Determining Stellar Properties

Spec mapping: OCR H556 Module 5.5 — Astrophysics and Cosmology (combining Stefan–Boltzmann $L = 4\pi r^{2}\sigma T^{4}$L=4πr2σT4, Wien's displacement law $\lambda_{\max}T = 2.90 \times 10^{-3}$λmax​T=2.90×10−3 m K and the inverse-square law $F = L/(4\pi d^{2})$F=L/(4πd2) to determine stellar properties — radius, surface temperature and luminosity — from observations of spectra, fluxes and parallax distances). Refer to the official OCR H556 specification document for exact wording.

So far we have met three independent pieces of physics:

• The inverse-square law for the observed intensity of a star: F = L/(4πd²) (Lesson 1).
• The Stefan–Boltzmann law for a black body's luminosity: L = 4πr²σT⁴ (Lesson 2).
• Wien's displacement law for the peak wavelength of a black body's spectrum: λ_max × T = 2.90 × 10⁻³ m K (Lesson 3).

Individually, each law relates two or three quantities. Combined, they allow us to determine the full set of basic stellar properties — radius, temperature, and luminosity — from observations of a star's spectrum and brightness. This lesson shows how to put the laws together in a single worked-example-driven exercise, which is exactly the kind of problem that OCR loves to set in A-Level Physics A papers.

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The Chain Of Inferences

To determine a star's radius, temperature and luminosity, an astronomer typically uses the following chain of measurements:

graph TD
    A[Measure spectrum] -- Wien's law --> B[Temperature T]
    A -- absolute flux --> C[Observed F at Earth]
    C -- distance d --> D[Luminosity L]
    B -- Stefan-Boltzmann --> E[Radius r from L and T]
    D -- Stefan-Boltzmann --> E

• Temperature comes from Wien's law applied to the peak of the spectrum.
• Luminosity comes from the inverse-square law applied to the observed intensity, provided we know the distance d.
• Radius follows from the Stefan–Boltzmann law, given L and T.

Each of these steps requires two inputs and produces one output. The question on an exam paper will typically give you exactly what you need — no more, no less — and expect you to identify which laws to apply in what order.

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Worked Example 1: A Sun-Like Star

A nearby star has observed peak wavelength λ_max = 480 nm and observed intensity at Earth F = 1.20 × 10⁻⁸ W m⁻². Its distance from Earth is known to be d = 3.26 light years (= 3.08 × 10¹⁶ m). Find:

(a) the surface temperature; (b) the luminosity; (c) the radius; (d) express the radius in solar radii.

(a) Temperature from Wien's law.

$$\begin{aligned} T = (2.90 \times 10^{-3}) / \lambda _max \\ = (2.90 \times 10^{-3}) / (480 \times 10^{-9}) \\ = 6040 K \end{aligned}$$T=(2.90×10−3)/λm​ax=(2.90×10−3)/(480×10−9)=6040K​

This is about 240 K hotter than the Sun — a slightly hotter late-F or early-G star.

(b) Luminosity from inverse-square law.

$$\begin{aligned} L = 4 \pi d^{2} F \\ = 4 \pi \times (3.08 \times 10^{16})^{2} \times (1.20 \times 10^{-8}) \\ = 4 \pi \times 9.49 \times 10^{32} \times 1.20 \times 10^{-8} \\ = 1.43 \times 10^{26} W \end{aligned}$$L=4πd2F=4π×(3.08×1016)2×(1.20×10−8)=4π×9.49×1032×1.20×10−8=1.43×1026W​

(c) Radius from Stefan–Boltzmann.

Rearrange L = 4πr²σT⁴:

$$r^{2} = L / (4 \pi \sigma T^{4})$$r2=L/(4πσT4)

Compute T⁴:

$$\begin{aligned} T^{4} = (6040)^{4} \\ = 1.33 \times 10^{15} K^{4} \end{aligned}$$T4=(6040)4=1.33×1015K4​

Denominator:

$$\begin{aligned} 4 \pi \sigma T^{4} = 4 \pi \times (5.67 \times 10^{-8}) \times (1.33 \times 10^{15}) \\ = 4 \pi \times 7.54 \times 10^{7} \\ = 9.47 \times 10^{8} W m^{-2} \end{aligned}$$4πσT4=4π×(5.67×10−8)×(1.33×1015)=4π×7.54×107=9.47×108Wm−2​

Numerator:

$$\begin{aligned} r^{2} = (1.43 \times 10^{26}) / (9.47 \times 10^{8}) \\ = 1.51 \times 10^{17} m^{2} \\ r = 3.89 \times 10^{8} m \end{aligned}$$r2=(1.43×1026)/(9.47×108)=1.51×1017m2r=3.89×108m​

(d) In solar radii:

$$\begin{aligned} r / R_☉ = (3.89 \times 10^{8}) / (6.96 \times 10^{8}) \\ = 0.559 \end{aligned}$$r/R☉​=(3.89×108)/(6.96×108)=0.559​

So this star has radius about 0.56 R_☉ and temperature about 6000 K. It is intrinsically less luminous than the Sun (\approx 0.37 L_☉) because its smaller size more than compensates for its hotter temperature. This star is likely to be a young low-mass main-sequence star.

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Worked Example 2: A Red Giant

A star is observed with λ_max = 1.10 × 10⁻⁶ m (1100 nm, near infrared) and intensity at Earth F = 4.50 × 10⁻¹⁰ W m⁻². Its distance is d = 200 light years (= 1.89 × 10¹⁸ m). Find (a) temperature, (b) luminosity, (c) radius.

(a) Temperature.

$$\begin{aligned} T = (2.90 \times 10^{-3}) / (1.10 \times 10^{-6}) \\ = 2640 K \end{aligned}$$T=(2.90×10−3)/(1.10×10−6)=2640K​

A cool red star — consistent with a red giant or red dwarf, but we need luminosity to distinguish them.

(b) Luminosity.

$$\begin{aligned} L = 4 \pi d^{2} F \\ = 4 \pi \times (1.89 \times 10^{18})^{2} \times (4.50 \times 10^{-10}) \\ = 4 \pi \times 3.57 \times 10^{36} \times 4.50 \times 10^{-10} \\ = 2.02 \times 10^{28} W \end{aligned}$$L=4πd2F=4π×(1.89×1018)2×(4.50×10−10)=4π×3.57×1036×4.50×10−10=2.02×1028W​

In solar units: L/L_☉ = (2.02 × 10²⁸)/(3.83 × 10²⁶) \approx 53. This is 50 solar luminosities — unmistakably a giant. Red dwarfs are 10⁻² L_☉ or less; a red star 50 times more luminous than the Sun must be a red giant.

(c) Radius.

$$\begin{aligned} T^{4} = (2640)^{4} = 4.86 \times 10^{13} K^{4} \\ 4 \pi \sigma T^{4} = 4 \pi \times (5.67 \times 10^{-8}) \times (4.86 \times 10^{13}) \\ = 4 \pi \times 2.76 \times 10^{6} \\ = 3.46 \times 10^{7} W m^{-2} \\ r^{2} = L / (4 \pi \sigma T^{4}) \\ = (2.02 \times 10^{28}) / (3.46 \times 10^{7}) \\ = 5.84 \times 10^{20} m^{2} \\ r = 2.42 \times 10^{10} m \end{aligned}$$T4=(2640)4=4.86×1013K44πσT4=4π×(5.67×10−8)×(4.86×1013)=4π×2.76×106=3.46×107Wm−2r2=L/(4πσT4)=(2.02×1028)/(3.46×107)=5.84×1020m2r=2.42×1010m​

In solar radii:

$$r / R_☉ = (2.42 \times 10^{10}) / (6.96 \times 10^{8}) = 34.7$$r/R☉​=(2.42×1010)/(6.96×108)=34.7

So this star is around 35 solar radii. That is about the size of Aldebaran — a textbook red giant. The star is cooler per unit area than the Sun (which reduces σT⁴), but its surface area is \approx 1200 times larger, so its total luminosity is much greater. This is the classic signature of a red giant: cooler but vastly larger, hence more luminous.

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Worked Example 3: A White Dwarf

A white dwarf has observed intensity F = 2.0 × 10⁻¹³ W m⁻² at a distance d = 2.5 pc = 7.72 × 10¹⁶ m. Its observed spectrum peaks at λ_max = 300 nm (ultraviolet). Find temperature, luminosity, and radius.

(a) Temperature.

$$\begin{aligned} T = (2.90 \times 10^{-3}) / (300 \times 10^{-9}) \\ = 9670 K \end{aligned}$$T=(2.90×10−3)/(300×10−9)=9670K​

(b) Luminosity.

$$\begin{aligned} L = 4 \pi \times (7.72 \times 10^{16})^{2} \times (2.0 \times 10^{-13}) \\ = 4 \pi \times 5.96 \times 10^{33} \times 2.0 \times 10^{-13} \\ = 1.50 \times 10^{22} W \end{aligned}$$L=4π×(7.72×1016)2×(2.0×10−13)=4π×5.96×1033×2.0×10−13=1.50×1022W​

In solar units: L/L_☉ \approx 4 × 10⁻⁵. A very faint object.

(c) Radius.

$$\begin{aligned} T^{4} = (9670)^{4} = 8.75 \times 10^{15} K^{4} \\ 4 \pi \sigma T^{4} = 4 \pi \times (5.67 \times 10^{-8}) \times (8.75 \times 10^{15}) \\ = 4 \pi \times 4.96 \times 10^{8} \\ = 6.23 \times 10^{9} W m^{-2} \\ r^{2} = (1.50 \times 10^{22}) / (6.23 \times 10^{9}) \\ = 2.41 \times 10^{12} m^{2} \\ r = 1.55 \times 10^{6} m \end{aligned}$$T4=(9670)4=8.75×1015K44πσT4=4π×(5.67×10−8)×(8.75×1015)=4π×4.96×108=6.23×109Wm−2r2=(1.50×1022)/(6.23×109)=2.41×1012m2r=1.55×106m​

In solar radii: r/R_☉ \approx 2.2 × 10⁻³. So this is an object roughly 0.002 solar radii — or about one quarter the size of Earth. Despite having a surface temperature almost twice that of the Sun, its tiny size makes it intrinsically faint.

This is the signature of a white dwarf: hot but small, therefore hot per unit area but low in total luminosity. In Lessons 5 and 6 we shall see how white dwarfs fit into the Hertzsprung–Russell diagram and where they come from in stellar evolution.

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Pattern: Three Classes Of Star At A Glance

The three worked examples above span the main categories of stars you will encounter on the HR diagram (Lesson 5):

Star type	T (K)	L (W)	r (m)	r in R_☉
Sun-like (Ex. 1)	6040	1.43 × 10²⁶	3.89 × 10⁸	0.56
Red giant (Ex. 2)	2640	2.02 × 10²⁸	2.42 × 10¹⁰	34.7
White dwarf (Ex. 3)	9670	1.50 × 10²²	1.55 × 10⁶	0.0022

Notice:

• The red giant is the coolest but the most luminous.
• The white dwarf is hotter than the Sun-like star but far fainter.
• The temperature alone does not tell you luminosity; you also need the radius. The Stefan–Boltzmann law is the bridge.

This is exactly why the HR diagram (Lesson 5) plots luminosity against temperature: the combination reveals the type of star, not either quantity alone.

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Worked Example 4: Combining Laws With Unknown Distance

A star is observed with λ_max = 290 nm and F = 1.00 × 10⁻¹¹ W m⁻². Its radius has been measured (e.g. from interferometry or binary-star observations) to be r = 1.50 × 10⁹ m. Find the distance to the star.

Step 1. Temperature from Wien's law:

$$\begin{aligned} T = (2.90 \times 10^{-3}) / (2.90 \times 10^{-7}) \\ = 10 000 K \end{aligned}$$T=(2.90×10−3)/(2.90×10−7)=10000K​

Step 2. Luminosity from Stefan–Boltzmann:

$$\begin{aligned} L = 4 \pi r^{2} \sigma T^{4} \\ = 4 \pi \times (1.50 \times 10^{9})^{2} \times (5.67 \times 10^{-8}) \times (10^{4})^{4} \\ = 4 \pi \times 2.25 \times 10^{18} \times 5.67 \times 10^{-8} \times 10^{16} \\ = 4 \pi \times 1.28 \times 10^{27} \\ = 1.60 \times 10^{28} W \end{aligned}$$L=4πr2σT4=4π×(1.50×109)2×(5.67×10−8)×(104)4=4π×2.25×1018×5.67×10−8×1016=4π×1.28×1027=1.60×1028W​

Step 3. Distance from inverse-square law:

$$\begin{aligned} F &= \dfrac{L}{4\pi d^{2}} \\ d^{2} &= \dfrac{L}{4\pi F} \\ &= \dfrac{1.60 \times 10^{28}}{4\pi \times 10^{-11}} \\ &= \dfrac{1.60 \times 10^{28}}{1.26 \times 10^{-10}} \\ &= 1.27 \times 10^{38}\,\text{m}^{2} \\ d &= 1.13 \times 10^{19}\,\text{m} \approx 1200 \text{ light years} \end{aligned}$$Fd2d​=4πd2L​=4πFL​=4π×10−111.60×1028​=1.26×10−101.60×1028​=1.27×1038m2=1.13×1019m≈1200 light years​

This is called a spectroscopic distance — the distance is deduced from assumed luminosity rather than from geometric parallax. It is one of the workhorses of stellar astronomy beyond the few hundred parsecs where trigonometric parallax fails.

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A Visual Anchor: Decision Tree For Stellar Property Problems

flowchart TD
  A[Stellar property question] --> B{What is given?}
  B -- λ_max only --> C[Wien: T = b/λ_max]
  B -- F and d --> D["Inverse-square: L = 4π d² F"]
  B -- λ_max and F and d --> E[Full chain — most common]
  B -- λ_max and r --> F["Wien → T; SB → L = 4π r² σ T⁴"]
  E --> G[Wien for T → IS-law for L → SB for r]
  C --> H[Quote T to 3 s.f.]
  D --> H
  F --> H
  G --> H
  H --> I[Sanity-check vs Sun: T ≈ 5800 K, L_⊙ = 3.83e26 W, R_⊙ = 6.96e8 m]

This decision tree compresses the entire toolkit. Identify which two or three of the four key observables you have ($\lambda_{\max}$λmax​, $F$F, $d$d, $r$r) and read off the route. OCR questions are almost always designed so that exactly the right inputs are provided — your job is to recognise the pattern.

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Synoptic Links

• Lessons 1, 2, 3 combined: The full inference chain $\lambda_{\max} \to T$λmax​→T (Wien) $\to L = 4\pi d^{2}F$→L=4πd2F (inverse-square) $\to r = \sqrt{L/(4\pi\sigma T^{4})}$→r=L/(4πσT4)​ (Stefan–Boltzmann). Each link is a single equation; the art is recognising which link comes next.
• Parallax distance (Module 5.5 background): For nearby stars, $d$d comes from trigonometric parallax — the apparent shift of a star against the background as Earth orbits the Sun. The Hipparcos and Gaia missions have measured parallaxes for $\sim 10^{9}$∼109 stars; the resulting distances feed directly into the inverse-square law of Lesson 1.
• Hertzsprung–Russell diagram (Lesson 5): Once $T$T and $L$L are known for many stars, plotting them produces the HR diagram — the master visualisation of stellar physics. Lesson 5 explores what its structure tells us about stellar populations and evolution.

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Specimen question modelled on the OCR H556 paper format

Question (12 marks): A star has measured peak wavelength $\lambda_{\max} = 4.83 \times 10^{-7}$λmax​=4.83×10−7 m, observed intensity at Earth $F = 1.50 \times 10^{-9}$F=1.50×10−9 W m$^{-2}$−2, and known distance (from parallax) $d = 4.00 \times 10^{17}$d=4.00×1017 m. Take $\sigma = 5.67 \times 10^{-8}$σ=5.67×10−8 W m$^{-2}$−2 K$^{-4}$−4, $b = 2.90 \times 10^{-3}$b=2.90×10−3 m K, $L_{\odot} = 3.83 \times 10^{26}$L⊙​=3.83×1026 W, $R_{\odot} = 6.96 \times 10^{8}$R⊙​=6.96×108 m.

(a) Calculate the surface temperature $T$T of the star. [2]

(b) Calculate the luminosity $L$L of the star. [2]

(c) Calculate the radius $r$r of the star. [3]

(d) Express $L$L and $r$r as multiples of $L_{\odot}$L⊙​ and $R_{\odot}$R⊙​. Identify the most likely stellar type. [3]

(e) The student measures the star's parallax angle as $p = 0.005$p=0.005 arcseconds. If the true parallax was actually $0.006$0.006 arcseconds, by what factor was the inferred luminosity in (b) in error? [2]

AO breakdown