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How do we measure stars and the expanding universe?

Stellar luminosity and Wien's law and Stefan's law, the use of standard candles and parallax for distance, the redshift of galaxies, and Hubble's law and the expanding universe.

A focused answer to the Edexcel 9PH0 astrophysics content, covering stellar luminosity with Wien's and Stefan's laws, parallax and standard candles for distance, the redshift of galaxies, and Hubble's law and the expanding universe.

Generated by Claude Opus 4.812 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

Edexcel wants you to relate a star's luminosity to its temperature using Wien's and Stefan's laws, measure astronomical distances by parallax and standard candles, understand the redshift of galaxies, and use Hubble's law as evidence for an expanding universe.

The answer

Black-body radiation: Wien and Stefan

A star behaves like a black body, so its colour reveals its temperature: red stars are cooler, blue-white stars hotter. Stefan's law shows luminosity depends very strongly on temperature (the fourth power) and on size (the radius squared), which is why a hot, large star is overwhelmingly luminous.

Measuring distance: parallax and standard candles

Parallax works only for relatively nearby stars, where the angle is large enough to measure. For galaxies far beyond, astronomers use standard candles: comparing the known luminosity with the observed (much fainter) brightness gives the distance, since brightness falls off as the inverse square of distance.

Redshift and the expanding universe

The spectral lines of distant galaxies are shifted towards longer (redder) wavelengths, showing they move away. Crucially, the more distant the galaxy, the greater its redshift and recession speed.

Hubble's law

This proportionality is exactly what a uniformly expanding universe predicts: as space itself stretches, every galaxy sees all others receding, with the farthest receding fastest. Running the expansion backwards points to a hot, dense origin, the Big Bang, and the reciprocal of the Hubble constant gives a rough estimate of the age of the universe.

Examples in context

The Gaia space telescope has measured parallaxes for over a billion stars, mapping the Milky Way in three dimensions. Type Ia supernovae as standard candles revealed in the late 1990s that the expansion of the universe is accelerating, evidence for dark energy. Cepheid variables let Hubble first establish his law in the 1920s. Wien's law is used everywhere from classifying stars by colour to designing infrared cameras that detect the peak emission of warm bodies.

Try this

Q1. State Wien's displacement law in words. [1 mark]

  • Cue. The peak emission wavelength of a black body is inversely proportional to its absolute temperature.

Q2. A galaxy recedes at 3.0×1063.0 \times 10^{6} m per second with H0=2.3×10−18H_0 = 2.3 \times 10^{-18} per second. Find its distance. [2 marks]

  • Cue. d=vH0=3.0×1062.3×10−18=1.3×1024d = \frac{v}{H_0} = \frac{3.0 \times 10^{6}}{2.3 \times 10^{-18}} = 1.3 \times 10^{24} m.

Q3. Explain why parallax cannot be used to measure the distance to very distant galaxies. [2 marks]

  • Cue. The parallax angle becomes far too small to measure, so standard candles are used instead.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20194 marksA star emits radiation with a peak wavelength of 500500 nm. Use Wien's displacement law to calculate its surface temperature. Take the Wien constant as 2.9×10−32.9 \times 10^{-3} m K.
Show worked answer →

Wien's law: λmax⁡T=2.9×10−3\lambda_{\max} T = 2.9 \times 10^{-3} m K, so T=2.9×10−3λmax⁡T = \frac{2.9 \times 10^{-3}}{\lambda_{\max}}.

T=2.9×10−3500×10−9=2.9×10−35.0×10−7=5.8×103T = \frac{2.9 \times 10^{-3}}{500 \times 10^{-9}} = \frac{2.9 \times 10^{-3}}{5.0 \times 10^{-7}} = 5.8 \times 10^{3} K.

Markers reward λmax⁡T=\lambda_{\max} T = constant, conversion of nm to metres, and the value about 58005800 K.

Edexcel 20225 marksA distant galaxy shows a redshift corresponding to a recession speed of 1.5×1071.5 \times 10^{7} m per second. Using Hubble's law with H0=2.3×10−18H_0 = 2.3 \times 10^{-18} per second, determine the distance to the galaxy and explain how this supports an expanding universe.
Show worked answer →

Hubble's law: v=H0dv = H_0 d, so d=vH0d = \frac{v}{H_0}.

d=1.5×1072.3×10−18=6.5×1024d = \frac{1.5 \times 10^{7}}{2.3 \times 10^{-18}} = 6.5 \times 10^{24} m.

Support for expansion: almost all galaxies are redshifted, and the recession speed is proportional to distance (Hubble's law). This is exactly what is expected if space itself is expanding uniformly, carrying galaxies apart, with more distant galaxies receding faster.

Markers reward d=vH0d = \frac{v}{H_0}, the value about 6.5×10246.5 \times 10^{24} m, and the link from proportional recession to uniform expansion.

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