I've heard it said (on the YouTube channel vsauce) that the earth is smoother than a billiards ball if it's scaled down.
Is this true?
Of course the earth is relatively smooth:
But a billiards ball feels as smooth as a mirror!
This strongly depends on definition of what smoothness is.
The Discover Magazine blog addressed this in 2008
OK, first, how smooth is a billiard ball? According to the World Pool-Billiard Association, a pool ball is 2.25 inches in diameter, and has a tolerance of +/- 0.005 inches. In other words, it must have no pits or bumps more than 0.005 inches in height. That’s pretty smooth. The ratio of the size of an allowable bump to the size of the ball is 0.005/2.25 = about 0.002.
The Earth has a diameter of about 12,735 kilometers (on average, see below for more on this). Using the smoothness ratio from above, the Earth would be an acceptable pool ball if it had no bumps (mountains) or pits (trenches) more than 12,735 km x 0.00222 = about 28 km in size.
The highest point on Earth is the top of Mt. Everest, at 8.85 km. The deepest point on Earth is the Marianas Trench, at about 11 km deep.
Hey, those are within the tolerances! So for once, an urban legend is correct. If you shrank the Earth down to the size of a billiard ball, it would be smoother.
I disagree with definition of smoothness used by Discovery Magazine. By that definition, medium sandpaper (grit particle size of 0.005 in) is also smooth, which doesn't quite go with my definition of smoothness. In fact I find claim that sandpaper is smooth to be ridiculous.
With mountains reaching in excess of 8,000m, scaled down that would be 0.0015 in. which means, that scaled down Earth's "smoothness" is equivalent to that of 320 grit sandpaper.
How does it compare with actual billiard ball, woliveirajr's answer is helpful:
What does ball surface look like:
Note, that variation is about 0.55μm, while 0.005 inches official tolerance for shape is 127μm. 0.55μm scaled up to Earth size would be less than 125 meters.
As for shape, which is really what the ±0.005 inches regulation is about, Earth is non-spherical, it's oblate spheroid with:
Just the non-spherical shape already disqualifies scaled down Earth as official billiard ball, allowable tolerance in diameter would be 28,326 m while difference between Earth's polar diameter and mean diameter is 28,513 m. Although it is quite close call.
I think vartec has the best answer so far. The quoted tolerance (specifications link) of 0.005 is for total size, not smoothness. The spec says 2.25+.005, not +/-, is that a typo or does it mean the balls must be at least 2.25 but not more than 2.255"? Most balls are actually manufactured to a higher tolerance, with good ones being under 0.001".
The pic from the site woliveirajr found is shows 1mm of an actual ball. That equates to about 220 km on the surface of the Earth, here's the pic compared to part of the grand canyon and everest:
And while the scaled-down grand canyon would be 8.2 micrometers deep, the variation in the marks is less than 1 micrometer (about 0.87).
So while I've seen billiard balls with scratches and chips that might be larger than mountains on Earth might be at that scale, that is not what you think of when you think of how smooth a billiard ball is.
Everest is different than the Grand Canyon though, Mount McKinley in Alaska is actually taller base-to-peak as Everest has a higher base. So while Everest would rise to a point further from the center of the billiard ball, Mount McKinley would be the highest bump at about 26 micrometers from the surrounding surface.
I don't compare Mauna Kea because I would argue that under sea level should not be taken into account. After all, looking at the Earth from space, you cannot see the Mariana Trench. You run into all sorts of problems thinking about a giant being trying to feel how smooth the Earth is, so I would just just use how it looks from space, water or not:
60 foot Ocean waves would be about 0.08 microns, but as that is far from the norm and waves would be so tightly packed that they would seem to be almost one surface, most of the planet would be much smoother than a billiard ball. Much of the rest of the world would be about as smooth as well, it would just be the large mountain ranges that would really be much rougher.
I'm sorry to rain on vartec's parade, but his answer is conceptually incorrect and falls into the fallacy of comparing apples and oranges and tries to appeal to familiarity with every day objects to make the case (incorrectly). The quoted article is correct (of course, requires one to make the assumption that the earth is round, which is not a terrible assumption for the purposes of the article) and below, I offer a reasoning of why vartec's answer is grossly misleading.
To understand why, one should understand dimensional comparisons. The billiards ball example compared the maximum bump to the radius (or diameter). Both these directions are comparable. As the quoted article shows, doing the same for the "bumps" on the Earth led to a smaller ratio of max bump size to radius and hence "smoother".
Now the part that's misleading (and upon which the rest of the argument is based) is that scaling this down to "normal size" makes it comparable to sand paper! How? How is this scaling done? To scale properly, one would have to take the dimensionless ratio of bump size to radius (see crude figure below) of the billiard ball (or the Earth) and then multiply it by the corresponding dimension of another object you wish to compare. This dimensionless ratio is the definition of "smoothness" (0.0022 in the image), not the actual number 0.005".
In the comparison with the sand paper, he "scales" Mt.Everest down to 0.0015 inches. How was this "rescaling" done? What is the base line that this is a "deviation" from? He doesn't tell you. And this is where the mistake lies. In the second half of my crude hand drawn image, I give an example. Let's assume that 0.0015 grit size sand paper is chosen. We now need to consider the dimension for which this grit can be considered a "bump". No, it's not the length, but the thickness of the backing paper that matters. Let's now assume a thickness of 1/32" (don't know the actual number, but somewhere between 1/32" and 1/16" is my guess). Calculate the dimensionless "smoothness" factor again – turns out it is 0.048, which is much greater than 0.0022. Well, duh! No wonder it is rough — it is about 20 times rougher, in fact.
If you've ever worked on auto detailing, you might have come across 2500 grit sand paper. These are very smooth to the touch (seriously, smooth as talc), and still they're about 5 times rougher than a billiards ball.
So in summary, the entire answer is based on false math and I'm concerned that it has as many as 50+ votes on a Skeptics site.
