Answer
$\Delta K=-5\times 10^{14}J$
Work Step by Step
Kinetic energy is defined as $$K=\frac{1}{2}mv^2$$ Therefore, the change in kinetic energy can be calculated as $$\Delta K=\frac{1}{2}(v_f^2-v_o^2)$$ The meteorite would be at rest once it collides, therefore, $v_f=0.0m/s$. Substitute this value and other known values of $v_o=15km/s=15000m/s$, $m=4\times 10^6kg$ yields a change in kinetic energy of $$\Delta K=\frac{1}{2}(4\times 10^6m/s)((0m/s)^2-(15000m/s)^2)$$ $$\Delta K=-5\times 10^{14}J$$
