question_answer

An asteroid of mass m is approaching earth, initially at a distance of \[10\,{{R}_{e}}\] with speed \[{{v}_{1}}\].  It hits the earth with a speed \[{{v}_{f}}({{R}_{e}}\,and\,\,{{M}_{e}}\] are radius and mass of earth), then

A) \[\nu _{f}^{2}=\nu _{i}^{2}+\frac{2Gm}{{{M}_{e}}R}\,\left( 1-\frac{1}{10} \right)\]          

B) \[\nu _{f}^{2}=\nu _{i}^{2}+\frac{2G{{M}_{e}}}{{{\operatorname{R}}_{e}}}\,\left( 1+\frac{1}{10} \right)\]

C) \[\nu _{f}^{2}=\nu _{i}^{2}+\frac{2G{{M}_{e}}}{{{\operatorname{R}}_{e}}}\,\left( 1-\frac{1}{10} \right)\]

D) \[\nu _{f}^{2}=\nu _{i}^{2}+\frac{2Gm}{{{\operatorname{R}}_{e}}}\,\left( 1-\frac{1}{10} \right)\]

Correct Answer: C

Solution :

Applying law of conservation of energy for asteroid at a distance \[10\,\,{{\operatorname{R}}_{e}}\] and at earth?s surface, \[{{K}_{i}}\,\,+\,\,{{U}_{i}}\,\,=\,\,{{K}_{f}}+{{U}_{f}}\]                                            (i) Now \[{{K}_{f}}=\frac{1}{2}m\nu _{f}^{2}\,and\,{{U}_{f}}=-\frac{G{{M}_{e}}m}{{{\operatorname{R}}_{e}}}\] Substituting these values in Eq. (i), we get \[\frac{1}{2}m\nu _{i}^{2}-\frac{G{{M}_{e}}m}{10{{\operatorname{R}}_{e}}}\,=\,\,\frac{1}{2}m\nu _{f}^{2}-\frac{G{{M}_{e}}m}{{{\operatorname{R}}_{e}}}\] \[\Rightarrow \,\,\,\,\frac{1}{2}m\nu _{f}^{2}=\frac{1}{2}m\nu _{i}^{2}+\frac{G{{M}_{e}}m}{{{\operatorname{R}}_{e}}}\,\,-\,\,\frac{G{{M}_{e}}m}{10\,{{\operatorname{R}}_{e}}}\] \[\Rightarrow \,\,\,\,\nu _{f}^{2}=\nu _{i}^{2}+\frac{2G{{M}_{e}}}{{{\operatorname{R}}_{e}}}\,\,-\,\,\frac{2G{{M}_{e}}}{10\,{{\operatorname{R}}_{e}}}\] \[\therefore \,\,\,\,\,\,\nu _{f}^{2}=\,\,\nu _{i}^{2}+\frac{2G{{M}_{e}}}{{{\operatorname{R}}_{e}}}\left( 1-\frac{1}{10} \right)\]