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. $v_f^2=v_i^2+\frac{2Gm}{M_{e}R}(1-\frac{1}{10})$
• B. $v_f^2=v_i^2+\frac{2G{M}_e}{R_e}(1+\frac{1}{10})$
• C. $v_f^2=v_i^2+\frac{2G{m}_E}{R_E}\left(1-\frac{1}{10}\right)$
• D. $v_f^2=v_i^2+\frac{2Gm}{R_e}(1-\frac{1}{10})$

Answer 1

Answer: (C)

Applying law of conservation of energy for asteroid at a distance $10R_e$ and at earth's surface,
$K_i+U_i=K_f+U_f...(i)$
Now, $K_i=\frac{1}{2}m{v}_i^2$ and $U_i=\frac{G{M}_{e}m}{10R_e}$
$K_f=\frac{1}{2}m{v}_f^2$ and $U_f=\frac{G{M}_{e}m}{R_e}$
Substituting these values in Eq. (i), we get
$\frac{1}{2}m{v}_i^2-\frac{G{M}_{e}m}{10R_e}=\frac{1}{2}m{v}_f^2-\frac{G{M}_{e}m}{R_e}$
$\Rightarrow \frac{1}{2}m{v}_f^2=\frac{1}{2}m{v}_i^2+\frac{G{M}_{e}m}{R_e}-\frac{G{M}_{e}m}{10R_e}$
$\Rightarrow v_f^2=v_i^2+\frac{2G{M}_e}{R_e}-\frac{2G{M}_e}{10R_e}$
$\therefore v_f^2=v_i^2+\frac{2G{M}_e}{R_e}\left(1-\frac{1}{10}\right)$