☰ Contents · Astronomy

Celestial coordinates, maps and magnitudes

Lessons 6–8 · 3 lessons · M. Mamadazimov. Astronomy Grade 11, 1st edition. DAVR NASHRIYOTI, Tashkent, 2018
8

Apparent magnitudes of stars

Textbook: p. 14
GoalExplain the apparent magnitude scale and Pogson’s relation, and calculate ratios of brightness.
New words
magnitude · yulduz kattaligiapparent magnitude · ko‘rinma kattalikilluminance (flux) · yoritilganlikPogson’s relation · Pogson formulasi
Explanation

In antiquity Hipparchus grouped stars by magnitude from 1 to 6: the brightest were 1st magnitude and the faintest the eye can barely see 6th. In 1856 Pogson made the scale exact: a difference of 5 magnitudes corresponds to exactly a factor of 100 in illuminance, so 1 magnitude is a factor of 100^(1/5) ≈ 2.512. The formula is m₁ – m₂ = –2.5 · lg(E₁/E₂), where E is the illuminance from the star at the Earth’s surface. The brightest bodies have negative magnitudes: Sirius ≈ –1.46, the full Moon ≈ –12.7, the Sun ≈ –26.7; Vega is taken as about 0 and Polaris ≈ 2. Under a dark sky the eye reaches about magnitude 6, binoculars about 9, and large telescopes around 30.

Worked examples
Stars of magnitude 3 and 8 differ by 5 magnitudes, so the magnitude-3 star is 100 times brighter than the magnitude-8 star.
If the magnitude difference is 10, the ratio of brightness is 100 · 100 = 10,000 (two steps of 5 magnitudes).
Class activity

A daytime classroom task: with a lamp and a lux meter (or phone app) measure illuminance at 1 m and 2 m from the lamp (inverse-square law). Never look at the Sun; an adult switches on the lamp.

Practice
1
How many times brighter is a magnitude-1 star than a magnitude-6 star?
2
What is the brightness ratio for a magnitude difference of 15? (100 · 100 · 100)
3
Which is brighter: Sirius (–1.46) or Polaris (2)? Say why.
4
Why is the Sun’s magnitude negative and as large as –26.7?