Intensity, parallax and standard candles

Measuring distances in space is hard. Parallax works for nearby stars; standard candles reach much further.

Intensity and the inverse square law

Radiation from a source of luminosity L spreads out over a sphere of area 4πd². The intensity (power per unit area) at distance d is

I = L / 4πd² (W/m²)

Doubling the distance quarters the intensity.

Trigonometric parallax

As the Earth orbits the Sun, a nearby star appears to shift against the very distant background stars. Measure its position six months apart. The parallax angle p is half the total angular shift, and the baseline is the Earth-Sun distance (1 AU = 1.50 x 10¹¹ m):

tan p = 1 AU / d, so d ≈ 1 AU / p for small p (in radians)

A star with a parallax angle of 1 arcsecond is 1 parsec away (3.09 x 10¹⁶ m): d (pc) = 1 / p (arcsec). Parallax only works for nearby stars, as distant stars' angles are too small to measure.

Standard candles

A standard candle is an object whose luminosity L is known, such as a Cepheid variable star (its period of pulsation reveals its luminosity) or a type Ia supernova. Measure its intensity I at Earth, then

d = sqrt(L / 4πI)

This method reaches far further than parallax, out to other galaxies.

Worked example

A Cepheid variable has luminosity 1.0 x 10²⁹ W and its measured intensity at Earth is 3.0 x 10^-9 W/m². Find its distance.

d = sqrt(L / 4πI) = sqrt(1.0 x 10²⁹ / (4π x 3.0 x 10^-9))
= sqrt(2.65 x 10³⁶)

Answer: d = 1.6 x 10¹⁸ m

Key idea: I = L/4πd²; parallax: d = 1 AU / p, d (pc) = 1/p (arcsec), nearby stars only; standard candles: known L, measure I, d = sqrt(L/4πI).

Check yourself

The Sun has luminosity 3.85 x 10²⁶ W and is 1.50 x 10¹¹ m from Earth. Find the intensity of its radiation at Earth, in W/m², to 3 s.f.
Show answer
1360 W/m². I = L / 4πd² = 3.85 x 10²⁶ / (4π x 2.25 x 10²²) = 3.85 x 10²⁶ / 2.83 x 10²³ = 1360 W/m².
A star has a parallax angle of 0.25 arcseconds. How far away is it in parsecs?
Show answer
4 pc. d = 1 / p = 1 / 0.25 = 4.0 pc.
A star has a parallax angle of 4.0 x 10^-6 rad. Using 1 AU = 1.50 x 10¹¹ m, find its distance, in units of 10¹⁶ m.
Show answer
3.75 x 10¹⁶ m. d = 1 AU / p = 1.50 x 10¹¹ / 4.0 x 10^-6 = 3.75 x 10¹⁶ m.
A standard candle of luminosity 1.0 x 10²⁹ W has a measured intensity of 3.0 x 10^-9 W/m². Find its distance, in units of 10¹⁸ m, to 3 s.f.
Show answer
1.63 x 10¹⁸ m. d = sqrt(L / 4πI) = sqrt(1.0 x 10²⁹ / 3.77 x 10^-8) = sqrt(2.65 x 10³⁶) = 1.63 x 10¹⁸ m.
Why can trigonometric parallax only be used for relatively nearby stars?
- For distant stars the parallax angle is too small to measure
- Distant stars do not move
- Distant stars are too bright
- The Earth's orbit is not circular
Show answer
For distant stars the parallax angle is too small to measure. The angle gets smaller as d increases, until it is below the measurement resolution.
What is a standard candle?
- A star with a parallax of 1 arcsecond
- A star exactly 1 parsec away
- An astronomical object of known luminosity
- Any star on the main sequence
Show answer
An astronomical object of known luminosity. Knowing L and measuring I gives d.
A star is moved to twice its original distance. What happens to its intensity at Earth?
- It becomes a quarter as large
- It halves
- It doubles
- It stays the same
Show answer
It becomes a quarter as large. I is proportional to 1/d².

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Last updated 7 November 2026. Spotted a mistake? Tell us.