A) 11.2 km/s
B) 112 km/s
C) 1.12 km/s
D) 1120 km/s
Correct Answer: B
Solution :
The escape velocity at earth is given by \[{{v}_{e}}=\sqrt{\frac{2\,G{{M}_{e}}}{{{R}_{e}}}}\] where G is gravitational constant, \[{{M}_{e}}\]and \[{{R}_{e}}\]are mass and radius of earth respectively. \[\therefore \] \[{{v}_{p}}=\sqrt{\frac{2\,G{{M}_{p}}}{{{R}_{p}}}}\] (planet) Given, \[{{M}_{p}}=1000\,{{M}_{e}},\,{{R}_{p}}=10\,{{R}_{e}}\] \[\therefore \] \[\frac{{{v}_{p}}}{{{v}_{e}}}=\sqrt{\frac{2G\times 1000\,{{M}_{e}}}{10\,{{R}_{e}}}\times \frac{{{R}_{e}}}{2G{{M}_{e}}}}\] \[=\sqrt{100}=10\] \[\Rightarrow \] \[{{v}_{p}}=10\,{{v}_{e}}\] \[\Rightarrow \] \[{{v}_{p}}=10\times 11.2\] \[\Rightarrow \] \[{{v}_{p}}=112\,\,km/s\]
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