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Expert-verified Found in: Page 170 ### Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922 # 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?

a) The destructive energy released when the asteroid embeds itself in the Earth is $$1.50 \times {10^{21}}\;{\rm{J}}$$.

b) The number of equivalent nuclear bombs is 37500.

See the step by step solution

## Identification of the given data

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 $${E_{{\rm{nuclear}}}} = 4 \times {10^{16}}\;{\rm{J}}$$.

## (a) Definition of kinetic energy

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.

## (a) Determination of the mass of the asteroid

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}

## (a) Determination of the destructive energy released

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}

## (b) Determination of the number of equivalent nuclear bombs

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} ### Want to see more solutions like these? 