A) 56.63 \[km{{s}^{-1}}\]
B) 33 \[km{{s}^{-1}}\]
C) 39 \[km{{s}^{-1}}\]
D) 31.7 \[km{{s}^{-1}}\]
Correct Answer: D
Solution :
According to the principle of conservation of energy, initial kinetic energy + initial potential energy = final kinetic energy + final potential energy \[\Rightarrow \] \[\frac{1}{2}m{{v}^{2}}-\frac{GMm}{R}=\frac{1}{2}mv{{}^{2}}+0\] \[\Rightarrow \] \[\frac{1}{2}mv{{}^{2}}=\frac{1}{2}m{{v}^{2}}-\frac{GMm}{R}\] ... (i) Also consider, \[{{v}_{e}}\]= escape velocity \[\frac{1}{2}mv_{e}^{2}=\frac{GMm}{R}\] ?. (ii) 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, \[\left. \begin{align} & {{v}_{e}}=11.2\,km{{s}^{-1}} \\ & v=3\,\,{{v}_{e}} \\ \end{align} \right\}\] ?. (iv) From Eqs. (iii) and (iv), we get \[v{{}^{2}}={{(3{{v}_{e}})}^{2}}-v_{e}^{2}\] \[v{{}^{2}}=9v_{e}^{2}-v_{e}^{2}=8v_{e}^{2}=8\times {{(11.2)}^{2}}\] \[\Rightarrow \] \[v=\sqrt{8\times {{(11.2)}^{2}}}=\sqrt{8}\times 11.2\] \[v=2\times 1.414\times 11.2=31.68\,\,km{{s}^{-1}}\] \[\therefore \] Speed of the body far away from the Earth, \[v=31.68\,\,km{{s}^{-1}}=31.7\,km{{s}^{-1}}\]
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