A star's apparent magnitude $m$ depends on two things: how much light it gives out and how far away it is. To compare stars fairly, astronomers define the absolute magnitude $M$: the apparent magnitude the star would have if it sat at a standard distance of 10 parsecs. Combining magnitude arithmetic with the inverse-square law gives a short, powerful formula for distance.
💡 Why it works
Light spreads over a sphere, so a star of luminosity $L$ at distance $d$ delivers flux $F = L/(4\pi d^2)$. Compare the same star at its real distance $d$ and at 10 pc: $$m - M = -2.5\log_{10}\frac{F(d)}{F(10\text{ pc})} = -2.5\log_{10}\left(\frac{10\text{ pc}}{d}\right)^{2}.$$ The power of 2 comes out of the log as a factor, and $-2.5 \times (-2) = 5$, so $m - M = 5\log_{10}(d/10\text{ pc})$. The luminosity cancels, which is the whole point.
🔑 Key idea
The distance modulus is $$m - M = 5\log_{10}\frac{d}{10\text{ pc}}, \qquad d = 10^{(m - M + 5)/5}\text{ pc}.$$ It depends only on distance (when there is no dust in the way).
Each factor of 10 in distance adds 5 to the distance modulus.
Distance
$m - M$
1 pc
$-5$
10 pc
$0$
100 pc
$5$
1 kpc
$10$
10 kpc
$15$
1 Mpc
$25$
Light from a source spreads over an area that grows as $d^2$. Twice as far means a quarter of the flux, which is $5\log_{10}2 \approx 1.5$ magnitudes fainter.
Worked examples
✏️ Worked example: The Sun's absolute magnitude
The Sun has $m = -26.74$ at a distance of $1\text{ AU} = 4.848 \times 10^{-6}$ pc. Find its absolute magnitude.
Answer: $M_\odot \approx 4.83$. From 10 pc the Sun would be a faint naked-eye star.
✏️ Worked example: A Cepheid in a nearby galaxy
A Cepheid variable with $M = -4.0$ appears at $m = 16.0$. How far away is it?
$m - M = 16.0 - (-4.0) = 20.0$.
$d = 10^{(20.0 + 5)/5}\text{ pc} = 10^{5}$ pc.
Answer: $d = 1.0 \times 10^{5}$ pc $= 100$ kpc, well outside the disk of the Milky Way.
✏️ Worked example: Predicting how bright a star looks
A star with $M = 1.5$ lies 250 pc away. What is its apparent magnitude?
$m = M + 5\log_{10}(250/10) = 1.5 + 5\log_{10}25$.
$\log_{10}25 = 1.398$, so $5\log_{10}25 = 6.99$.
$m = 1.5 + 6.99 = 8.49$.
Answer: $m \approx 8.5$, too faint for the naked eye but easy in binoculars.
⚠️ Common mistake
Putting the distance in light-years. The 10 in the formula is 10 parsecs, so $d$ must be in parsecs too. A star 100 pc away is 326 ly away; plugging in 326 gives $m - M = 7.6$ instead of the correct $5.0$, an error of 2.6 magnitudes. Convert first ($1\text{ pc} = 3.26$ ly). A second slip is writing $5\log_{10}d$ and forgetting the 10 pc, which shifts every answer by 5. Check: a star at exactly 10 pc must come out with $m = M$.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.