I'll invite you to help me with this challenge, namely by providing the challenges. Enrico Fermi was famous for being able to estimate physical quantities even in situations where he did not (and, possibly, nobody) knew what the actual answer was. This is enormously helpful in science. One of the things we need to know is whether the answer we got back from our observing system or calculation was reasonable. When you start working in brand new areas, it's much harder to know what is reasonable. A Fermi estimate gives you that first guess. As Fermi also was a Nobel Laureate and did a lot of creative, original work, it might be a good thing if we practiced this skill ourselves.
Fortunately, you don't have to be a Nobel Laureate to do it, and the subject needn't be one on the frontiers of human knowledge. I made some use of it, for instance, regarding the company 'Joe the Plumber' wanted to buy. (After doing so: it isn't a small company.) The classic example is to estimate how many piano tuners there are in New York city. But I've seen that one a bunch of times, and don't have a piano, so it lacks something for meaning. The reason I need your help is that anything that leaps to my mind to create an example from will be in some area that I know something about already. So, what number -- about the world -- would you like to see estimated? No 'guess what number I'm thinking of' or 'what is the tangent of a trillion'? But something about how many, how big, how hot, ... of something in the observable universe.
25 November 2008
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11 comments:
Hi! It's Grace! (hugs)
Order of magnitude estimate?
What's the approximate density of the nucleus of a medium-sized atom? (no using hydrogen, it's unfair)
Have a look at my new blog, by the way, if you get the chance. I'm sorry, I borrowed your naming template even though it's not entirely accurate :)
http://www.evenmoregrumbinescience.blogspot.com/
Happy Thanksgiving!
Hugs to you too! Glad to see you're out and about and joining the blogosphere.
(Grace is my daughter; stepdaughter if you get really technical, which is why her 'not entirely accurate' comment. Still, after this many years, I think she's been thoroughly enough infected with 'grumbine-ness' to join the group.)
Do have a look at her blog. It'll be quite different than mine as her interests and talents are different. But it'll be good (you can see some samples already) and she does welcome comments.
Now, for her question. Of course it's not one of the easier ones to do, even for hydrogen. (Of course, I say, because she knows a fair amount about what I do and don't know, and didn't go closer to my areas. And of course because she's sharp and knows what'll be challenging.)
First, as in any Fermi estimate, I'll back off to the really simple parts of the question. The density is the mass divided by the volume. How much stuff is packed in to how small a volume. At the start of a trip, my suitcase is filled at pretty low density, but after I've thrown the many books and such from touristing, the density is pretty sizeable. Suitcase isn't any larger, but there's more mass present.
So I need to estimate two things: the mass of an atomic nucleus, and the volume it occupies. The second is the tougher one.
For the mass, I go back to my high school chemistry and recall that the density of middle-of-the-table metals (also that the middle of the table is indeed metals) is something like 10 grams per cubic centimeter. Might be as high as about 20, or as low as maybe 3. But 10 is a good middling number and easy to do my later math with. (In doing Fermi estimates, feel free to use numbers that make the math easier!) Now that's a density itself. I also know, however, that the atomic mass (a.k.a. gram-molecular weight) in the middle of the table is something like, oh, 50. Meaning that if we had 1 Avagadro's number of atoms of molecular weight 50, they'd weigh 50 grams. Ok, didn't need that density figure. We'll take the 50 grams and divide by Avagadro's number to find the mass of 1 mid-table atom. Avagadro's number is about 6*10^23 as I recall it. So let's make our middling element have atomic weight 60 (told you, make the math easy; and there are such atoms, so we're not really cheating). 1 atom then masses about 10^(-22) grams.
All atoms are about 1 Angstrom unit (1^(-8) cm) across. (This is a strange and interesting fact, and one whose variations tells you a fair amount about the elements involved.) So, if the nucleus were to span the whole size of the atom, and we take the atom to be a sphere, the volume of the atom would be about 4*r^3, about 4*10^(-24) cm^3 (note: the real formula for the volume of a sphere is 4/3*pi*r^3. But pi is about equal to 3, so we're left with about 4 after we do the multiplication of 4/3*pi). Now dividing the mass by the volume gives us 25 g/cm^3. So we're back to the sort of number we started with. Still in the Fermi ballpark.
Now to the difficult part: How much of that space does the nucleus occupy? The electrons have far less mass than the protons and neutrons that make up the nucleus (about 1800 times less mass), so we can ignore them for making a Fermi estimate. We know from Rutherford's experiments (lobbing alpha particles at thin sheets of metals) that the nucleus occupies a very small fraction of the volume (or area if we were shooting things at it). If I remembered the fraction of the time that the alpha particles got bounced back, that would tell me the fraction of the area they accounted for. But I don't, so on to another method. (Note, though: there are usually many different ways to make a Fermi estimate. It's a good idea to make several such estimates if you can and the problem matters.)
I'll hold to that thought about most of the volume of an atom is empty. One thing that I do know is that the outer electrons of mid-table metals can be removed by about 1 electron-Volt (eV) of energy. I also know that the inner ones take something like 30,000 eV. And I know that the energy required is proportional to 1/distance. So, those inner electrons are 30,000 times closer to the nucleus than the outer ones. The outer ones are about 1 Angstrom unit away (hence the size of the atom being 1 Angstrom unit). This says that the sphere the nucleus occupies is not larger than 1/30,000 angstrom units in radius. If I run that through the formula above, I get a density of about 10^15 g/cm^3. (6.75*10^14 with just this change, but I've done a bunch of simplifying and rounding and such, so I'll do it again here).
So there's my Fermi estimate: 10^15 g/cm^3. I expect it to be perhaps low, since there's probably some distance between inner electrons and the nucleus itself. But I have no further ideas on how to refine the estimate at the moment.
Now let's see what an outside source has to say. Wikipedia under 'nuclear density' gives 10^18 kg/m^3. So I have to convert units. When I do so, I get ... 10^15 g/cm^3. My Fermi estimate gave the right answer! (I'm a little surprised; even though I did know the answer from my astrophysics -- this is the density of neutron stars -- but I hadn't tried to estimate it before.)
Some things I used in making the estimate:
Approximate Avagadro's number
Rough density of common metals
Rough values for atomic masses of common elements
Rough size of atoms
Formula for volume of a sphere
Rough ionization potential for metals, outer and inner electron levels
Shape of the formula for potential.
How to convert units
The key: Not a single thing on that list required advanced scientific knowledge, nor did I ever use math beyond arithmetic! Knowing a few simple numbers and how things relate (more or less -- I never used exact formulas either) was all.
In this particular case, it would have been easy to look up the answer. The strength of Fermi estimation is that it works whether you can look up the answer or not, and whether anybody even knows the answer. And it works even if you don't know the gory details of a subject (ok, I did take a course on nuclear and particle physics, but that was ... quite a few years ago; more to the point -- I didn't use any of that here.)
One I've often wondered on train journeys, and never got round to working on, is the number of trees (any limit will be arbitrary - eg what's a tree not a bush, etc., so pick your own) in the UK, and secondly in the world.
I like your way much better :) The way I'd figured it out was circular reasoning, in a sense, as it used derived quantities (the approximate diameter of a proton) and involved modeling them as closely-packed, spheres, which they're really not (more of a quark soup as I understand it?). I hadn't thought to use the ionization energies.
I remember a school physics lesson discussion about what would happen if all of the ships in the world sailed east on the equator. How much would it affect the Earth's rotation?
For this, of course, we needed the mass of the world's shipping...
Let's see. Each tree needs a certain amount of distance to its neighbor, or else it gets shaded out. That means, equally, that they need a certain amount of area. There's also only so much area in the UK (or world). So if we knew both figures, we can just divide square meters (in UK, or world) by square meters per tree, and we're set.
I don't recall the area of the UK offhand, but curtesy of my recent look at population, do know that it has about 60 million people. Now, at large scale, a high population density, and the UK is pretty densely populated, is about 1000 per square km, so that's something like 60,000 square km for the UK. (Now that I see that figure, 50,000 is coming to mind as the actual. But let's continue.)
A lot of those 60,000 square km are occupied by people, farms, factories, etc., or are not suitable for trees in the first place for one reason or another. I'll go with 5,000 square km as tree-worthy and tree-available.
Now, how far apart are the trees? If I'd been paying more attention while on my trail-running, I'd know right off. Must be at least 3 steps (about 3 meters), and not more than 30, as a rule, between trees. Let's go to the shading business. If the trees are 10 meters (30 feet, 3 stories) tall, then they'll cast a shadow of about 10 meters when the sun is middling high in the sky. Part of the day the sun's higher than that, part of the day less so. So, a separation about equal to the height of the tree makes some sense. (Maybe it's half or a quarter instead, but we're dealing with Fermi estimation, so won't be that fussy.)
But how tall are trees? A young hardwood tree is, say, 10 feet tall (3 meters), or the size of a mature fruit or other short tree. (I get this height by comparison to basketball rims and 1 story houses.) A tall tree is about 30 meters (100 feet, 10 stories), which I get from thinking about modestly tall buildings (apartment complexes in Chicago, for instance) versus some mature Sycamores I know. If I take 10 meters, then, it's about 3 times taller than a short tree and 3 times shorter than a tall tree. Good middling estimate.
So, back to 10 meters tall per tree. Each tree occupies a circular disk 10 meters in radius. The area of that disk is about 300 m^2. (Again, keep the math easy, so I make pi = 3.) 1 square km is 1000 m by 1000 m, so I have 1,000,000 m^2 to put trees in.
That gives me about 3,000 trees per km^2. Then multiply by 5,000 km^2 in the UK, and I guess about 15,000,000 trees in the UK.
For the world at large, much of which doesn't get as much rain as the UK, and parts of which are under ice sheets, etc. ... Well, start with 150 million km^2. I took about 1/10th of the UK as being tree-available and tree-friendly. The world at large is a lot less so, I think, so I'll take 1/30th (makes for a round number and 1/30th is a good bit smaller than 1/10th). So 5 million (instead of 5 thousand) square km to put trees on. So something like 15,000,000,000 trees worldwide. (15 billion in US, but not UK, usage.)
So there are my estimates:
15 million trees in the UK
15 billion trees in the world
Facts used:
Trees need light and cast shadows
Shadows are proportional to height
Trees range between 10 and 100 feet commonly (3-30 meters)
Area of a circle.
Rough value of a moderately high population density
Rough population of the UK
Can we check this answer in any way? I did a search on 'number of trees in the world', and one site that was helpful was the UNEP Billion tree campaign. Now, if there were only 1 million trees in the world, trying to plant a billion would be absurd. Conversely if there were a trillion, then adding such a small number is nothing very significant (and why bother with a global campaign to do something trivial?). But if the estimate of 15 billion is reasonable, then the goal of adding 1 billion seems significant. It's not a small part of what's already present, and not huge compared to what's already there. So sounds good.
Then I encountered
http://understory.ran.org/2008/04/22/how-many-trees-are-cut-down-every-year/, which mentions that primary forests have a tree density of 50,000-100,000 trees per square km. That instead of my estimate of 3,000. Oops, off by a factor of about 20. That's high for a Fermi estimate. It suggests that the separation is more like 1/4th what I took, so 2.5 meter radius, 5 meters between trees.
This makes my estimates:
300 million in the UK
300 billion in the world
It's still subject to a lot of variation, depending on what fraction of the UK and the world we're getting primary forest in (and exactly what 'primary forest' means as compared to our question).
We don't have an accurate number, obviously, but we know a few things. One is that if we wanted a good number for the trees in the UK, we would absolutely not want to do it by walking from tree to tree and counting them. We'll have to use some method that lets us, say, use a satellite to sample an area and estimate how many trees are in the area. Then repeat for the country or region we want. Also tells us that anybody who says they did count every tree by hand is lying. Not a concern here, but I have seen cases where people claimed to do things that a Fermi estimate told me were impossible.
Reading the working is fascinating!
I fly a lot. Recently, while immersed in the constant hum and bland air of another aeroplane cabin, I started to wonder just how many people, at any given moment, are 35,000 feet or so off the planet?
I've made my own estimate. I think the likely error would be enormous. If you have the time or inclination I'd be very interested to see another estimate!
That was really interesting, many thanks.
Before I got to the end I knew I would disagree with your assumption about separation as some trees happily grow in shade of others.
I might have assumed a higher proportion to have trees as they tend to be planted alongside non-tree areas, eg around buildings, roads, railways, etc.
However that was not because I knew/know any better.
There is also the arbitrary cut off of what isn't a tree we would probably disagree on - but as I gave you free rein, I wouldn't use that to disagree with your result.
Anyway, asking you the question meant I finally looked up the UK answer. All I could find was the answer for Great Britain, which is smaller than the UK (doesn't include NI for one).
The answer is "3,814 million" according to the Forestry Commission.
See: http://tinyurl.com/6yxhq9
I've not looked for the totals for the rest of the UK not included in that.
I'll also get round to reading their survey technique when I get the chance.
Actually, here's a summary:
"In the latest 1995-99 inventory, data were collected for one hectare squares, covering a wide variety of topics, include ownership type, species and age. The results were uprated to total woodland areas from a digital map based on aerial photography."
Here's one that puzzles me. How many years before the next abrupt climate change? I wonder what Fermi's answer would be to that one.
Alastair.
Alastair, Adam: Sorry about the delay.
Alastair: We'd need some definition of what was an 'abrupt' climate change. I'll take the Younger Dryas event as a type case. In the last 110,000 years, there's been 1 such event. There was also a volcanic eruption which had a comparable magnitude effect about 70,000 years ago (Toba, if I remember correctly).
Volcanoes erupt essentially randomly in time, and one Toba's size is believed to have a recurrance time of 100,000 years. This does not mean set your clock on a 100,000 year countdown, unfortunately. It means that every year, there is a 1 in 100,000 chance of such an eruption.
The Younger Dryas, on the other hand, has been considered until very recently to have been a process internal to the climate system caused by some details of how you melt the major northern hemisphere ice sheets and run its deep ocean circulation. In that case, we're looking at the time being sometime during the next major northern hemisphere deglaciation -- not before at least 100,000 years have passed.
On the other hand, there are some recent mentions of possibly the event being impact-driven. I haven't followed them up to see how credible the claims are. But since they are in the professional literature, will grant them provisional plausibility. In this case, we're looking at a random event with, again, about a 100,000 year recurrence time.
Adam: Since I did correct for my great overestimate of how much space trees like, the remaining corrections are due to my underestimates of how large the UK is, and what fraction of the UK is forested.
So ... Fermi estimates can be pretty good, or pretty bad. Even where pretty bad, as for the trees, they can be a big help. Even my very low estimate more than sufficed to show us that we didn't want to go out counting trees one by one. Some kind of remote sensing or automated method was needed.
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