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An asteroid of mass \(m\) is approaching earth, initially at a distance of \(10\ R_e\), with speed \(v_{i}\), it hits the earth with a speed \(v_{f}\)(\(R_{e}\) and \(M_{e}\) are radius and mass of earth), then ( )
A. \(v^{2}_{f}=v^{2}_{i}+\frac {2Gm}{M_{e}R}(1-\frac {1}{10})\)
B. \(v^{2}_{f}=v^{2}_{i}+\frac {2Gm_{e}}{R_{e}}(1+\frac {1}{10})\)
C. \(v^{2}_{f}=v^{2}_{i}+\frac {2Gm_{e}}{R_{e}}(1-\frac {1}{10})\)
D. \(v^{2}_{f}=v^{2}_{i}+\frac {2Gm}{R_{e}}(1-\frac {1}{10})\)

Answer

C
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Solution

Initial energy of the asteroid is given by
\(K_i+U_i\)
\(\frac {1}{2}mv^{2}_{i}-\frac {GM_{e}m}{(10R_{e})}\)
Final energy of asteroid is given as
\(K_f+U_f\)
\(\Rightarrow \frac {1}{2}mv^{2}_{f}-\frac {GM_{e}m}{R_{e}}\)
By law of conservation of energy,
Initial energy\(=\)final energy
\(\Rightarrow \frac {1}{2}mv^{2}_{i}-\frac {GM_{e}m}{(10R_{e})}=\frac {1}{2}mv^{2}_{f}-\frac {GM_{e}m}{R_{e}}\)
\(\Rightarrow v^{2}_{f}=v^{2}_{i}+\frac {2GM_{e}}{R_{e}}(1-\frac {1}{10})\)
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