}},=2\times 11.2= 22.4 km/s\]
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