Estimation and Orders of Magnitude

Spec mapping: OCR H556 Module 2.2 — Making measurements and analysing data, specifically estimation of physical quantities, order-of-magnitude reasoning, and sanity-checking numerical results. (Refer to the official OCR H556 specification document for exact wording.)

One of the most valuable — and least drilled — skills in A-Level Physics is the ability to make a rough-but-reliable estimate of a physical quantity. A trained physicist can tell you, to within a factor of ten, the kinetic energy of a running human, the number of atoms in a teaspoon of water, or the frequency of a typical human heartbeat, without looking anything up. This is not a party trick. It is the single most powerful method a working scientist has to (i) sanity-check the output of any calculation; (ii) decide whether a proposed experiment is feasible; (iii) catch unit-conversion errors that the calculator cannot see; and (iv) resist the false precision that an over-eager scientific calculator encourages.

OCR examines this explicitly in Module 2.2 of the H556 specification under "Estimation of physical quantities". You are expected to be able to give an order-of-magnitude estimate, use approximate values of common constants, sanity-check a numerical answer, and perform simple dimensional analysis to infer the form of an equation. This lesson trains all four skills, with a particular focus on Fermi-style decomposition — the method named after Enrico Fermi who could routinely arrive at numerical estimates within a factor of two using nothing more than a pencil and a sense of physical scales.

Key definition: The order of magnitude of a physical quantity is the power of ten closest to it. If the quantity is written as $a \times 10^{n}$a×10n with $1 \le a < 10$1≤a<10, then the order of magnitude is $n$n. (Some textbooks round when $a > \sqrt{10} \approx 3.16$a>10​≈3.16 and instead quote $n+1$n+1; OCR accepts either convention as long as it is applied consistently.)

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What an order of magnitude means

When two quantities differ by more than one order of magnitude, one is unambiguously much bigger than the other; they live on different rungs of the power-of-ten ladder. When they differ by less than an order of magnitude, they are in the same "ballpark" — within a factor of about ten — and a finer comparison requires more careful measurement.

Quantity	Value	Order of magnitude
Planck length	$1.6 \times 10^{-35}\,\text{m}$1.6×10−35m	$-35$−35
Diameter of a proton	$1 \times 10^{-15}\,\text{m}$1×10−15m	$-15$−15
Diameter of a hydrogen atom	$1 \times 10^{-10}\,\text{m}$1×10−10m	$-10$−10
Wavelength of visible light	$5 \times 10^{-7}\,\text{m}$5×10−7m	$-7$−7
Thickness of a sheet of paper	$1 \times 10^{-4}\,\text{m}$1×10−4m	$-4$−4
Height of a human	$1.7\,\text{m}$1.7m	$0$0
Height of Mount Everest	$8.8 \times 10^{3}\,\text{m}$8.8×103m	$4$4
Radius of the Earth	$6.4 \times 10^{6}\,\text{m}$6.4×106m	$7$7
Distance to the Moon	$3.8 \times 10^{8}\,\text{m}$3.8×108m	$8$8
Distance Earth–Sun (1 AU)	$1.5 \times 10^{11}\,\text{m}$1.5×1011m	$11$11
Distance to the nearest star	$4 \times 10^{16}\,\text{m}$4×1016m	$16$16
Diameter of the Milky Way	$1 \times 10^{21}\,\text{m}$1×1021m	$21$21
Observable universe	$9 \times 10^{26}\,\text{m}$9×1026m	$27$27

The full span from Planck length to observable universe is sixty-two orders of magnitude. No other quantitative discipline routinely deals with such a range. Estimation is the cognitive tool that lets us think across it.

Exam tip: If OCR asks you to "estimate the order of magnitude" of a quantity, you only need to give the power of ten. Quoting "7.31 × 10²⁷" on an estimation question loses marks for false precision — the estimate is uncertain at the leading-digit level, so claiming three sig figs is dishonest about the uncertainty.

Inline scale ladder — the universe in powers of ten

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Quantities every A-Level physicist should know

You should know the following approximate values to within an order of magnitude without consulting a data sheet. These are the building blocks of every estimation question.

Quantity	Typical value (order of magnitude)
Mass of an electron	$\sim 10^{-30}\,\text{kg}$∼10−30kg ($9.11 \times 10^{-31}$9.11×10−31 exact)
Mass of a proton	$\sim 10^{-27}\,\text{kg}$∼10−27kg ($1.67 \times 10^{-27}$1.67×10−27)
Mass of a human	$\sim 70\,\text{kg}$∼70kg
Mass of a car	$\sim 10^{3}\,\text{kg}$∼103kg
Mass of the Earth	$\sim 6 \times 10^{24}\,\text{kg}$∼6×1024kg
Mass of the Sun	$\sim 2 \times 10^{30}\,\text{kg}$∼2×1030kg
Radius of an atom	$\sim 10^{-10}\,\text{m}$∼10−10m
Radius of a nucleus	$\sim 10^{-15}\,\text{m}$∼10−15m
Radius of the Earth	$\sim 6 \times 10^{6}\,\text{m}$∼6×106m
Radius of the Sun	$\sim 7 \times 10^{8}\,\text{m}$∼7×108m
Speed of sound in air	$\sim 340\,\text{m s}^{-1}$∼340m s−1
Speed of light	$3 \times 10^{8}\,\text{m s}^{-1}$3×108m s−1 (exact, defines metre)
Acceleration of free fall (UK)	$9.81\,\text{m s}^{-2}$9.81m s−2
Atmospheric pressure	$\sim 10^{5}\,\text{Pa}$∼105Pa ($1.013 \times 10^{5}$1.013×105)
Density of water	$\sim 10^{3}\,\text{kg m}^{-3}$∼103kg m−3
Density of air	$\sim 1.2\,\text{kg m}^{-3}$∼1.2kg m−3
Avogadro constant	$\sim 6 \times 10^{23}\,\text{mol}^{-1}$∼6×1023mol−1
Boltzmann constant	$\sim 1.4 \times 10^{-23}\,\text{J K}^{-1}$∼1.4×10−23J K−1
Planck constant	$\sim 6.6 \times 10^{-34}\,\text{J s}$∼6.6×10−34J s
Elementary charge	$\sim 1.6 \times 10^{-19}\,\text{C}$∼1.6×10−19C
Age of universe	$\sim 4 \times 10^{17}\,\text{s}$∼4×1017s ($\approx 14\,\text{Gyr}$≈14Gyr)
Distance Earth–Sun	$1.5 \times 10^{11}\,\text{m}$1.5×1011m (1 AU)

You are not expected to recite Planck's constant to twelve significant figures — the data sheet provides it. But you should be able to say "about $10^{-34}\,\text{J s}$10−34J s" and use that in a Fermi estimate without consulting the sheet.

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Fermi estimation — break it down, then multiply

The Italian-American physicist Enrico Fermi was legendary for his ability to make good estimates with minimal information. On the morning of the Trinity nuclear test in 1945, the story goes, Fermi stood at the observation post, dropped small scraps of paper before the shockwave arrived, and watched how far the blast wind blew them. From that observation alone he reportedly estimated the yield to within a factor of two of the true value, before any instruments had reported in. The technique he used has come to be called Fermi estimation: decompose a hard question into a small number of easier sub-questions, estimate each to within a factor of ten, and combine.

The mathematical reason Fermi estimation works is that the errors tend to cancel. If you over-estimate one quantity by 30% and under-estimate another by 30%, the product is correct to within 10%. With enough sub-quantities, individual errors of order 30% combine to a total error of maybe 50%, which is well within an order of magnitude.

Worked Example 1 — A classic Fermi question

Estimate the number of breaths a typical human takes in a lifetime.

Decompose:

• Breaths per minute (resting): $\approx 15$≈15
• Minutes per hour: $60$60
• Hours per day: $24$24
• Days per year: $\approx 365 \approx 4 \times 10^{2}$≈365≈4×102
• Human lifespan: $\approx 80\,\text{years}$≈80years

Multiply:

$N \approx 15 \times 60 \times 24 \times 400 \times 80$N≈15×60×24×400×80

Group the powers of ten:

$N \approx (15 \times 60 \times 24) \times (400 \times 80) = 21600 \times 32000 \approx 2 \times 10^{4} \times 3 \times 10^{4} \approx 7 \times 10^{8}$N≈(15×60×24)×(400×80)=21600×32000≈2×104×3×104≈7×108

About a billion breaths — the order of magnitude is $10^{8}$108 to $10^{9}$109. An OCR examiner will accept anything in that range.

Worked Example 2 — Falling apple energy

An apple falls from a tree of height $3\,\text{m}$3m. Estimate the kinetic energy at the bottom.

Decompose. Mass of apple $\approx 0.2\,\text{kg}$≈0.2kg, height $3\,\text{m}$3m, $g \approx 10\,\text{m s}^{-2}$g≈10m s−2 (round for estimation). By energy conservation:

$E_k \approx mgh = 0.2 \times 10 \times 3 = 6\,\text{J}$Ek​≈mgh=0.2×10×3=6J

Order of magnitude: $10^{0}$100–$10^{1}\,\text{J}$101J. The 6 J figure agrees with the kinetic energy of a slow-walking pedestrian, which is the right comparison.

Worked Example 3 — Atoms in a human body

Estimate the number of atoms in a 70 kg human.

A human is roughly 65% water by mass, but for an order-of-magnitude estimate we treat the human as if entirely water. Water has molar mass $M_{\text{r}} = 18\,\text{g mol}^{-1}$Mr​=18g mol−1.

Number of moles of water in 70 kg:

$n \approx \frac{70\,000\,\text{g}}{18\,\text{g mol}^{-1}} \approx 4000\,\text{mol}$n≈18g mol−170000g​≈4000mol

Number of water molecules:

$N_{\text{H}_2\text{O}} \approx 4000 \times 6 \times 10^{23} \approx 2.4 \times 10^{27}$NH2​O​≈4000×6×1023≈2.4×1027

Each water molecule contains 3 atoms (2 H + 1 O):

$N_{\text{atoms}} \approx 3 \times 2.4 \times 10^{27} \approx 7 \times 10^{27}$Natoms​≈3×2.4×1027≈7×1027

Order of magnitude: $10^{28}$1028 atoms. This is the figure often quoted in popular science.

Worked Example 4 — Light travel time from Sun to Earth

Estimate the time for light to travel from the Sun to Earth.

Distance $\approx 1.5 \times 10^{11}\,\text{m}$≈1.5×1011m; speed of light $c \approx 3 \times 10^{8}\,\text{m s}^{-1}$c≈3×108m s−1.

$t = \frac{d}{c} = \frac{1.5 \times 10^{11}}{3 \times 10^{8}} = 5 \times 10^{2}\,\text{s} \approx 500\,\text{s} \approx 8\,\text{min}$t=cd​=3×1081.5×1011​=5×102s≈500s≈8min

The famous "sunlight is 8 minutes old" result, derived from a single back-of-envelope calculation.

Worked Example 5 — Energy a heavyweight boxer can deliver

Estimate the kinetic energy in the punch of a heavyweight boxer.

Decompose: fist mass $\approx 0.5\,\text{kg}$≈0.5kg (effective, the boxing arm contributes mass to the strike), fist speed $\approx 10\,\text{m s}^{-1}$≈10m s−1 at impact (about 22 mph).

$E_k = \tfrac{1}{2}mv^{2} = \tfrac{1}{2} \times 0.5 \times 10^{2} = 25\,\text{J}$Ek​=21​mv2=21​×0.5×102=25J

About 25 J. Order of magnitude: $10^{1}$101–$10^{2}\,\text{J}$102J. By comparison, lifting a kilogram one metre against gravity takes 9.81 J; the boxer's punch is roughly 2.5 times this — physically reasonable, and the right ballpark for the figure reported in sports-science measurements.

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Sanity-checking — when an answer is wrong

Estimation is not just for problems where the question explicitly demands an estimate. The most important use of estimation is to sanity-check answers obtained via more precise calculation. If your full algebraic solution gives a number wildly different from the estimate, one of them is wrong — usually the algebraic one, because the estimate is robust.

Worked Example 6 — Catching a kinetic-energy slip

A $1.0\,\text{kg}$1.0kg apple falls from a tree $4\,\text{m}$4m high. A student calculates the KE on impact as $78.5\,\text{J}$78.5J. Is this reasonable?

Sanity check:

$E_k \approx mgh = 1.0 \times 10 \times 4 = 40\,\text{J}$Ek​≈mgh=1.0×10×4=40J

The student's answer is twice this. Something is wrong — perhaps they forgot to divide by 2 in $\tfrac{1}{2}mv^{2}$21​mv2, used $g = 19.6$g=19.6, doubled the height, or substituted the velocity instead of the velocity-squared. Whatever the error, the sanity check has caught it. The correct answer, using $g = 9.81$g=9.81, is $E_k = mgh = 1.0 \times 9.81 \times 4 \approx 39\,\text{J}$Ek​=mgh=1.0×9.81×4≈39J.

When estimation reveals a missing prefix

If your algebraic answer is wrong by a factor of $10^{3}$103, $10^{6}$106, or $10^{9}$109, the most likely cause is a missing prefix conversion. Estimation acts as a unit-conversion sanity check.

For example: if a calculation of the power dissipated in a $4.7\,\text{k}\Omega$4.7kΩ resistor at $5\,\text{mA}$5mA current gives $P = 588\,\text{W}$P=588W, your estimation alarm should fire instantly — a small resistor at milliamp current cannot dissipate household-electric-fire wattage. The actual answer (using $P = I^{2}R$P=I2R with $I = 5 \times 10^{-3}\,\text{A}$I=5×10−3A, $R = 4700\,\Omega$R=4700Ω) is

$P = (5 \times 10^{-3})^{2} \times 4700 = 25 \times 10^{-6} \times 4700 = 0.12\,\text{W}$P=(5×10−3)2×4700=25×10−6×4700=0.12W