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Numbers, Units and Estimation · Lesson 6 of 6

Estimating by Orders of Magnitude

13 min read2 ideas6 checks2 figures to playBy Asheshanand Acharya
What you will learn 0 / 2
Estimating by orders of magnitude
Look around
How much fuel does a reactor burn?

The Kudankulam power station on the coast of Tamil Nadu makes electricity from uranium. A coal station of the same size burns several trainloads of coal every day. How much uranium does a reactor use in a day? A kilogram, a tonne, a trainload? No table in this book lists the answer, yet you can find it before the end of this lesson.

you will estimate the uranium a reactor uses in a day with four numbers and one line of arithmetic. You will also see why so rough a method comes so close.

Break the question into factors

Enrico Fermi liked to ask his students questions that seemed impossible to answer, such as the number of piano tuners in Chicago. The trick is to write the unknown as a product of factors, each of which you can guess or already know. Then round each factor and multiply. Such a calculation is called a Fermi estimate.

Take a simple case: the mass of the air in your classroom. A classroom is about 10 m long, 8 m wide and 3 m high. Air has a density of about \(1.2\,\text{kg/m}^3\). So the mass is \(10 \times 8 \times 3 \times 1.2 \approx 290\,\text{kg}\), about 300 kg. The air in a room weighs as much as five students.

definition 1.4
A Fermi estimate

A Fermi estimate finds the order of magnitude of a quantity. It writes the quantity as a product of factors, each guessed or known to within a factor of about two or three. Each factor is rounded to one figure or to a power of ten, and the factors are multiplied. The answer is trusted only to its order of magnitude.

RememberA Fermi estimate
\[ Q \approx f_1 \times f_2 \times \cdots \times f_n \]
Holds whenwhen each factor is known to within a factor of about two or three, and the errors of the factors are independent. The product is then usually right within a factor of ten.
Symbols\(Q\) the quantity estimated, \(f_1, f_2, \ldots, f_n\) the factors, each rounded to one figure or a power of ten
Unitsthe units of the factors must multiply to the unit of \(Q\)
EstimateUnanswered

Estimate the mass of air in the physics laboratory of your school, in kilograms. Choose its length, breadth and height yourself, and take the density of air as \(1.2\,\text{kg/m}^3\).

kgrough value is enough

Why so rough a method works

Each factor of an estimate may be wrong by a factor of two. So why is the product not wrong by \(2 \times 2 \times 2 \times 2 = 16\)? Because the errors are independent. Some roundings are too large and some are too small, and they partly cancel. The product is usually much closer than the worst case. Fig 6.1 shows this cancelling in the estimate of the phenomenon.

ExampleThis is the estimate of The Nucleus, Binding Energy and Nuclear Reactions, lesson 5, done with the tool

A reactor produces \(1000\,\text{MW}\) of heat. Each fission of uranium-235 gives \(200\,\text{MeV} = 3.2 \times 10^{-11}\,\text{J}\). Estimate the mass of uranium-235 that it uses in one day.

Try it on paper first. The solution is one tap away when you are ready.
The fuel of a reactor as a chain of factorsA reactor makes 10 to the 9 joules of heat each second. On a ruler of powers of ten, from 1 to 10 to the 25, four arrows follow the estimate of the uranium-235 it uses in a day, one row per step. Dividing by 3.2 times 10 to the minus 11 joule per fission is a long step up, to 3.1 times 10 to the 19 fissions a second. Multiplying by 86 400 seconds gives 2.7 times 10 to the 24 fissions a day. Dividing by 6.02 times 10 to the 23 is a long step down, to 4.5 moles. Multiplying by 235 grams gives 1054 grams. With the second chip every factor is rounded to a power of ten: 10 to the minus 11, 10 to the 5, 10 to the 24 and 10 to the 2. The arrows change a little, but the chain ends at 10 cubed grams, 1 kilogram, almost on top of the exact 1.05 kilograms, because the roundings up and down nearly cancel. 1 10⁵ 10¹⁰ 10¹⁵ 10²⁰ 10²⁵ start: 10⁹ J/s ÷ 3.2 × 10⁻¹¹ J × 86 400 s ÷ 6.02 × 10²³ × 235 g step 1 of 4
Fig 6.1The estimate above as a chain of factors on a ruler of powers of ten, one row per step. Press play: dividing by the tiny energy of one fission is a long step up, and dividing by Avogadro's number a long step down. The second chip rounds every factor to a power of ten, \(10^{-11}\), \(10^5\), \(10^{24}\) and \(10^2\). The chain still ends at \(10^3\,\text{g}\), almost on top of the exact 1054 g.

With every factor rounded to a power of ten, the arithmetic is only a sum of exponents: \(9 + 11 + 5 - 24 + 2 = 3\), so the answer is \(10^3\,\text{g}\). Replacing \(3.2 \times 10^{-11}\) by \(10^{-11}\) made the answer 3.2 times too large. Replacing 235 by 100 and \(6.02 \times 10^{23}\) by \(10^{24}\) made it about 3.9 times too small. Replacing 86 400 by \(10^5\) added 16 %. Together the errors nearly cancelled.

The phenomenon is answered. A 1000 MW reactor uses about one kilogram of uranium-235 a day. A coal station giving the same heat burns thousands of tonnes of coal in the same day. One fission releases tens of millions of times the energy of burning one atom of carbon, so the nuclear fuel is millions of times lighter.

Checkpoint · numericUnanswered

One unit of the Kudankulam station makes about \(3000\,\text{MW}\) of heat. Each fission gives \(3.2 \times 10^{-11}\,\text{J}\). How many fissions happen each second? Write the answer as \(a \times 10^{19}\) and enter \(a\).

× 10¹⁹ per second± 2 %
CheckpointUnanswered

An estimate multiplies four factors, and each factor may be wrong by up to a factor of two. Why is the product usually wrong by much less than a factor of 16?

EstimateUnanswered

Estimate the number of water molecules in a glass of water. Take the glass to hold \(200\,\text{g}\), the molar mass of water as \(18\,\text{g}\), and Avogadro's number as \(6 \times 10^{23}\). Give the answer in units of \(10^{24}\) molecules.

× 10²⁴ moleculesrough value is enough
Checking an answer by its size

Is the answer the right size?

A quick estimate is also the best check of a long calculation. Before you calculate, guess the order of magnitude of the answer. After you calculate, compare. If a photon of light is found to have \(10^{-5}\,\text{J}\), or a cricket ball with \(10^{8}\,\text{J}\), something has gone wrong. Most such errors are a lost power of ten or a unit left unconverted. Fig 6.2 gives a few energies of the book to compare with.

Energies of the book on one ladderA vertical ruler of energies in joules, one step for each factor of ten, from 10 to the minus 20 joule at the bottom to 10 to the 15 joules at the top. Marked on it: a photon of visible light, 3.2 times 10 to the minus 19 joule; one fission of uranium-235, 3.2 times 10 to the minus 11 joule; a cricket ball of 0.16 kilogram bowled at 40 metres per second, 1.3 times 10 squared joules; one unit on the electricity bill, 3.6 times 10 to the 6 joules; and the electrical output of a 1000 megawatt reactor unit for a day, 8.6 times 10 to the 13 joules. The reactor's day is about 10 to the 32 times the energy of the photon. 10⁻²⁰ J 10⁻¹⁵ J 10⁻¹⁰ J 10⁻⁵ J 1 J 10⁵ J 10¹⁰ J 10¹⁵ J a photon of visible light: 3.2 × 10⁻¹⁹ J one fission of uranium-235: 3.2 × 10⁻¹¹ J a ball bowled at 40 m/s: 1.3 × 10² J one unit on the bill: 3.6 × 10⁶ J a reactor unit for a day: 8.6 × 10¹³ J
Fig 6.2Energies of the book on one ruler of powers of ten. A photon of visible light has \(3.2 \times 10^{-19}\,\text{J}\) and one fission \(3.2 \times 10^{-11}\,\text{J}\). A cricket ball of 0.16 kg bowled at 40 m/s has 128 J. One unit on the bill is \(3.6 \times 10^6\,\text{J}\), and a 1000 MW reactor unit gives \(8.6 \times 10^{13}\,\text{J}\) in a day.

Knowing a few such sizes is worth more than any rule. A wave on a stretched string moves at tens or hundreds of metres per second. Sound in air travels at about \(340\,\text{m/s}\). Light travels at \(3 \times 10^8\,\text{m/s}\). An answer far outside these ranges deserves a second look before it is written down.

CheckpointUnanswered

A student finds the speed of a wave on a sitar string as \(2 \times 10^5\,\text{m/s}\). What should the student conclude?

CheckpointUnanswered

In a quick estimate, the number of seconds in a day, \(86\,400\), is rounded to a power of ten. Which power is nearest, by the rule of lesson 1?

Common mistake

An estimate fails when a single factor is wrong by a large amount. Rounding cannot rescue a factor guessed a hundred times too large, or a model that omits the main physics. An estimate also fails for a small difference of large numbers, where a small error in either number changes the answer completely. Trust an estimate to its order of magnitude, and never to its second figure.

Story · Enrico Fermi, 16 July 1945
Strips of paper in the desert

At the first test of an atomic bomb, in the desert of New Mexico, Enrico Fermi lay about 16 km from the explosion. When the blast wave reached him, about forty seconds after the flash, he dropped small strips of paper from his hand. They were carried about two and a half metres. From that distance alone he estimated the energy of the explosion as about 10 kilotons of TNT. The careful measurements, made later, gave about 20 kilotons. His estimate was right within a factor of two.

In the world
Where this tool works in physics

Estimation runs through the book. The nuclear chapter estimates the fuel of a reactor and the energy of a fission, about 200 MeV. The chapter on radiation estimates the sunlight on a roof and the temperature of the Earth. The chapter on projectiles estimates how far the round Earth drops below a long flight. Many chapters close with an estimation question of their own.

Exam traps
  1. Guessing the whole answer at once.
    Break it into factors you can guess or know, and multiply them.
  2. Giving an estimate to three significant figures.
    An estimate is good to its order of magnitude. Write about 1 kg, not 1.054 kg.
  3. Mixing units among the factors.
    Put every factor in SI units before multiplying, or the powers of ten will be wrong.

Where this tool works in the book

The physics that leans on this lesson. Each card opens the topic where the tool is at work.

Motion in Two Dimensions
Lesson 6 · The Trajectory as a ToolLimits of the parabolaThe parabola assumes no air, a flat ground and a gravity fixed in size and direction.
Radiation, Convection and Cooling
Lesson 4 · The Sun and the EarthTemperature of the earthA planet that absorbs sunlight on a disc and radiates from its whole sphere settles at \(T = ((1 - A)S/4\sigma)^{1/4}\): 278 K for the Earth with no reflection, 255 K with 30 % reflected.
The Nucleus, Binding Energy and Nuclear Reactions
Lesson 3 · The Binding Energy CurveEnergy from moving toward the peakAny reaction that moves nucleons towards the peak of the curve releases energy, \(Q = B(\text{after}) - B(\text{before})\).Lesson 5 · Fission and the ReactorThe nuclear reactorA thermal reactor holds \(k = 1\).