Respuesta :
Answer:
[tex]v_f = 11.35 km/s[/tex]
Explanation:
Given data:
Earth mass [tex]= 5.972 \times 10^{24} kg[/tex]
distance between earth and asteroid is [tex]5.0\times 10^7 km[/tex]
asteroid mass [tex]= 2 \times 10^{13} kg[/tex]
speed of asteroid = 2 km/s
potential energy is [tex]U = - \frac{GM_e m}{R}[/tex]
[tex]U =- \frac{ 6.67 \times 10^{-11} \times 5.972 \times 10^{24} 2 \times 10^{13}}{5.0 \times 10^{10}}[/tex]
[tex]U = -0.0159 \times 10^{19} J[/tex]
Kinetic energy [tex]= \frac{1}{2} mv^2 = \frac{1}{2} \times 2\times 10^{13} \times (2\times 10^3)^2 = 4 \times 10^[19} J[/tex]
Final potential energy of asteroid is
[tex]U_f = potential energy is U = - \frac{GM_e m}{R}[/tex]
[tex]=- \frac{ 6.67 \times 10^{-11} \times 5.972 \times 10^{24} 2 \times 10^{13}}{6.371 \times 10^6}[/tex]
[tex]U_f = -125\times 10^{19} J[/tex]
from conservation of energy principle
[tex]U + K = U_f + K_f[/tex]
solving for K_f
[tex]K_f = U +K - U_F[/tex]
[tex]k_f = Kinetic energy = \frac{1}{2} mv_f^2[/tex]
[tex]v_f = \sqrt{\frac{2(U +K - U_F)}{m}}[/tex]
[tex]v_f = \sqrt{\frac{2\times (-0.0159 + 4+125) \times 10^{19}}{2\times 10^{13}}[/tex]
[tex]v_f = 11.35 km/s[/tex]
