Absolute Magnitude Calculator
Pick which of the three quantities you want, apparent magnitude m, absolute magnitude M or distance d, enter the other two, and the calculator solves the distance modulus m − M = 5 log₁₀(d ÷ 10 pc). It also gives the distance in light-years, an optional correction for dust, the luminosity in suns, and the brightness ratio between any two magnitudes. Free, no sign-up, and nothing leaves your browser.
Dimming by interstellar dust. Leave at 0 if unknown.
Results
1.43
Absolute magnitude M
-2.89
Distance modulus m − M
2.6371
Distance in parsecs
8.6011
Distance in light-years
21.002
Luminosity in solar units
Meaningful only when M is a bolometric magnitude (or you compare in the same band as the Sun's).
m − M = 5 log₁₀(d ÷ 10 pc) + A
Compare two magnitudes
100.00 ×
Object 1 is 100.00 times brighter than object 2.
5.00
Difference in magnitudes
Five magnitudes is a factor of exactly 100, so one magnitude is 10^0.4 ≈ 2.512. A smaller number is brighter.
Apparent magnitudes here are for one band (usually V); M uses the same band. Real stars also have a bolometric correction between the visual and total output, which this calculator does not apply.
How to Use the Absolute Magnitude Calculator
- 1. Choose what to find. Absolute magnitude M is the usual question: how bright a star really is. Choose Apparent magnitude m to predict how bright an object will look from a given distance, or Distance d when you know both magnitudes.
- 2. Enter the other two quantities. Magnitudes can be negative, and the distance can be in parsecs, light-years, kiloparsecs, megaparsecs or astronomical units. The Start from an object menu loads the Sun, Sirius, Vega, Proxima Centauri and the Andromeda Galaxy.
- 3. Add extinction if you know it. Dust between you and the object dims it by A magnitudes. Leave it at 0 for nearby stars.
- 4. Read the result. You get the solved quantity, the distance modulus, the distance in parsecs and light-years and the luminosity in suns. The second panel compares any two magnitudes.
Apparent and Absolute Magnitude
The magnitude scale ranks brightness backwards: the brighter the object, the lower the number, and the Sun is −26.74. Apparent magnitude m is how bright an object looks from where you stand. Absolute magnitude M is how bright it would look from a fixed 10 parsecs (32.6 light-years), which removes distance from the comparison and so measures intrinsic luminosity. Because light dims with the square of distance and five magnitudes is exactly a factor of 100 in brightness, the two are linked by the distance modulus:
Distance modulus
m − M = 5 log₁₀(d ÷ 10 pc) = 5 log₁₀ d − 5
With extinction
m − M = 5 log₁₀ d − 5 + A
Distance
d = 10^((m − M + 5) ÷ 5) parsecs
At exactly 10 pc, m = M. Ten times farther, 100 pc, makes an object 5 magnitudes fainter, and a modulus of 25 puts it a megaparsec away. Distances here are in parsecs, so 1 AU is 4.8481 × 10⁻⁶ pc.
Worked Examples
| Object | m | Distance | M | Modulus m − M |
|---|---|---|---|---|
| The Sun | −26.74 | 1 AU | 4.83 | −31.57 |
| Sirius | −1.46 | 2.6371 pc (8.6011 ly) | 1.43 | −2.89 |
| Vega | 0.03 | 7.6787 pc (25.045 ly) | 0.60 | −0.57 |
| Proxima Centauri | 11.13 | 1.302 pc (4.2466 ly) | 15.56 | −4.43 |
| Andromeda Galaxy | 3.44 | 0.78 Mpc | −21.02 | 24.46 |
The apparent V magnitudes are SIMBAD values, and the distances come from parallax, except the Sun's 1 AU and Andromeda's rounded 0.78 Mpc. Sirius is the brightest star in the night sky, yet at M = 1.43 it is about 22 times as luminous as the Sun in V light. Proxima Centauri, at M = 15.56, has less than a ten-thousandth of the Sun's visual luminosity, so even though it is the nearest star it is invisible without a telescope. The Andromeda Galaxy at M = −21.02 shines as roughly twenty billion Suns.
From Magnitude to Luminosity
Luminosity in solar units follows from comparing absolute magnitudes: L ÷ L☉ = 10^(0.4 × (M☉ − M)). The answer is only meaningful when both magnitudes are in the same band. The calculator offers the Sun's nominal bolometric magnitude of 4.74, set by the IAU in 2015 for total output, and its V-band magnitude of 4.81 from Willmer (2018). Feeding Sirius's V-band 1.43 into the V comparison gives 22.4 suns, but into the bolometric one 21.0; neither is its true bolometric luminosity, because stars radiate outside the visible band and that needs a bolometric correction.
Why the Sun's M is 4.83 and 4.81
Using the classic apparent magnitude of −26.74 and a distance of 1 AU, the calculator returns 4.83, the value long quoted for the Sun in V. Willmer's 2018 synthetic-spectrum estimate is 4.81 ± 0.02. The difference matters only if you compare luminosities to better than a few percent.
Comparing Magnitudes
A difference of Δm magnitudes is a brightness ratio of 10^(0.4 Δm). One magnitude is 2.512, five is exactly 100, ten is 10,000. Sirius at −1.46 outshines Vega at 0.03 by a factor of 3.9446, and the Sun is about 400,000 times brighter than the full Moon, whose magnitude is −12.74. The second panel does this arithmetic for any pair.
Limits
- Single band. Apparent and absolute magnitudes must be in the same filter, usually V.
- Extinction. Dust makes distant objects look fainter and redder. Without A, the distance you solve for is too large.
- Variable stars. A pulsating or eclipsing star has no single magnitude; use a mean value.
- Cosmological distances. For galaxies at redshifts above about 0.1, the luminosity distance, not the simple modulus, applies.
Frequently Asked Questions
- What is absolute magnitude?
- It is the apparent magnitude an object would have if it were exactly 10 parsecs (32.6 light-years) from the observer. Because distance is removed, absolute magnitude measures how luminous the object really is. The Sun's is about 4.8, which would make it a faint but visible star from 10 parsecs.
- How do you convert apparent magnitude to absolute magnitude?
- Use M = m − 5 log₁₀(d) + 5, with d in parsecs. For Sirius, with m = −1.46 and d = 2.6371 pc, M = −1.46 − 2.106 + 5 = 1.43. Add the extinction A to the subtracted part when interstellar dust dims the light.
- How do you find a star's distance from its magnitudes?
- Rearrange the distance modulus: d = 10^((m − M + 5) ÷ 5) parsecs. A star with m = 15 and M = 5 has a modulus of 10 and is 1000 pc away. This is the standard-candle method used for Cepheid variables and supernovae.
- Why do brighter stars have lower magnitudes?
- The scale comes from the ancient Greek ranking of stars from first magnitude, the brightest, to sixth, the faintest. In 1856 Norman Pogson fixed it so that five magnitudes equal a factor of 100 in brightness, which is why the numbers still run backwards and the brightest objects are negative.
- What is the distance modulus?
- It is the difference m − M between apparent and absolute magnitude, which depends only on distance (and extinction). A modulus of 0 means 10 pc, 5 means 100 pc, 10 means 1 kpc and 25 means 1 Mpc. Astronomers often quote the distance to a galaxy or star cluster as its modulus.
- How many times brighter is magnitude 1 than magnitude 6?
- Exactly 100 times, because five magnitudes is defined as a factor of 100. Each single magnitude step is a factor of 100^(1/5) = 2.512, so the ratio for any difference Δm is 2.512^Δm.