The energy yield of a nuclear weapon is often defined in terms of the equivalent mass of a conventional explosive. 1 ton of a conventional explosive releases 4.2 GJ. A typical nuclear warhead releases 250,000 times more, so the yield is expressed as 250 kilotons. That is a staggering explosion, but the asteroid impact that wiped out the dinosaurs was significantly greater. Assume that the asteroid was a sphere 10 km in diameter, with a density of 2500 kg/m3 and moving at 30 km/s. Part A: What energy was released at impact, in joules? Assume all kinetic energy of the asteroid to be released. Part B: What energy was released at impact, in kilotons? Assume all kinetic energy of the asteroid to be released.

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skyluke89 skyluke89 · Answer 1 · 2019-04-19T08:45:54+02:00

A) [tex]5.88\cdot 10^{23}J[/tex]

The kinetic energy of an object is given by

[tex]K=\frac{1}{2}mv^2[/tex]

where

m is the mass of the object

v is the speed of the object

For the asteroid in the problem, we have

[tex]d=2500 kg/m^3[/tex] is the density

[tex]r=\frac{10 km}{2}=5 km=5000 m[/tex] is the radius

We can find its volume:

[tex]V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi (5000 m)^3=5.23\cdot 10^{11} m^3[/tex]

And so its mass

[tex]m=d V =(2500 kg/m^3)(5.23\cdot 10^{11} m^3)=1.31\cdot 10^{15} kg[/tex]

while its speed is

[tex]v=30 km/s=30,000 m/s[/tex]

So its kinetic energy is

[tex]K=\frac{1}{2}(1.31\cdot 10^{15} kg)(30,000 m/s)^2=5.88\cdot 10^{23}J[/tex]

so, this is the energy released in the impact of the asteroid.

B) [tex]1.4\cdot 10^{11} kton[/tex]

The conversion factor is

[tex]1 ton = 4.2 GJ = 4.2 \cdot 10^9 J\\1 kton = 4.2 \cdot 10^{12}J[/tex]

So the energy released in kilotons will be

[tex]E=\frac{5.88\cdot 10^{23} J}{4.2\cdot 10^{12} J}=1.4\cdot 10^{11} kton[/tex]