Question

An asteroid known as 7482 (1994 PCI ) is expected to pass near the Earth this week. It is estimated to have a mass of 700 million kg and a speed of 19000 m/s . If it were to actually hit the Earth, how much kinetic energy would be released when it struck?

Answer 1

Take into account that the kinetic energy is given by the following formula:

$K=\frac{1}{2}mv^2$

where,

m: mass of the asteroid = 700*10^6 kg = 7*10^8 kg

v: speed of the asteroid = 19000 m/s

Replace the previous values of the parameters into the formula for K and simplify:

$\begin{gathered} K=\frac{1}{2}(7\cdot10^8kg)(19000\frac{m}{s})^2 \\ K\approx2.5\cdot10^{17}J \end{gathered}$

Hence, approximately 2.5*10^17 J of energy would be released when the asteroid impacts the Earth