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=(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=(1)/(2)(7\cdot10^8kg)(19000(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