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9702 · 25.1

Standard candles — practice questions

Practice and worked examples for 9702 Standard candles. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A distant star is observed to have a flux (F) of 2.5×10−10 W m−22.5 \times 10^{-10} \text{ W m}^{-2}. If it is identified as a standard candle with a known luminosity (L) of 3.8×1026 W3.8 \times 10^{26} \text{ W} (similar to our Sun), calculate its distance from Earth. Give your answer in light-years, where 1 light-year ≈9.46×1015 m\approx 9.46 \times 10^{15} \text{ m}.

Show solution outline
  1. Start with the inverse square law rearranged for distance: d=L4πFd = \sqrt{\frac{L}{4\pi F}}.
  2. Substitute the given values: d=3.8×1026 W4π(2.5×10−10 W m−2)d = \sqrt{\frac{3.8 \times 10^{26} \text{ W}}{4\pi (2.5 \times 10^{-10} \text{ W m}^{-2})}}.
  3. Calculate the distance in meters: d=3.8×10263.14159×10−9=1.21×1035≈3.48×1017 md = \sqrt{\frac{3.8 \times 10^{26}}{3.14159 \times 10^{-9}}} = \sqrt{1.21 \times 10^{35}} \approx 3.48 \times 10^{17} \text{ m}.
  4. Convert the distance from meters to light-years: dly=3.48×1017 m9.46×1015 m/light-yeard_{\text{ly}} = \frac{3.48 \times 10^{17} \text{ m}}{9.46 \times 10^{15} \text{ m/light-year}}.
  5. Final answer: dly≈36.8 light-yearsd_{\text{ly}} \approx 36.8 \text{ light-years}.

Worked example 2

A supergiant star has a surface temperature of 3600 K and a radius of 6.0 x 10¹¹ m. The radiant flux received at Earth from this star is measured to be 1.1 x 10⁻⁷ W m⁻². Calculate the distance of the star from Earth. (The Stefan-Boltzmann constant σ is 5.67 x 10⁻⁸ W m⁻² K⁻⁴).

Show solution outline
  1. Calculate the star's luminosity (L) using the Stefan-Boltzmann Law: L=4πr2σT4L = 4\pi r^2 \sigma T^4 Substitute the given values: L=4π(6.0×1011)2(5.67×10−8)(3600)4L = 4\pi (6.0 \times 10^{11})^2 (5.67 \times 10^{-8}) (3600)^4 L=4π(3.6×1023)(5.67×10−8)(1.6796×1014)L = 4\pi (3.6 \times 10^{23}) (5.67 \times 10^{-8}) (1.6796 \times 10^{14}) L≈4.30×1031 WL \approx 4.30 \times 10^{31} \text{ W}
  2. Calculate the distance (d) using the inverse square law: The formula relating flux, luminosity, and distance is F=L4πd2F = \frac{L}{4\pi d^2}. Rearrange for distance: d=L4πFd = \sqrt{\frac{L}{4\pi F}} Substitute the calculated luminosity and given flux: d=4.30×10314π(1.1×10−7)d = \sqrt{\frac{4.30 \times 10^{31}}{4\pi (1.1 \times 10^{-7})}} d=4.30×10311.382×10−6d = \sqrt{\frac{4.30 \times 10^{31}}{1.382 \times 10^{-6}}} d=3.11×1037≈5.58×1018 md = \sqrt{3.11 \times 10^{37}} \approx 5.58 \times 10^{18} \text{ m}
  3. Final Answer: The distance to the star is approximately 5.6×1018 m5.6 \times 10^{18} \text{ m} (to 2 significant figures).
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