Question

An average-sized asteroid located 2.9 × 107 km from Earth with mass 2.5 1013 kg is detected headed directly toward Earth with speed of 1.2 km/s, what will its speed be Cin km/s) just before it hits our atmosphere? (You may ignore the size of the asteroid. Assume the height of our atmosphere is 100 km.) km/s

Answer 1

m = 2.5 times 10^3 k g mu_c = 5.99 times 10^24 k g R_c = 6.371 times 10^6 M R = 2.3 times 10^7 km = 2.9 times 10^10 m v_c = 1.2 times 10^3 m/sec using conservation of energy U_i + K E_i = K E_i + U_f 1/2 mv_f^2 + (- G M_e m/R_L) = 1/2 mv_i^2 + (- G mu_c/R) 1/2 m v_f^2 = 1/2 m v_i^2 + G mu m/R_c - G mu_c m/R v_f^2 = v_i^2 + 2 G M_c [1/R_c - 1/R] v_f^2 = v_i^2 + 2 G mu_c/R R_l (R - R_c) Therefore V_f^2 = (1.2 times 10^3)^2 + 2 times 6.67 times 0^-11 times 5.99 times 10^24/6.371 times 10^6 times 10^10 times 2.9 (2.9 times 10^&10 - 6.371 times 10^6) = 1.44 times 10^6 + 4.3249 times 10^6 (2.9 times 10^4 - 6.3 H) times 10^- 3 = 1.44 times 10^6 + 4.3245 times 10^3 (28.9936 times 10^6) v_f^2 = (1.44 + 125.3945) times 10^6 v_f = squareroot 126.8345 times 10^6 v_f = 11.262 times n/sec v_f = 11.262 km/sec