A) \[\text{56}.\text{63 km}{{\text{s}}^{\text{-1}}}\]
B) \[\text{33 km}{{\text{s}}^{\text{-1}}}\]
C) \[~\text{39 km}{{\text{s}}^{\text{-1}}}\]
D) \[\text{31}.\text{7 km}{{\text{s}}^{\text{-1}}}\]
Correct Answer: D
Solution :
According to the principle of conservation of energy, initial kinetic energy + initial potential energy = final kinetic energy + final potential energy \[\Rightarrow \] \[\frac{1}{2}m{{v}^{2}}-\frac{GMm}{R}=\frac{1}{2}m{{v}^{'2}}+0\] \[\Rightarrow \] \[\frac{1}{2}m{{v}^{'2}}-=\frac{1}{2}m{{v}^{2}}-\frac{GMm}{R}\] ? (i) Also consider, \[{{v}_{e}}=\]escape velocity \[\frac{1}{2}mv_{e}^{2}=\frac{GMm}{R}\] ? (ii) \[\therefore \]From Eqs. (i) and (ii), we get \[\frac{1}{2}mv{{'}^{2}}=\frac{1}{2}m{{v}^{2}}-\frac{1}{2}mv_{e}^{2}\] ? (iii) \[\Rightarrow \] \[{{v}^{'2}}={{v}^{2}}-v_{e}^{2}\] Now, \[\left. \begin{matrix} {{v}_{e}}=11.2\text{km}{{\text{s}}^{\text{-1}}} \\ v=3{{v}_{e}} \\ \end{matrix} \right\}\] ?(iv) From Eqs. (iii) and (iv), we get \[{{v}^{'2}}={{(3{{v}_{e}})}^{2}}-v_{e}^{2}\] \[{{v}^{'2}}=9v_{e}^{2}-v_{e}^{2}=8v_{e}^{2}=8\times {{(11.2)}^{2}}\] \[=8\times {{(11.2)}^{2}}\] \[\Rightarrow \] \[v'=\sqrt{8\times {{(11.2)}^{2}}}=\sqrt{8}\times 11.2\] \[v'=2\times 1.414\times 11.2=31.68\text{km}{{\text{s}}^{\text{-1}}}\] \[\therefore \]Speed of the body far away from the Earth, \[\text{v }\!\!'\!\!\text{ = 31}\text{.68 km}{{\text{s}}^{\text{-1}}}\text{=31}\text{.7 km}{{\text{s}}^{\text{-1}}}\]
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