Question

An average-sized asteroid located 5.0 x 10^7 km from Earth with mass 2.0 x 10^13 kg is detected headed directly toward Earth with speed of 2.0 km/s.
1. What will its speed be just before it hits the surface of the Earth? Assume the thickness of the atmosphere to be zero. (Ignore the size of the asteroid.) [average radius of Earth = 6.3781 x 10^6 m, mass of Earth = 5.972 x 10^24 kg]

Answer 1

Answer:

$v_f = 11.35 km/s$

Explanation:

Given data:

Earth mass $= 5.972 \times 10^{24} kg$

distance between earth and asteroid is $5.0\times 10^7 km$

asteroid mass $= 2 \times 10^{13} kg$

speed of asteroid = 2 km/s

potential energy is $U = - \frac{GM_e m}{R}$

$U =- \frac{ 6.67 \times 10^{-11} \times 5.972 \times 10^{24} 2 \times 10^{13}}{5.0 \times 10^{10}}$

$U = -0.0159 \times 10^{19} J$

Kinetic energy $= \frac{1}{2} mv^2 = \frac{1}{2} \times 2\times 10^{13} \times (2\times 10^3)^2 = 4 \times 10^[19} J$

Final potential energy of asteroid is

$U_f = potential energy is U = - \frac{GM_e m}{R}$

$=- \frac{ 6.67 \times 10^{-11} \times 5.972 \times 10^{24} 2 \times 10^{13}}{6.371 \times 10^6}$

$U_f = -125\times 10^{19} J$

from conservation of energy principle

$U + K = U_f + K_f$

solving for K_f

$K_f = U +K - U_F$

$k_f = Kinetic energy = \frac{1}{2} mv_f^2$

$v_f = \sqrt{\frac{2(U +K - U_F)}{m}}$

$v_f = \sqrt{\frac{2\times (-0.0159 + 4+125) \times 10^{19}}{2\times 10^{13}}$

$v_f = 11.35 km/s$