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AS & A-Level Physics 25 — Stars and the Expanding Universe

AS & A-Level Physics 25 — Stars and the Expanding Universe

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Independent Deckloop revision aligned with the Cambridge International AS & A Level Physics (9702) syllabus, 2025–2027. Not affiliated with or endorsed by Cambridge International Education. Chapter 25 of 26: Stars and the expanding Universe. Concepts, worked applications and misconception checks.

Physics EN A-Level
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Radiant Flux Intensity and the Inverse Square Law

Radiant flux intensity, FF, also known as apparent brightness, is the radiant power received per unit area at a given distance from a source. As light from a star spreads out uniformly in all directions, it illuminates an ever-increasing spherical area. Assuming the star is a point source and emits radiation isotropically (equally in all directions), the total luminosity LL is spread over the surface area of a sphere of radius dd (the distance from the star). The surface area of this sphere is 4πd24\pi d^2. Therefore, the radiant flux intensity FF at distance dd is given by the inverse square law: F=L4πd2F = \frac{L}{4\pi d^2}. The SI unit for radiant flux intensity is watts per square metre (W m−2^{-2}). This law is fundamental for determining distances to celestial objects.

Key points

  • Radiant flux intensity (FF) is the power received per unit area from a source.
  • It follows the inverse square law: F=L4πd2F = \frac{L}{4\pi d^2}.
  • Assumes isotropic emission from a point source.
  • The SI unit is watts per square metre (W m−2^{-2}).
  • It describes how apparent brightness decreases with distance.

Worked example

Question

In a model, an exoplanet orbits a star with a luminosity of 7.5×1026 W7.5 \times 10^{26} \text{ W}. If the radiant flux intensity measured at the planet’s orbital distance, perpendicular to the incoming light and outside any atmosphere is 1200 W m−21200 \text{ W m}^{-2}, calculate the distance between the exoplanet and its star. Give your answer to 2 significant figures.

Solution

1. Recall the inverse square law for radiant flux intensity: F=L4πd2F = \frac{L}{4\pi d^2}.
2. Rearrange the formula to solve for distance dd: d2=L4πFd^2 = \frac{L}{4\pi F}, so d=L4πFd = \sqrt{\frac{L}{4\pi F}}.
3. Substitute the given values:
4. Luminosity L=7.5×1026 WL = 7.5 \times 10^{26} \text{ W}
5. Radiant flux intensity F=1200 W m−2F = 1200 \text{ W m}^{-2}
6. d=7.5×10264π×1200d = \sqrt{\frac{7.5 \times 10^{26}}{4\pi \times 1200}}
7. d=7.5×102615079.6d = \sqrt{\frac{7.5 \times 10^{26}}{15079.6}}
8. d=4.973×1022d = \sqrt{4.973 \times 10^{22}}
9. d=2.23×1011 md = 2.23 \times 10^{11} \text{ m}
10. Round to 2 significant figures: d=2.2×1011 md = 2.2 \times 10^{11} \text{ m}.

The distance between the exoplanet and its star is 2.2×1011 m2.2 \times 10^{11} \text{ m}.

Common pitfalls

  • Forgetting the 4π4\pi in the denominator: The area is that of a sphere, not a circle.
  • Not converting units consistently: Ensure all distances are in metres and power in watts before calculation.

Prerequisites

  • Basic algebraic manipulation and scientific notation.
  • Understanding of luminosity.
Further resources