A Short History of Nearly Everything - Some Corrections

I have just finished the book “A Short History of Nearly Everything” by Bill Bryson and I thought it was absolutely fantastic. There were a couple of mistakes that I noticed (and I am sure others that I did not - by ‘noticed’ I mean ‘didn’t believe the numbers and so checked them out’) that I thought I should correct. This post just provides a few scientific corrections, and is meant in no way as disrespect to what I believe to be a truly brilliant book.

On a diagram of the solar system to scale, with Earth reduced to about the diameter of a pea, Jupiter would be over a thousand feet away and Pluto would be a mile and a half distant (and about the size of a bacterium, so you wouldn’t be able to see it anyway).

Okay, assume that the radius of a pea is about \(4\) millimetres, or \(4 \times 10^{-3}\) metres. The mean radius of the Earth is \(6.371 \times 10^3\) metres, or the peas radius is \(\frac{1}{1592750000}\) of the Earth’s. We will use this fraction as the scale for the rest of the system. The distance between the Earth and Jupiter is \(5.808\) AU so on our scale that turns out to be \(1790\) feet. Pluto (which is \(32.15\) AU away) would be \(1.88\) miles away. So in this sense Bryson underestimated their distances (slightly). However, as for the size of Pluto, it has a mean radius of \(1.173 \times 10^6\) metres, so on our scale that would become \(7.4 \times 10^{-4}\) metres, i.e. a total diameter of 1.5 millimetres, very visible and about 75 times larger than a median bacterium!

Protons are so small that a little dib of ink like the dot on this i can hold something in the region of 500,000,000,000 of them, rather more than the number of seconds contained in half a million years.

I think the approximate size of a ‘dot’ is about \(0.5\) millimetres, so that means the area, in metres, of such a dot is: \(\pi\cdot 0.00025^2=1.963\times 10^{-7}\). The size of a proton is \(8.77 \times 10^{-16}\) metres, and so the number of protons that would fit in such a dot is \(223,887,732\). Or \(2233\) times less than the Bryson stated number. As for the number of seconds in half a million years: \(1 \mbox{ year } = 31556926 \mbox{ seconds}\) and \(500,000 \mbox{ years } = 15,800,000,000,000 \mbox{ seconds}\). Quite a lot more than what was stated, in fact, \(500,000,000,000\) seconds is only 15844 years.

Even so, it [the Voyager spacecraft] took them nine years to reach Uranus and a dozen to cross the orbit of Pluto.

Actually, Voyager 1 never got to Uranus nor cross the orbit of Pluto, only Voyager 2 did.

[Betelgeuse is] fifty thousand light years away.

Betelgeuse is around \(643 \pm 143\) light years away.

It is thought that our entire planet may contain, at any given moment, fewer than twenty francium atoms.

There is almost certainly a lot more than twenty francium atoms, according to Adloff and Kauffman, there is 550 grams in the Earth’s crust.

It took thousands of workers to clear 1.8 billion tonnes of debris from the 6.5 hectares of the World Trade Center site.

1.8 million tonnes.

It rises exponentially, so that a 7.3 quake is fifty ties more powerful than a 6.3 earthquake and 2,500 time more powerful than a 5.3 earthquake.

Exactly what the mistake is here depends slightly on what you interpret the word powerful means. The size of an earthquake increases by a factor of 10 as magnitude increases by one whole number. So a magnitude 7.3 earthquake is 10 times larger than a 6.3, and 100 times larger than 5.3. The amount of energy released increases by a factor of 31.6 \(\left(10^{1.5}\right)\). So a 7.3 earthquake releases roughly 32 times more energy than a magnitude 6.3; and about 1,000 times more energy than a 5.3.

It has been calculated that if you sunk a well to the center [of the Earth] and dropped a brick into it, it would take only 45 minutes for it to hit the bottom.

The terminal velocity of an object falling through the air is the maximum speed that the object will reach when the resistance of the medium through which it falls (in this case, air) prevents further acceleration. This velocity depends on various factors, including the mass and aerodynamic properties of the object (such as its shape and surface area), as well as the density of the air. For a brick, calculating its terminal velocity involves considering its drag coefficient, cross-sectional area perpendicular to the direction of motion, and mass. Using sensible estimates for this value we could calculate the terminal velocity of a house brick to be approximately 45 meters per second. Given that the Earth’s radius is approximately 6371 km, the brick would take about 39 hours to fall to the centre (given all the other difficulties with this thought experiment!).