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# Question and Answer

## QuestionPhysicsClass 12

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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