🎯 Hook – the Fermi estimate
Enrico Fermi famously asked his students to estimate the number of piano tuners in Chicago with no data at all — just population, households, pianos per household, tuning frequency, and how long a tuning takes. Every step is a rough, rounded guess, yet the final answer lands within a factor of two or three of the true value. Order-of-magnitude thinking is one of a physicist's most-used tools — it catches wrong answers before a calculator ever gets involved.
🔢 Scientific notation and significant figures
Write a number as \(a \times 10^{n}\) with \(1 \le a < 10\) and \(n\) an integer: it shows size and precision at a glance.
Significant figures (s.f.) are all the digits in a number that carry meaning. All non-zero digits are significant (452 → 3 s.f.); zeros between non-zero digits are significant (4.052 → 4 s.f.); leading zeros are never significant (0.004 50 → 3 s.f.); trailing zeros after a decimal point are always significant (4.50 → 3 s.f.). Trailing zeros in a whole number are ambiguous (4500 could be 2, 3 or 4 s.f.) — scientific notation removes the doubt: \(4.50\times10^{3}\) is unambiguously 3 s.f.
Rules for calculating: multiplying or dividing — give the answer to the fewest significant figures of the data used. Adding or subtracting — give the answer to the fewest decimal places of the data used. Keep extra digits in intermediate steps and round only the final answer; in practical work the number of significant figures in a result should match its uncertainty (9.8 ± 0.2 m s⁻², not 9.8137 ± 0.2).
📏 Orders of magnitude and estimation
The order of magnitude of a quantity is the power of ten closest to it. IB convention rounds the logarithm: \(6.32\times10^{8}\) → \(10^{9}\) (because \(\log 6.32 \approx 0.8\), which rounds up); \(1.60\times10^{-19}\) → \(10^{-19}\). "Two orders of magnitude larger" means about 100 times larger.
Estimating means reaching a sensible value with rounded inputs and simple reasoning (a Fermi estimate). State assumptions, round to 1 s.f., and check the answer is plausible. Knowing when an effect can be neglected (air resistance on a dense ball over a short drop) simplifies problems — always check the neglected effect is small by an order-of-magnitude estimate.
📐 Proportion, logs and trigonometry
| Relationship | Written as | If x doubles, y becomes | Straight-line graph |
|---|---|---|---|
| Direct | \(y \propto x\) | \(2y\) | y vs x |
| Inverse | \(y \propto 1/x\) | \(y/2\) | y vs 1/x |
| Square | \(y \propto x^2\) | \(4y\) | y vs x² |
| Inverse square | \(y \propto 1/x^2\) | \(y/4\) | y vs 1/x² |
| Square root | \(y \propto \sqrt{x}\) | \(1.41y\) | y vs √x |
Logarithms: \(\log_{10}(10^n)=n\); \(\ln(e^n)=n\) (e ≈ 2.718). \(\log(ab)=\log a+\log b\); \(\log(a/b)=\log a-\log b\); \(\log(a^n)=n\log a\). Logs turn multiplication into addition and powers into multiplication — which is why power laws and exponentials become straight lines on log graphs (Lesson 08).
Trigonometry: in a right-angled triangle, \(\sin\theta=\)opposite/hypotenuse, \(\cos\theta=\)adjacent/hypotenuse, \(\tan\theta=\)opposite/adjacent; Pythagoras \(a^2+b^2=c^2\). Angles in radians: \(\theta = s/r\); \(2\pi\) rad = 360°. For small angles in radians, \(\sin\theta \approx \tan\theta \approx \theta\) and \(\cos\theta \approx 1\).
📊 Graph mastery – proportionality sketches
| Graph type | Axes (X, Y) | Shape | What to read |
|---|---|---|---|
| Direct proportion | x · y | Straight line through the origin | Gradient = constant of proportionality |
| Inverse square | 1/x² · y | Straight line through the origin | Confirms \(y\propto 1/x^2\); curved if plotted as y vs x |
| Exponential decay | t · N | Curve falling towards (never reaching) zero | Half-life read where N halves; straightens on a ln N vs t plot |
| ln N vs t | t · ln N | Straight line, negative gradient | Gradient = −λ; intercept = ln N₀ |
✍️ Worked example – significant figures and adding
✍️ Worked example – proportion and logs
\(t = \dfrac{1}{0.050}\ln(1/0.10) = 20\times2.303\)
🌍 Real-world: scaling and body size
Length × k → area × k² → volume (and mass, for fixed density) × k³. This is why small animals lose heat faster (surface area/volume ∝ 1/k) and must eat far more per kilogram of body mass than large animals, and why the critical mass of a nuclear fuel depends on its size.
🚀 Going further – why logs linearise graphs
Taking logs of \(y=kx^n\) gives \(\log y = \log k + n\log x\) — a straight line of gradient \(n\) on a \(\log y\) vs \(\log x\) plot, however large the power \(n\). This is the method behind testing power laws from experimental data (Lesson 06, Lesson 08).
Challenge: a graph of \(\log y\) against \(\log x\) has gradient 3 and intercept 0.70. What is the relationship between y and x? \(y = 10^{0.70}x^3 \approx 5.0x^3\).
📋 Quick practice — answers in italics
(a) Write 0.000 047 25 m in scientific notation to 3 s.f. 4.73×10⁻⁵ m.
(b) State the number of significant figures in 3.050×10⁴. 4 s.f.
(c) Calculate 45.67 − 3.1 to the correct precision. 42.6 (1 d.p.).
(d) State the order of magnitude of the mass of an adult human (≈70 kg). 10² kg.
(e) Solve \(5.0 = 20e^{-0.30t}\) for t. \(t=\ln4/0.30\approx4.6\ \text{s}\).
(f) Rearrange \(T=2\pi\sqrt{L/g}\) to make g the subject. \(g = 4\pi^2L/T^2\).
⚠️ Trap corner – common mistakes
| Misconception | Correct view |
|---|---|
| Leading zeros are significant | Leading zeros are never significant; trailing zeros after a decimal point always are |
| Rounding early in a multi-step calculation is fine | Rounding too early changes final answers — carry extra digits and round only the last line |
| Adding/subtracting uses significant figures | Adding/subtracting uses decimal places, not significant figures |
| Degrees and radians are interchangeable | Small-angle approximations and s = rθ need radians — check the calculator mode |
| Order of magnitude = the first digit | \(6\times10^{8}\) is of order \(10^{9}\), not \(10^{8}\) |
| ln and log (base 10) are the same function | ln is base e; log on most calculators is base 10 |
Complete the statements:
✓ (b) \(d=ct=(3.00\times10^{8})(3.2\times10^{7})=9.6\times10^{15}\ \text{m} = 9.6\times10^{12}\ \text{km}\).
✓ (c) \(9.6\times10^{15}\ \text{m} \to 10^{16}\ \text{m}\) (log 9.6 ≈ 0.98, rounds up).
✓ (d) \(4\times10^{16}/1.5\times10^{11} \approx 2.7\times10^{5}\) — about 5 orders of magnitude further.
✓ (b) \(t = \ln(800/100)/0.0231 = \ln8/0.0231 \approx \mathbf{90\ min}\).
✓ (c) Taking logs: \(\ln A = \ln A_0 - \lambda t\) — linear in t with gradient \(-\lambda\) and intercept \(\ln A_0\).
✓ (b) Mass ∝ density × volume ∝ r³ at constant density, so mass ratio \(=2^3=\mathbf{8}\).
✓ (c) Assumes the planet is a uniform sphere of the same density throughout (not just the same average density).
| L / m (±0.002) | Time for 20 oscillations / s |
|---|---|
| 0.200 | 17.9 |
| 0.400 | 25.4 |
| 0.600 | 31.1 (mean) |
| 0.800 | 35.9 |
| 1.000 | 40.1 |
✓ (b) Human reaction time (~0.1–0.3 s) dominates over the stopwatch's 0.01 s resolution; timing 20 oscillations spreads this fixed reaction-time error over many periods, reducing its fractional effect on a single period.
✓ (c) T = time/20: 0.895, 1.27, 1.56, 1.80, 2.01 s for the five lengths.
✓ (d) T²/L ≈ 4.01, 4.03, 4.03, 4.03, 4.02 — all close to 4.0, confirming \(T\propto\sqrt{L}\).
✓ (e) \(g = 4\pi^2/(T^2/L) = 4\pi^2/4.02 \approx \mathbf{9.82\ m\,s^{-2}}\).
✓ It catches gross errors — a misplaced decimal point or wrong unit — that a calculator will not flag, because the calculator trusts whatever is typed in.
✓ Example: a calculated mass of the Earth of \(5.97\times10^{48}\) kg is wrong by 24 orders of magnitude compared with the known \(\approx6\times10^{24}\) kg, visible without checking any arithmetic.
✓ Example: computing \(\sqrt{(1.03)^2-(1.00)^2}\) directly gives 0.245; rounding \(1.03^2\to1.06\) and \(1.00^2\to1.00\) first gives \(\sqrt{0.06}=0.245\) too here, but with closer values (e.g. 1.003 and 1.000) early rounding can erase the entire difference and give zero instead of the correct small non-zero answer.
✓ The rule: carry at least one extra significant figure through every intermediate line and round only the final answer.
✓ (b) A feather has a huge surface area relative to its weight, so drag is comparable to or larger than weight almost immediately — cannot be neglected.
✓ (c) A skydiver accelerates until drag balances weight (terminal velocity), so over any realistic fall time drag is essential to the motion and must be included.
✓ So per kilogram of body mass a mouse loses heat far faster than an elephant, and must eat proportionally more food to replace it.
✓ For graphing: taking logs of a power law or exponential relationship turns a curve into a straight line, so the gradient and intercept directly give the power or decay constant — the basis of linearising graphs in Lesson 08.