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The night sky has always been one of humanity's great laboratories. Long before the invention of the telescope, people knew that some stars looked brighter than others. With modern instruments we now understand that the brightness of a star as seen from Earth depends on two quite different things: the true power output of the star, and its distance from us. A dim star close by can look brighter than a brilliant supergiant thousands of light years away.
Module 5.5 of the OCR A-Level Physics A specification (H556) — Astrophysics and Cosmology — is where you learn to distinguish these two quantities and, using a handful of physical laws, measure the properties of stars from the light they send us. OCR makes astrophysics compulsory for all candidates (unlike AQA, where astrophysics is an optional module). You cannot avoid it, but you also do not need to — the physics is satisfying, and it builds on things you already know from earlier modules on waves, thermal physics and quantum physics.
This first lesson introduces the most important idea in stellar physics: luminosity. We shall define it carefully, distinguish it from apparent brightness, and see why stars have such enormous power outputs in the first place.
The luminosity L of a star is the total power radiated by the star, across all wavelengths, in every direction. It is measured in watts (W), just like any other power.
Luminosity is an intrinsic property of the star. A star has a definite luminosity regardless of where you happen to be standing; a Martian looking at the Sun and an Earthling looking at the Sun measure the same L, even though they see the Sun at different angular sizes and different apparent brightnesses. Only the star's physical properties — its mass, composition, age and evolutionary state — determine its luminosity.
For comparison, the luminosity of the Sun is
L_☉ = 3.83 × 10²⁶ W
This is the value given on the OCR data sheet. Astronomers often quote stellar luminosities in solar luminosities, as a convenient ratio L/L_☉. A star with L = 10 L_☉ radiates ten times the power of the Sun. A red supergiant like Betelgeuse has L ≈ 10⁵ L_☉; a dim red dwarf might have L ≈ 10⁻⁴ L_☉. Stellar luminosities therefore span nine orders of magnitude — which is one reason the Hertzsprung–Russell diagram (Lesson 5) uses a logarithmic scale.
The apparent brightness (or intensity, or flux) F of a star as seen from Earth is the power received per unit area at our location. It is measured in W m⁻². If you place a solar panel outside and point it at the star, the power absorbed per square metre is F (minus atmospheric losses).
Apparent brightness depends not only on the star's luminosity but also on how far away it is. Imagine a star as a point source emitting power L uniformly in all directions. At distance d, this power is spread over the surface of a sphere of radius d and area 4πd². Therefore
F = L / (4πd²)
This is the inverse-square law for light, and it follows directly from energy conservation. It is not a separate empirical law — it is a geometric consequence of the fact that energy is conserved and that a sphere of radius d has area 4πd².
graph LR
Star((Star<br/>L)) -- radiates --> S1[Sphere at d1<br/>area 4πd1²]
Star -- radiates --> S2[Sphere at 2d1<br/>area 16πd1²]
S1 -- "F1 = L/4πd1²" --> O1[Observer A]
S2 -- "F2 = F1/4" --> O2[Observer B]
If observer B is twice as far from the star as observer A, the observed brightness is not half — it is a quarter. Double the distance, one quarter the brightness. Ten times the distance, one hundredth the brightness.
Exam Tip: OCR specification 5.5.1(a) distinguishes "luminosity" (total power emitted) from "observed intensity" (power per unit area at the observer). You must use the correct term — examiners will sometimes deliberately mix them in a question to see whether you confuse them.
At the top of Earth's atmosphere, the intensity of sunlight is known as the solar constant:
F_Earth = 1.36 × 10³ W m⁻²
The Earth–Sun distance is d = 1.50 × 10¹¹ m (one astronomical unit). Using F = L/(4πd²):
L_☉ = 4π d² F
= 4π × (1.50 × 10¹¹)² × 1.36 × 10³
= 4π × 2.25 × 10²² × 1.36 × 10³
= 3.84 × 10²⁶ W
This agrees with the tabulated value L_☉ = 3.83 × 10²⁶ W to three significant figures — all we had to do was measure the solar constant and know the Earth–Sun distance, and we could compute the total power of the Sun. This is the essence of astrophysical measurement: geometry turns a local observation into a global statement about a distant object.
Suppose a star has luminosity L = 2.00 × 10²⁸ W (around 50 times the Sun) and lies at a distance d = 100 light years from Earth. Given that 1 light year = 9.46 × 10¹⁵ m, compute the observed intensity at Earth.
Step 1. Convert distance to metres:
d = 100 × 9.46 × 10¹⁵ = 9.46 × 10¹⁷ m
Step 2. Apply the inverse-square law:
F = L / (4π d²)
= (2.00 × 10²⁸) / (4π × (9.46 × 10¹⁷)²)
= (2.00 × 10²⁸) / (4π × 8.95 × 10³⁵)
= (2.00 × 10²⁸) / (1.12 × 10³⁷)
= 1.78 × 10⁻⁹ W m⁻²
Although this star emits 50 times as much power as the Sun, its observed brightness is about 10⁻¹² times that of the Sun, because it is roughly 6.3 × 10⁶ times further away, and the 1/d² factor suppresses the flux by a factor of 4 × 10¹³.
Why do stars shine in the first place? In the nineteenth century, before the discovery of nuclear physics, astronomers struggled to explain the Sun's power output. Chemical burning (coal, say) would exhaust the Sun in a few thousand years. Gravitational contraction — the Kelvin–Helmholtz mechanism, in which the Sun releases gravitational potential energy by shrinking — could only sustain the Sun for tens of millions of years. Yet geological evidence from fossils and strata showed the Earth (and therefore the Sun) to be billions of years old.
The resolution came with the discovery of the atomic nucleus and of Einstein's mass–energy equivalence E = mc². Stars are powered by nuclear fusion in their cores, where very high temperatures (around 1.5 × 10⁷ K in the Sun) and densities drive hydrogen nuclei together against their mutual electric repulsion until the strong nuclear force takes over and fuses them. In the Sun, the dominant reaction is the proton–proton chain, whose net effect is
4 ¹H → ⁴He + 2 e⁺ + 2 ν_e + 2 γ
The mass of four protons slightly exceeds the mass of one helium nucleus (plus two positrons and two neutrinos); the tiny mass difference, about 0.7% per reaction, is released as energy via E = mc². In each second, the Sun converts around 4.3 × 10⁹ kg of mass into energy, radiating that energy as photons, neutrinos and the kinetic energy of ejected particles. This prodigious mass-to-energy conversion is what makes stars shine for billions of years rather than thousands.
You met the basic ideas of fusion in Module 6 (Nuclear and Particle Physics). Here we are interested in the astronomical consequences: fusion sets the power output, and hence the luminosity, of any main-sequence star. A more massive star can squeeze its core to higher temperatures and therefore burns fuel far faster than a lower-mass star — so massive stars are both more luminous and shorter lived.
For main-sequence stars, the luminosity is a steep function of the mass. Observationally,
L ∝ M^α with α ≈ 3.5 for intermediate-mass main-sequence stars
This relation is not required by OCR for explicit calculation, but it is useful background for understanding why the Hertzsprung–Russell diagram (Lesson 5) has the shape it does, and why high-mass stars evolve so much faster than low-mass ones. A 10-solar-mass star is roughly 10^3.5 ≈ 3000 times more luminous than the Sun but contains only ten times more fuel; it therefore exhausts its core hydrogen in about 10⁷ years, compared with 10¹⁰ years for the Sun.
Specification 5.5.1 asks you to:
(a) understand that stars are sources of electromagnetic radiation across a wide range of wavelengths, powered by nuclear fusion;
(b) know and use luminosity as the total power output of a star in watts;
(c) distinguish luminosity from observed intensity (often called flux or apparent brightness), and apply the inverse-square law F = L/(4πd²);
(d) prepare the ground for the Stefan–Boltzmann law (Lesson 2) and Wien's displacement law (Lesson 3), which together allow you to calculate stellar properties from observed spectra.
You will not be asked to derive the proton–proton chain or carry out mass-energy calculations for fusion in this module — those come in Module 6. In 5.5, you treat luminosity as a given quantity and use it to work out distances, radii and temperatures.
A particular star is known (from spectral analysis) to have L = 4.0 × 10²⁷ W. The observed intensity at Earth is F = 2.0 × 10⁻⁸ W m⁻². Find the distance to the star.
Rearranging F = L/(4πd²):
d² = L / (4π F)
= (4.0 × 10²⁷) / (4π × 2.0 × 10⁻⁸)
= (4.0 × 10²⁷) / (2.51 × 10⁻⁷)
= 1.59 × 10³⁴ m²
d = 1.26 × 10¹⁷ m
≈ 13 light years
This is roughly the distance to Tau Ceti. Notice that we had to know L independently — from the shape of the spectrum, combined with Stefan–Boltzmann and Wien's law, or from some other "standard candle" argument. Measuring F is easy (you just point a photometer); measuring d the hard way (via parallax) is separate. Knowing L from the spectrum and then extracting d is one of the fundamental techniques of astrophysics.
A quick reminder of the units you will need throughout this course:
| Quantity | Symbol | SI unit | Typical stellar value |
|---|---|---|---|
| Luminosity | L | W | 10²³ to 10³² W |
| Observed intensity | F | W m⁻² | 10⁻¹⁵ to 10³ W m⁻² |
| Distance | d | m | 10¹⁶ to 10²⁵ m |
| Stellar radius | r | m | 10⁶ to 10¹² m |
| Surface temperature | T | K | 2500 to 50 000 K |
| Peak wavelength | λ_max | m | 10⁻⁸ to 10⁻⁶ m |
Always convert to SI units before substituting into formulas. A frequent source of error is leaving a distance in light years or astronomical units — both are valid astronomical units, but neither is an SI unit, and the Stefan–Boltzmann and inverse-square laws are both dimensional.
Throughout this course, the Sun serves as a useful reference. Its tabulated properties are:
| Quantity | Symbol | Value |
|---|---|---|
| Luminosity | L_☉ | 3.83 × 10²⁶ W |
| Radius | R_☉ | 6.96 × 10⁸ m |
| Surface temperature | T_☉ | 5800 K |
| Mass | M_☉ | 1.99 × 10³⁰ kg |
| Peak wavelength | λ_max | approx 500 nm |
Even if the numbers differ slightly from what is given on your actual exam data sheet, remember these to one or two significant figures; they anchor your intuition when you encounter larger or smaller stars.
4πd².4π in the inverse-square law. F = L/(4πd²), not F = L/d². The 4π comes from the surface area of a sphere, and is essential.F = L/(4πd²).10⁻⁹ or so, because the Earth subtends a tiny solid angle from the Sun. The rest goes off into space in every other direction.L of a star is its total radiated power in watts, independent of observer.F is the power per unit area at Earth, related to luminosity by the inverse-square law F = L/(4πd²).E = mc².L_☉ = 3.83 × 10²⁶ W serves as a convenient unit for expressing stellar power outputs.