Determining Stellar Properties

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.

T = (2.90 × 10⁻³) / λ_max
  = (2.90 × 10⁻³) / (480 × 10⁻⁹)
  = 6040 K

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

(b) Luminosity from inverse-square law.

L = 4π d² F
  = 4π × (3.08 × 10¹⁶)² × (1.20 × 10⁻⁸)
  = 4π × 9.49 × 10³² × 1.20 × 10⁻⁸
  = 1.43 × 10²⁶ W

(c) Radius from Stefan–Boltzmann.

Rearrange L = 4πr²σT⁴:

r² = L / (4π σ T⁴)

Compute T⁴:

T⁴ = (6040)⁴
   = 1.33 × 10¹⁵ K⁴

Denominator:

4π σ T⁴ = 4π × (5.67 × 10⁻⁸) × (1.33 × 10¹⁵)
         = 4π × 7.54 × 10⁷
         = 9.47 × 10⁸ W m⁻²

Numerator:

r² = (1.43 × 10²⁶) / (9.47 × 10⁸)
   = 1.51 × 10¹⁷ m²
r = 3.89 × 10⁸ m

(d) In solar radii:

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.

T = (2.90 × 10⁻³) / (1.10 × 10⁻⁶)
  = 2640 K

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

(b) Luminosity.

L = 4π d² F
  = 4π × (1.89 × 10¹⁸)² × (4.50 × 10⁻¹⁰)
  = 4π × 3.57 × 10³⁶ × 4.50 × 10⁻¹⁰
  = 2.02 × 10²⁸ W

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.

T⁴ = (2640)⁴ = 4.86 × 10¹³ K⁴
4π σ T⁴ = 4π × (5.67 × 10⁻⁸) × (4.86 × 10¹³)
        = 4π × 2.76 × 10⁶
        = 3.46 × 10⁷ W m⁻²

r² = L / (4π σ T⁴)
   = (2.02 × 10²⁸) / (3.46 × 10⁷)
   = 5.84 × 10²⁰ m²

r = 2.42 × 10¹⁰ m

In solar radii:

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.

T = (2.90 × 10⁻³) / (300 × 10⁻⁹)
  = 9670 K

(b) Luminosity.

L = 4π × (7.72 × 10¹⁶)² × (2.0 × 10⁻¹³)
  = 4π × 5.96 × 10³³ × 2.0 × 10⁻¹³
  = 1.50 × 10²² W