## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) We can use conservation of energy to find the meteoroid's speed as it hits the earth's surface. Let $M_e$ be the earth's mass and let $M_m$ be the meteoroid's mass. Let $R_1$ be the distance at the moon's orbit and let $R_2$ be the distance from the center of the earth at the moment of impact. Note that $R_2$ is equal to the earth's radius. $K_2+U_2 = K_1+U_1$ $\frac{1}{2}M_mv_2^2 = \frac{1}{2}M_m~v_1^2-\frac{G~M_e~M_m}{R_1}-(-\frac{G~M_e~M_m}{R_2})$ $v_2^2 = v_1^2-\frac{2~G~M_e}{R_1}+\frac{2~G~M_e}{R_2}$ $v_2^2 = v_1^2+(2~G~M_e)(\frac{1}{R_2}-\frac{1}{R_1})$ $v_2 = \sqrt{v_1^2+(2~G~M_e)(\frac{1}{R_2}-\frac{1}{R_1})}$ $v_2 = \sqrt{(2.0\times 10^3~m/s)^2+(2)(6.67\times 10^{-11}~m^3/kg~s^2)(5.98\times 10^{24}~kg)(\frac{1}{6.38\times 10^6~m}-\frac{1}{3.84\times 10^8~m})}$ $v_2 = 1.13\times 10^4~m/s$ $v_2 = 11.3~km/s$ The meteoroid's speed at impact is 11.3 km/s. (b) We can use conservation of energy to find the meteoroid's speed at the closest point. Let $M_e$ be the earth's mass and let $M_m$ be the meteoroid's mass. Let $R_1$ be the distance at the moon's orbit and let $R_2$ be the distance from the center of the earth to the closest point. Note that $R_2$ is equal to the earth's radius + 5000 km which is 11,380 km. $K_2+U_2 = K_1+U_1$ $\frac{1}{2}M_mv_2^2 = \frac{1}{2}M_m~v_1^2-\frac{G~M_e~M_m}{R_1}-(-\frac{G~M_e~M_m}{R_2})$ $v_2^2 = v_1^2-\frac{2~G~M_e}{R_1}+\frac{2~G~M_e}{R_2}$ $v_2^2 = v_1^2+(2~G~M_e)(\frac{1}{R_2}-\frac{1}{R_1})$ $v_2 = \sqrt{v_1^2+(2~G~M_e)(\frac{1}{R_2}-\frac{1}{R_1})}$ $v_2 = \sqrt{(2.0\times 10^3~m/s)^2+(2)(6.67\times 10^{-11}~m^3/kg~s^2)(5.98\times 10^{24}~kg)(\frac{1}{1.138\times 10^7~m}-\frac{1}{3.84\times 10^8~m})}$ $v_2 = 8.49\times 10^3~m/s$ $v_2 = 8.49~km/s$ The meteoroid's speed at the closest point is 8.49 km/s.