Question

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. $v^2_f = v^2_i + \frac{2Gm}{M_e R}\bigg(1 - \frac{1}{10}\bigg)$
• B. $v^2_f = v^2_i + \frac{2GM_e}{R_e}\bigg(1 +\frac{1}{10}\bigg)$
• C. $v^2_f = v^2_i + \frac{2GM_e}{M_e R}\bigg(1 - \frac{1}{10}\bigg)$
• D. $v^2_f = v^2_i + \frac{2Gm}{R_e}\bigg(1 - \frac{1}{10}\bigg)$

Answer 1

The Correct Option is C

Applying law of conservation of energy for asteroid at a distance $10 R_e$ and at earth's surface, $\hspace20mm K_i + U_i = K_f + U_f \hspace20mm ...(i)$ Now, $\hspace20mm K_i = \frac{1}{2}mv^2_i $ and $U_i = \frac{GM_e m}{10 R_e}$ $\hspace10mm K_f = \frac{1}{2}mv^2_f $ and $U_f = \frac{GM_e m}{R_e}$ Substituting these values in E (i), we get $\hspace10mm \frac{1}{2}mv_i^2 -\frac{GM_e m}{10R_e} =\frac{1}{2}mv^2_f - \frac{GM_e m}{R_e}$ $\Rightarrow\hspace20mm \frac{1}{2}mv_f^2 =\frac{1}{2}mv^2_i +\frac{GM_e m}{R_e}-\frac{GM_e m}{10R_e}$ $\Rightarrow\hspace25mm v_f^2 =v^2_i +\frac{2GM_e}{R_e}-\frac{2GM_e}{10R_e}$ $\therefore\hspace25mm v_f^2 =v^2_i +\frac{2GM_e}{R_e}\bigg(1-\frac{1}{10}\bigg)$