LearnYPH11WPH15 · Unit 5: Thermodynamics, Radiation, Oscillations and CosmologySpaceStandard candles

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Standard Candles

A star's luminosity $L$ is the total power it radiates. At distance $d$ , that power spreads over a sphere of area $4\pi d^2$ , so the radiant flux intensity received is $F=\dfrac L{4\pi d^2}$ . A standard candle is an object whose luminosity is known, such as a Cepheid variable star or a type Ia supernova. Measuring its flux then gives its distance: $d=\sqrt{\dfrac L{4\pi F}}$ .

Cambridge 9702 Paper 4Edexcel YPH11 WPH15AQA 7408 Paper 3B≈ 30 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Define luminosity and radiant flux intensity.
Use the inverse square law $F=\dfrac L{4\pi d^2}$ .
Explain what a standard candle is and give examples.
Use standard candles to find distances to galaxies.

Pause and think

Two stars look equally bright in the sky. Are they the same distance away?

Section 2 of 14

Core idea

Start hereHow far is a street light?

If you know a street lamp is 100 W, you can judge its distance by how dim it looks. Astronomers do the same with stars of known power: standard candles.

The big idea

$F=\frac L{4\pi d^2}$ . Known $L$ + measured $F$ → distance.

1. Inverse-square law

Light spreads over a sphere of area 4πd²

At twice the distance, the same power covers four times the area: a quarter of the flux.

F=\frac L{4\pi d^2}

$F$: radiant flux intensity (W m⁻²).
$L$: luminosity (W).

Worked exampleThe solar constant

$L=3.83\times10^{26}$ W, $d=1.5\times10^{11}$ m.

Goal: $F$ at Earth.
Strategy: Formula.

$4\pi(1.5\times10^{11})^2=2.83\times10^{23}\text{ m}^2$
Sphere area.
$F=1.36\times10^3\text{ W m}^{-2}$
Divide.

Answer

1.36

kW m

^{-2}

✓ CheckMatches the measured solar constant ✓.

2. Finding distance

Flux from a star against distance

Knowing

L

and measuring

F

d=\sqrt{\dfrac{L}{4\pi F}}

Worked exampleA Cepheid

$L=1.0\times10^{30}$ W, $F=2.0\times10^{-12}$ W m $^{-2}$ .

Goal: $d$ .
Strategy: $d=\sqrt{\frac L{4\pi F}}$ .

$4\pi F=2.51\times10^{-11}$
Denominator.
$d=\sqrt{\frac{10^{30}}{2.51\times10^{-11}}}=2.0\times10^{20}\text{ m}$
Square root.

Answer

2.0\times10^{20}

m (~21 000 light-years).

✓ CheckWithin our galaxy's scale ✓.

Common misconceptionForgetting the square root

d\propto\sqrt{L/F}

Pause and think

Two standard candles?

Summary

$F=L/(4\pi d^2)$ .
Known $L$ → distance.

Luminosity: Total power emitted.

HistoryLeavitt's law. In

1912

Henrietta Swan Leavitt, working as a "computer" at Harvard measuring photographic plates, discovered that brighter Cepheids pulse more slowly. Edwin Hubble later used her relationship to show that the Andromeda "nebula" was a separate galaxy far beyond the Milky Way.

Fun factBrighter than a galaxy. A type Ia supernova can briefly outshine its entire host galaxy of

100

billion stars. Astronomers in

1998

used them to discover that the expansion of the universe is speeding up, earning a Nobel Prize.

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