Inverse Square and Stefan's Law: Stellar Properties Revision Notes | A Level AQA Physics

1. Why Distance Changes Everything

Picture a single light bulb in a dark room. Stand close and it dazzles you; walk to the far wall and it looks dim. The bulb has not changed — you have moved. This is the core puzzle of stellar astrophysics: how do we work out the true power output of a star when all we can measure is how bright it appears from Earth?

The answer lies in two laws that connect what we observe to what is physically happening at the star's surface. The inverse square law tells us how brightness falls off with distance. Stefan's law tells us how a star's power output depends on its size and temperature. Together, they let astronomers classify stars with remarkable precision.

2. The Inverse Square Law

Imagine a star radiating energy equally in all directions. At a distance $d$ from the star, that energy is spread over the surface of a sphere of radius $d$. The surface area of that sphere is $4\pi d^2$, so the power arriving per square metre — the flux, $F$ — is the total luminosity $L$ divided by that area.

Notice the $d^2$ in the denominator. Double the distance and the flux drops to one quarter. Triple the distance and it falls to one ninth. This rapid decrease is why even relatively nearby stars appear faint compared with the Sun.

These assumptions matter in exam questions. If a question asks you to state an assumption, the two most credit-worthy answers are that the star radiates isotropically and that there is no absorption of radiation between the star and the Earth.

3. Stefan's Law: Connecting Power, Size, and Temperature

Now that we can link observed flux to luminosity, we need a way to connect luminosity to the physical properties of the star itself. This is where Stefan's law becomes essential. A star behaves approximately as a black body — a perfect emitter of radiation — and for a black body, the power radiated per unit surface area depends only on temperature.

The $T^4$ dependence is the key insight. Temperature has a far more dramatic effect on luminosity than size does. Double a star's radius and its luminosity increases by a factor of four (because area scales as $r^2$). Double its temperature and its luminosity increases by a factor of $2^4 = 16$. Hot stars are extraordinarily powerful even if they are not particularly large.

For a spherical star of radius $r$, the surface area is $A = 4\pi r^2$, so Stefan's law becomes $P = 4\pi r^2 \sigma T^4$. This combined form is the version you will use in almost every calculation.

4. Comparing Stars Using Stefan's Law

One of the most powerful uses of Stefan's law is comparing two stars without needing to know the value of $\sigma$ at all. By forming a ratio, the constant cancels and you are left with a clean relationship between the observable quantities.

For two stars labelled 1 and 2, dividing their Stefan's law expressions gives:

This ratio approach is a favourite in AQA exam questions. You are often given two of the three quantities (luminosity, radius, temperature) for each star and asked to find the third. Setting up the ratio first, before substituting numbers, keeps the algebra clean and avoids errors with large powers of ten.

5. Worked Calculations

Theory becomes exam-ready through practice. The following worked examples cover the two most common question types: using the inverse square law to find luminosity from flux, and using Stefan's law to find a stellar radius. Work through each step yourself before revealing the next one.

Worked example: Finding Stellar Luminosity from Observed Flux

The star Sirius A has an observed flux at Earth of $F = 1.13 \times 10^{-7}$ W m⁻². The distance from Earth to Sirius A is $d = 8.14 \times 10^{16}$ m. Calculate the luminosity of Sirius A. [3 marks]

1. $F = \dfrac{L}{4\pi d^2}$ — [M1] Stating the correct formula earns the method mark. Rearrange to make $L$ the subject: $L = 4\pi d^2 F$.
2. $d^2 = (8.14 \times 10^{16})^2 = 8.14 \times 8.14 \times 10^{32}$ $8.14 \times 8.14 = 66.26$ $d^2 = 6.626 \times 10^{33}$ m² — Expand the square explicitly: $8.14 \times 8.14 = 66.26$, then adjust the power: $(10^{16})^2 = 10^{32}$, giving $6.626 \times 10^{33}$ m².
3. $L = 4\pi \times 6.626 \times 10^{33} \times 1.13 \times 10^{-7}$ $= 4 \times 3.1416 \times 6.626 \times 10^{33} \times 1.13 \times 10^{-7}$ — [M1] Correct substitution with units. Keep all values in scientific notation to avoid arithmetic errors.
4. $4 \times 3.1416 = 12.566$ $12.566 \times 6.626 = 83.25$ $83.25 \times 1.13 = 94.07$ $L = 94.07 \times 10^{33-7} = 94.07 \times 10^{26}$ $L \approx 9.41 \times 10^{27}$ W — [A1] Final answer with correct unit. The Sun's luminosity is $3.85 \times 10^{26}$ W, so Sirius A is roughly 24 times more luminous — a sensible order-of-magnitude check.

Answer: $L \approx 9.4 \times 10^{27}$ W

Worked example: Finding a Star's Radius Using Stefan's Law

A star has a luminosity of $P = 3.0 \times 10^{29}$ W and a surface temperature of $T = 12\,000$ K. Calculate the radius of the star. ($\sigma = 5.67 \times 10^{-8}$ W m⁻² K⁻⁴) [4 marks]

1. $P = 4\pi r^2 \sigma T^4$ — [M1] The surface area of a spherical star is $4\pi r^2$. Substituting this into $P = \sigma A T^4$ gives the working formula.
2. $r^2 = \dfrac{P}{4\pi \sigma T^4}$ — [M1] Isolate $r^2$ before substituting numbers. This avoids carrying a square root through messy arithmetic.
3. $T^4 = (12\,000)^4 = (1.2 \times 10^4)^4$ $= 1.2^4 \times 10^{16}$ $1.2^2 = 1.44$ $1.44^2 = 2.0736$ $T^4 = 2.0736 \times 10^{16}$ K⁴ — Work out $1.2^4$ in two squaring steps to avoid errors: $1.2 \times 1.2 = 1.44$, then $1.44 \times 1.44 = 2.0736$.
4. $r^2 = \dfrac{3.0 \times 10^{29}}{4\pi \times 5.67 \times 10^{-8} \times 2.0736 \times 10^{16}}$ Denominator: $4 \times 3.1416 = 12.566$ $12.566 \times 5.67 \times 10^{-8} = 71.25 \times 10^{-8} = 7.125 \times 10^{-7}$ $7.125 \times 10^{-7} \times 2.0736 \times 10^{16} = 14.77 \times 10^{9} = 1.477 \times 10^{10}$ $r^2 = \dfrac{3.0 \times 10^{29}}{1.477 \times 10^{10}} = 2.031 \times 10^{19}$ m² — Build the denominator step by step, keeping track of powers of ten separately from the numerical coefficients.
5. $r = \sqrt{2.031 \times 10^{19}}$ $= \sqrt{2.031} \times \sqrt{10^{19}}$ $= 1.425 \times 10^{9.5}$ $= 1.425 \times 3.162 \times 10^{9}$ $\approx 4.5 \times 10^{9}$ m — [A1] $\sqrt{10^{19}} = 10^{9.5} = 10^9 \times \sqrt{10} \approx 10^9 \times 3.162$. The Sun's radius is $6.96 \times 10^8$ m, so this star has a radius roughly 6–7 times the Sun's — consistent with a hot, luminous star.

Answer: $r \approx 4.5 \times 10^9$ m

6. Combining Both Laws: Finding Stellar Properties

The real power of these two laws comes when you use them together. From an observation, you can measure the flux $F$ and, if the distance $d$ is known (from parallax or other methods), calculate the luminosity $L$ using the inverse square law. Then, if the surface temperature $T$ is known (from the star's spectrum and Wien's law), you can substitute $L$ and $T$ into Stefan's law to find the star's radius $r$.

This chain of reasoning — observation to luminosity to radius — is how astronomers determine the physical size of stars that are far too distant to resolve directly. The logical flow is worth memorising as a method, not just as individual equations.

• Measure flux $F$ and distance $d$ → calculate luminosity: $L = 4\pi d^2 F$
• Determine surface temperature $T$ from the star's spectrum (peak wavelength via Wien's law)
• Substitute $L$ and $T$ into $L = 4\pi r^2 \sigma T^4$ → solve for radius $r$
• Compare $r$ and $T$ with the Sun's values to classify the star (main sequence, giant, dwarf)

7. Exam Technique and Common Mistakes

AQA questions on this topic tend to follow predictable patterns. Knowing where marks are won and lost is as important as knowing the physics. The guidance below consolidates the most credit-relevant points from across this lesson.

Taken together, the inverse square law and Stefan's law form the quantitative backbone of stellar classification. The inverse square law connects what we observe (flux) to what the star actually produces (luminosity), provided we know the distance and accept the stated assumptions. Stefan's law then connects that luminosity to the star's physical size and surface temperature, allowing astronomers to place stars on the Hertzsprung–Russell diagram with confidence.