Specification: AQA A-level Physics (7408) 3.9.2.3 Astrophysics option | Paper: 3 Section B | Data and formulae booklet: P = σAT⁴ is given in the exam (and the area of a sphere, 4πr², with the geometrical equations); I = P ÷ 4πd² is not given, so learn it
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📌 KEY DEFINITIONS – learn them word for word
Stefan’s law: the total power emitted per unit surface area of a black body is directly proportional to the fourth power of its absolute temperature, so P = σAT⁴.
Luminosity: the total power output of a star, the energy it radiates per second across all wavelengths, in watts. For a spherical star of radius r the surface area is A = 4πr².
Intensity and the inverse square law: intensity is the power received per unit area perpendicular to the direction of the radiation. At a distance d from a star, I = P ÷ 4πd², assuming no absorption on the way. The solar constant, the intensity of the Sun’s radiation at the top of Earth’s atmosphere, is about 1.4 kW m−2.
The equation
power = Stefan constant × surface area × (absolute temperature)⁴ P = σAT⁴, with A = 4πr² for a star and intensity = power ÷ (4π × distance²) I = P ÷ 4πd²
| Quantity | Symbol | Unit |
|---|---|---|
| power output (luminosity) | P | watts (W) |
| Stefan constant, 5.67 × 10−8 | σ | W m−2 K−4 |
| surface area | A | square metres (m²) |
| absolute surface temperature | T | kelvin (K) |
| radius of the star | r | metres (m) |
| intensity received | I | W m−2 |
| distance from the star | d | metres (m) |
Rearranged: T = (P ÷ σA)¼ | r = √(P ÷ 4πσT⁴) | P = 4πd²I | d = √(P ÷ 4πI) | For ratios, P ∝ r²T⁴: doubling the temperature multiplies the power by 16, doubling the radius multiplies it by 4.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) The Sun has a radius of 6.96 × 108 m and a surface temperature of 5800 K. Calculate its luminosity, to 2 significant figures. (σ = 5.67 × 10−8 W m−2 K−4)
Show solution
Step 1 – surface area
A = 4πr² = 4π × (6.96 × 108)² = 6.087… × 1018 m²
Step 2 – power
F: P = σAT⁴
I: P = 5.67 × 10−8 × 6.087… × 1018 × 5800⁴
F: P is already the subject
A: 3.905… × 1026 W = 3.9 × 1026 W (2 s.f.)
Watch out: forgetting to raise T to the fourth power, or squaring the radius but forgetting the 4π.
Q2 (Core) The Sun’s luminosity is 3.85 × 1026 W. Earth is 1.50 × 1011 m from the Sun. Calculate the intensity of the Sun’s radiation at Earth’s distance (the solar constant), to 2 significant figures.
Show solution
F: I = P ÷ 4πd²
I: I = 3.85 × 1026 ÷ (4π × (1.50 × 1011)²) = 3.85 × 1026 ÷ 2.827… × 1023
F: I is already the subject
A: 1361.6… W m−2 = 1.4 kW m−2 (2 s.f.)
Q3 (Core) Sirius A has a radius of 1.19 × 109 m and a luminosity of 9.7 × 1027 W. Calculate its surface temperature, to 2 significant figures. (σ = 5.67 × 10−8 W m−2 K−4)
Show solution
Step 1 – surface area
A = 4π × (1.19 × 109)² = 1.779… × 1019 m²
Step 2 – temperature
F: P = σAT⁴
I: 9.7 × 1027 = 5.67 × 10−8 × 1.779… × 1019 × T⁴
F: T⁴ = 9.7 × 1027 ÷ (5.67 × 10−8 × 1.779… × 1019) = 9.613… × 1015 K⁴, so T = (9.613… × 1015)¼
A: 9901.9… K = 9900 K (2 s.f.)
Watch out: taking the square root instead of the fourth root. On a calculator use x0.25 or take √ twice.
Q4 (Core) Betelgeuse has a luminosity of 1.0 × 105 times that of the Sun and a surface temperature of 3500 K. Calculate its radius in metres and as a multiple of the Sun’s radius, both to 2 significant figures. (Luminosity of Sun = 3.85 × 1026 W, radius of Sun = 6.96 × 108 m, σ = 5.67 × 10−8 W m−2 K−4)
Show solution
Convert first: P = 1.0 × 105 × 3.85 × 1026 = 3.85 × 1031 W
F: P = σ × 4πr² × T⁴
I: 3.85 × 1031 = 5.67 × 10−8 × 4π × r² × 3500⁴
F: r = √(3.85 × 1031 ÷ (4π × 5.67 × 10−8 × 3500⁴)) = √(3.85 × 1031 ÷ 1.069… × 108)
A: 6.000… × 1011 m = 6.0 × 1011 m (2 s.f.), and 6.000… × 1011 ÷ 6.96 × 108 = 860 solar radii (2 s.f.)
Q5 (Exam) Sirius A has a luminosity of 9.7 × 1027 W and is 8.6 light years from Earth. Calculate the intensity of its radiation at Earth, to 2 significant figures. (1 ly = 9.46 × 1015 m)
Show solution
Convert first: d = 8.6 × 9.46 × 1015 = 8.1356 × 1016 m
F: I = P ÷ 4πd²
I: I = 9.7 × 1027 ÷ (4π × (8.1356 × 1016)²) = 9.7 × 1027 ÷ 8.317… × 1034
F: I is already the subject
A: 1.166… × 10−7 W m−2 = 1.2 × 10−7 W m−2 (2 s.f.), about 1010 times less than the solar constant
Watch out: leaving d in light years. The light year is a distance, so convert it to metres before squaring.
Q6 (Exam) Star X has a radius 3.0 times that of star Y, but its surface temperature is only 0.60 times that of star Y. Calculate the ratio of luminosities PX ÷ PY, to 2 significant figures, and state which star is more luminous.
Show solution
F: P = σ4πr²T⁴, so PX ÷ PY = (rX ÷ rY)² × (TX ÷ TY)⁴
I: PX ÷ PY = 3.0² × 0.60⁴ = 9.0 × 0.1296
F: the ratio is already the subject
A: 1.1664 = 1.2 (2 s.f.), so star X is slightly more luminous
Watch out: σ and 4π cancel in a ratio, so never substitute them. The T⁴ term matters much more than r²: 0.60⁴ is only 0.13.
Q7 (Exam) Alpha Centauri A has a luminosity 1.5 times that of the Sun. The intensity of its radiation measured above Earth’s atmosphere is 2.7 × 10−8 W m−2. Calculate its distance from Earth in light years, to 2 significant figures. (Luminosity of Sun = 3.85 × 1026 W, 1 ly = 9.46 × 1015 m)
Show solution
Convert first: P = 1.5 × 3.85 × 1026 = 5.775 × 1026 W
F: I = P ÷ 4πd²
I: 2.7 × 10−8 = 5.775 × 1026 ÷ 4πd²
F: d = √(5.775 × 1026 ÷ (4π × 2.7 × 10−8)) = 4.125… × 1016 m
A: 4.125… × 1016 ÷ 9.46 × 1015 = 4.36… ly = 4.4 ly (2 s.f.)
Q8 (Exam) A satellite measures the solar constant as 1360 W m−2 at 1.50 × 1011 m from the Sun. The Sun’s radius is 6.96 × 108 m. Calculate (a) the luminosity of the Sun, to 3 significant figures, and (b) the Sun’s surface temperature, to 2 significant figures, treating it as a black body. (σ = 5.67 × 10−8 W m−2 K−4)
Show solution
Step 1 – (a) luminosity
F: I = P ÷ 4πd²
I: 1360 = P ÷ (4π × (1.50 × 1011)²)
F: P = 1360 × 4π × (1.50 × 1011)²
A: 3.845… × 1026 W = 3.85 × 1026 W (3 s.f.)
Step 2 – (b) temperature
F: P = σAT⁴, with A = 4π × (6.96 × 108)² = 6.087… × 1018 m²
I: 3.845… × 1026 = 5.67 × 10−8 × 6.087… × 1018 × T⁴
F: T⁴ = 1.114… × 1015 K⁴, so T = (1.114… × 1015)¼
A: 5777.3… K = 5800 K (2 s.f.)
Watch out: using the Earth–Sun distance as the radius in A = 4πr². The area in P = σAT⁴ is the star’s own surface; d is only used for intensity.
Q9 (Stretch) The spectrum of Rigel peaks at a wavelength of 240 nm. Its luminosity is 4.6 × 1031 W. Calculate (a) its surface temperature, to 3 significant figures, and (b) its radius in metres and as a multiple of the Sun’s radius, both to 2 significant figures. (Wien constant = 2.9 × 10−3 m K, σ = 5.67 × 10−8 W m−2 K−4, radius of Sun = 6.96 × 108 m)
Show solution
Convert first: 240 nm = 2.40 × 10−7 m
Step 1 – (a) temperature from Wien’s law
F: λmaxT = 2.9 × 10−3
I: T = 2.9 × 10−3 ÷ 2.40 × 10−7
A: 12 083.3… K = 12 100 K (3 s.f.)
Step 2 – (b) radius from Stefan’s law
F: P = σ4πr²T⁴
I: 4.6 × 1031 = 5.67 × 10−8 × 4π × r² × 12 083.3…⁴
F: r = √(4.6 × 1031 ÷ (4π × 5.67 × 10−8 × 12 083.3…⁴)) = √(4.6 × 1031 ÷ 1.518… × 1010)
A: 5.503… × 1010 m = 5.5 × 1010 m (2 s.f.), and 5.503… × 1010 ÷ 6.96 × 108 = 79 solar radii (2 s.f.)
Watch out: rounding T to 12 000 K before raising it to the fourth power. Keep the unrounded value in your calculator.
Q10 (Stretch) Sirius B, the white dwarf companion of Sirius A, is 2.64 pc from Earth. Its black-body spectrum peaks at 116 nm and the intensity of its radiation at Earth is 1.1 × 10−10 W m−2. (a) Calculate its luminosity, to 2 significant figures. (b) Calculate its radius, to 2 significant figures, and compare it with Earth’s radius of 6.37 × 106 m. (1 pc = 3.08 × 1016 m, Wien constant = 2.9 × 10−3 m K, σ = 5.67 × 10−8 W m−2 K−4)
Show solution
Convert first: d = 2.64 × 3.08 × 1016 = 8.1312 × 1016 m; 116 nm = 1.16 × 10−7 m
Step 1 – (a) luminosity from the inverse square law
F: I = P ÷ 4πd²
I: 1.1 × 10−10 = P ÷ (4π × (8.1312 × 1016)²)
F: P = 1.1 × 10−10 × 8.308… × 1034
A: 9.139… × 1024 W = 9.1 × 1024 W (2 s.f.)
Step 2 – temperature from Wien’s law
T = 2.9 × 10−3 ÷ 1.16 × 10−7 = 25 000 K
Step 3 – (b) radius from Stefan’s law
F: r = √(P ÷ 4πσT⁴)
I: r = √(9.139… × 1024 ÷ (4π × 5.67 × 10−8 × 25 000⁴)) = √(9.139… × 1024 ÷ 2.783… × 1011)
A: 5.730… × 106 m = 5.7 × 106 m (2 s.f.), about 0.90 times Earth’s radius. A white dwarf is very hot but very small, which is why it has low luminosity.
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