Astronomers estimate that a 2.0-km-diameter asteroid collides with the Earth once every million years. The collision could pose a threat to life on Earth. (a) Assume a spherical asteroid has a mass of 3200 kg for each cubic meter of volume and moves toward the Earth at\(15\;{\rm{km/s}}\). How much destructive energy could be released when it embeds itself in the Earth? (b) For comparison, a nuclear bomb could release about\(4.0 \times {10^{16}}\;{\rm{J}}\). How many such bombs would have to explode simultaneously to release the destructive energy of the asteroid collision with the Earth?

The diameter of the asteroid is\(d = 2\;{\rm{km}} = 2000\;{\rm{m}}\).

The density of the asteroid is\(D = 3200\;{\rm{kg/}}{{\rm{m}}^3}\).

The speed of the asteroid is\(v = 15\;{\rm{km/s}} = 15000\;{\rm{m/s}}\).

The energy released by a nuclear bomb is \({E_{{\rm{nuclear}}}} = 4 \times {10^{16}}\;{\rm{J}}\).

The energy possessed by a body by virtue of its motion is called kinetic energy. Mathematically, it is given by:

\(KE = \frac{1}{2}m{v^2}\) … (i)

Here, m is the mass and v is the velocity of the body.

The mass of the asteroid is determined from the formula of density. The volume of the asteroid is given as:

\(V = \frac{4}{3}\pi {r^3}\)

Since the asteroid is 2 km wide, its radius will be\(r = \frac{d}{2} = 1000\;{\rm{m}}\).

\(V = \frac{4}{3} \times \frac{{22}}{7} \times {\left( {1000\;{\rm{m}}} \right)^3}\) … (ii)

Therefore, the mass of the asteroid is:

\(\begin{aligned}{c}{\rm{Mass}} = {\rm{Density}} \times {\rm{Volume}}\\m = \left( {3200\;{\rm{kg/}}{{\rm{m}}^3}} \right) \times \frac{4}{3} \times \frac{{22}}{7} \times {\left( {1000\;{\rm{m}}} \right)^3}\\ = 1.34 \times {10^{13}}\;{\rm{kg}}\end{aligned}\)

Suppose the kinetic energy of the asteroid is the destructive energy released when it embeds itself in the Earth.

From equation (i), the kinetic energy of the asteroid is given by:

\(\begin{aligned}{c}KE = \frac{1}{2}\left( {1.34 \times {{10}^{13}}\;{\rm{kg}}} \right){\left( {15000\;{\rm{m/s}}} \right)^2}\\ = 1.50 \times {10^{21}}\;{\rm{J}}\end{aligned}\)

The number of equivalent nuclear bombs that have to explode simultaneously to release the destructive energy of the asteroid’s collision with the Earth is determined as:

\(\begin{aligned}{c}N = 1.50 \times {10^{21}}\;{\rm{J}} \times \frac{{1\;{\rm{bomb}}}}{{4 \times {{10}^{16}}\;{\rm{J}}}}\\ = 37500\;{\rm{bombs}}\end{aligned}\)