Question
An asteriod 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.
$$v_f^2 = v_i^2 + \frac{{2Gm}}{{{M_e}R}}\left( {1 - \frac{1}{{10}}} \right)$$
B.
$$v_f^2 = v_i^2 + \frac{{2G{M_e}}}{{{R_e}}}\left( {1 + \frac{1}{{10}}} \right)$$
C.
$$v_f^2 = v_i^2 + \frac{{2G{M_e}}}{{{R_e}}}\left( {1 - \frac{1}{{10}}} \right)$$
D.
$$v_f^2 = v_i^2 + \frac{{2Gm}}{{{R_e}}}\left( {1 - \frac{1}{{10}}} \right)$$
Answer :
$$v_f^2 = v_i^2 + \frac{{2G{M_e}}}{{{R_e}}}\left( {1 - \frac{1}{{10}}} \right)$$
Solution :
$$\eqalign{
& - \frac{{G{M_e}m}}{{10{R_e}}} + \frac{1}{2}mv_i^2 = - \frac{{G{M_e}m}}{{{R_e}}} + \frac{1}{2}mv_f^2 \cr
& \therefore v_f^2 = v_i^2 + \frac{{2G{M_e}}}{{{R_e}}}\left( {1 - \frac{1}{{10}}} \right). \cr} $$