Estimating physical quantitiesAQA A-Level Physics: Revision notes

The order of magnitude of a quantity is the power of ten nearest to its value. Write the number in standard form, A × 10ⁿ. If A is less than about 3 (strictly, √10 ≈ 3.16) the order of magnitude is 10ⁿ. If A is greater than about 3 it is 10ⁿ⁺¹.

• 2.5×10⁵ m is of order 10⁵ m
• 5.4×10⁵ J is of order 10⁶ J
• 4.2×10⁻⁵ m is of order 10⁻⁴ m

Two quantities that differ by a factor of 10ⁿ are said to be n orders of magnitude apart.

Learn a bank of typical values so you can start an estimate quickly:

• diameter of an atom 10⁻¹⁰ m; of a nucleus 10⁻¹⁵ m
• wavelength of visible light about 5×10⁻⁷ m, which is of order 10⁻⁶ m
• mass of a proton 10⁻²⁷ kg; of an electron 10⁻³⁰ kg
• mass of an adult 10² kg; of a family car 10³ kg
• height of an adult about 1.7 m, of order 10⁰ m
• speed of sound in air about 3×10² m s⁻¹; speed of light 3×10⁸ m s⁻¹
• density of water 10³ kg m⁻³; density of air about 1 kg m⁻³
• one year ≈ 3×10⁷ s

1. Break the problem into steps that use known physics.
2. Write down sensible round values and state your assumptions.
3. Convert to SI units.
4. Calculate, keeping only 1 or 2 significant figures.
5. State the answer to the nearest order of magnitude and say whether it is plausible.

Worked example. The mass of air in a 10 m × 8 m × 3 m room. Volume = 240 m³. Mass = ρV = 1.2 × 240 ≈ 290 kg, so about 10² kg.

Once one quantity is estimated, you can use physics equations to estimate others.

Worked example (heartbeats). 80 years ≈ 80 × 3×10⁷ = 2.5×10⁹ s. At 70 beats per minute (about 1.2 per second) that is about 3×10⁹ beats, of order 10⁹.

Worked example (kinetic energy and power). A 1.2×10³ kg car at 30 m s⁻¹ has E = ½mv² = 5.4×10⁵ J, of order 10⁶ J. Accelerating to this speed in 10 s needs an average power of 5.4×10⁵ ÷ 10 = 5.4×10⁴ W, of order 10⁵ W.

Ratios and scaling: lengths differ by 10ⁿ means areas differ by 10²ⁿ and volumes by 10³ⁿ. Atom to nucleus diameter ratio is 10⁵, so the volume ratio is 10¹⁵.