# Luis Alvarez's K-T Impactor Calculation

I am trying to perform the calculation that Luis Alvarez used to establish the size of the K-T impactor. I used the following information:

1. Assume that the clay layer with iridium was uniformly distributed around Earth by the impact.
2. On average, the layer had a concentration of iridium of 10 parts per billion (ppb) by weight.
3. On average, the layer was 4 cm thick.
4. The density of the layer was 2.5 g/cm3.
5. Assume the meteor was spherical, with a density of 6.0 g/cm3, and an iridium content of 0.5 parts per million (ppm) by weight.
6. The radius of Earth is 6378 km.

What is the diameter of the meteorite? The answer isn't exactly 10 km, as stated. By how much would you have to change the assumed thickness of the iridium layer to arrive at an asteroid diameter of exactly 10 km?

My attempt:

$$m_{layer} = \pi(2.5 g/cm^3)((6378 \cdot 10^5)^4 - (6378 \cdot 10^5 - 4)^4) = 3.26 \cdot 10^{28} g$$

$$m_{iridiuminlayer} = (3.26 \cdot 10^{28})(10^{-8}) = 3.26\cdot 10^{20} g$$

Here is where I get stuck. It seems there is not sufficient information to calculate the radius of the meteor, as D = m/V, and we are given density and a mass ratio. I tried doing the second portion, in which you assume the diameter of the asteroid is 10 km, calculating the mass of the meteor, the iridium in the meteor, and the average iridium distribution over Earth's surface. I would be able to solve the rest of the question if I knew how to calculate the Volume/mass of the meteor. Please help!

• You assumed the volume/mass of the meteor to 6 in your question. The interesting thing is that we probably don't know the iridium ratio of the meteor. Although it could be estimated by considering the meteor to be an ordinary stone (or metallic) meteor, enriched by iridium to get a density of 6, it leads to yet another problem: having anything in the space with so enriched with such a rare element is unthinkably unrealistic. – peterh Feb 25 '20 at 23:37
• ((Side side note: it would be funny is once a new technology would be found, enabling interstellar travel, whose drive somehow needs iridium...)) – peterh Feb 25 '20 at 23:42
• That was the homework question's assumption, I believe I have it answered now though – jorkz Feb 25 '20 at 23:45

The volume of a sphere is $$\small\sf{\frac4 3\pi R^3}$$, where $$\small\sf{R}$$ is the radius.

The volume of Earth with the 4 cm deep iridium rich layer is,

$$\small\sf{V_{Ei} = \frac 4 3\pi (6.378\cdot10^6)^3}$$ $$\small\sf{m^3}$$ = $$\small\sf{1.086 \ 781 \cdot 10^{21}}$$ $$\small\sf{m^3}$$

The volume of the 4 cm deep iridium rich layer is,

$$\small\sf{V_i = \frac 4 3\pi (6.378\cdot10^6)^3} -\frac 4 3\pi (6.378\cdot10^6 - 0.04)^3$$ $$\small\sf{m^3}$$ = $$\small\sf{2.044 \cdot 10^{13}}$$ $$\small\sf{m^3}$$

Now, density is mass divided by volume, $$\small\sf{\rho = m/v}$$, thus the mass of the iridium rich layer is,

$$\small\sf{m_i = 2.5 \cdot 2.044 \cdot 10^{13} = 5.110 \cdot 10^{13}}$$ tonnes

The proportion of iridium in the layer is 10 ppb, therefore the mass of iridium within the layer is,

$$\small\sf{5.110 \cdot 10^{13} \cdot 10 \cdot 10^{-9}} = 511 \ 000$$ tonnes

Now, the proportion of iridium in the meteor was 0.5 ppm, therefore the mass of the meteor was,

$$\small\sf{511 \ 000 / (0.5 \cdot 10^{-6}) = 1.0220 \cdot 10^{12}}$$ tonnes

The density of the meteor was $$\small\sf{6.0 \ g/cm^3}$$, which is the same as $$\small\sf{6.0 \ t/m^3}$$, therefore, the volume of the meteor was, $$\small\sf{1.0220 \cdot 10^{12}/6.0} = 1.703 \ 333 \cdot 10^{11} m^3$$

Using this and the equation for the volume of a sphere, the radius of the meteor was,

$$\small\sf{\sqrt[3](\frac 3{4\pi} \cdot 1.703 \ 333 \cdot 10^{11})} = 3438.774 \ m$$

and the diameter of the meteor was 6877.5 m or 6.878 km

Not the 10 km you wanted.

To get the thickness you want for a 10 km diameter meteor, do the calculation in reverse.

For diameter of 10 000 m, the volume of a spherical meteor would have been,

$$\small\sf{\frac 4 3 \cdot \pi \cdot 5000^3} = 523.598 \ 776 \cdot 10^9 \ m^3$$

With a density of $$\small\sf{6 \ t/m^3}$$, the mass of meteor would have been,

$$\small\sf{6(523.598 \ 776 \cdot 10^9)} = 3.141 \ 592 \ 6 \cdot 10^{12} t$$

with a metal grade of 0.5 ppm of iridium, the mass of iridium in the meteor would have been,

$$\small\sf{3.141 \ 592 \ 6 \cdot 10^{12} \cdot 0.5 \cdot 10^{-6} = 1 \ 570 \ 796 \ t}$$

The metal grade of iridium in the layer on the Earth is 10 ppb, therefore the mass of the iridium layer is,

$$\small\sf{(1 \ 570 \ 796)/ (10 \cdot 10^{-9}) = 1.570 \ 796\ 327 \cdot 10 ^{14} \ t}$$

With a density of $$\small\sf{2.5\ t/m^3}$$, the volume of the layer is, $$\small\sf{(1.570 \ 796\ 327 \cdot 10 ^{14})/2.5 = 6.283 \ 185 \cdot 10^{13} \ m^3}$$

From the previous calculation, the volume of layer is,

$$\small\sf{6.283 \ 185 \cdot 10^{13} = \frac 4 3\pi (6.378\cdot10^6)^3} -\frac 4 3\pi (6.378\cdot10^6 - x)^3$$ $$\small\sf{m^3}$$ = $$\small\sf{2.044 \cdot 10^{13}}$$ $$\small\sf{m^3}$$

Solving for $$\small\sf{x}$$, which is the thickness of the layer, gives $$\small\sf{x} = 0.123\ m\ or\ 12.3\ cm$$