Example: 2 stars (A and B) with the same absolute magnitude have apparent
magnitudes m=1 (A) and m=6 (B). How much more distant is B than A?
Solution: A difference of 5 in apparent magnitude corresponds to a factor of
10x10=100 in brightness. Since brightness drops off as the square of the
distance, the dimmer star (B) is 10 times farther away.
Object | app mag | distance (ly) | abs mag |
Sun |
-26.7 | 1/60,000 | 4.8 |
Moon (full) |
-12.5 | --- | --- |
Sirius (α Sco) |
-1.5 | 8.6 | 1.4 |
Venus (brightest) |
-4.4 | --- | --- |
Mars (brightest) | -2.9 | --- | --- |
Jupiter (brightest) |
-2.8 | --- | --- |
Vega (α Lyr) |
0 | 25 | 0.58 |
Betelgeuse | 0.45 | 430 | -5.1 |
Rigel | 0.12 | 800 | -6.7 |
Saturn (brightest) |
0.6 | --- | --- |
Polaris | 2.0 | 430 | -3.6 |
naked-eye limit |
6 | ||
Ceres | 6.7 | --- | |
binocular limit | 10 | ||
Pluto/20 cm telescope | 14 | --- | |
Hubble limit |
30 | ||
***Note that planets and the Moon can only reflect light, so assigning absolute magnitude in these cases is meaningless. |
A magnitude difference of: |
Equals a brightness ratio of: |
0.0 |
1.0 |
0.2 |
1.2 |
1.0 |
2.5 |
1.5 |
4.0 |
2.0 |
6.3 |
2.5 |
10.0 |
4.0 |
40.0 |
5.0 |
100.0 |
7.5 |
1000.0 |
10.0 |
10,000.0 |
Ceres is a bright asteroid but at magnitude 6.7, it would be a real trick to pick out with the eyes alone. Neptune at magnitude 7.7 and Pluto at 13 require optical aid. (Pluto, in particular, requires a large telescope.) There are also transient solar system objects that occasionally appear bright enough to be seen without optical aid, such as meteors or comets.