How bright a star looks is given by its apparent magnitude. This is different from its absolute magnitude. The absolute magnitude of a star is defined as the apparent magnitude that it would have if placed at a distance of 10 parsecs from the Earth.

Consider two stars A and B. Star A appears to be brighter than star B. In other words the intensity of the light reaching the observer from star A is greater than that from star B.

Let the apparent magnitude of star A = m

Referring back to the magnitude difference of 5 being a difference in intensity by a factor of 100 we can write:

Since if (m

Therefore taking logs of both side : lg(I

Therefore: m

Now let the magnitude of A
(m_{A}) be that at 10 parsecs, in other words the absolute magnitude of the star
(M) and let m_{B} be the magnitude (m) at some other distance d (also
measured in parsecs).

Therefore : m – M =
5/2[lg(I_{A}/I_{B})]

But from the inverse square law:
(I_{A}/I_{B}) = (d_{B}/d_{A})^{2} because the intensity is inversely
proportional to the square of the distance of the star.

m – M = 5/2lg(I

Therefore :

In our example if A appears to the observer to be brighter than B, and if we use m

The apparent and absolute magnitudes of a number of stars are given in the following table.

Object |
Apparentmagnitude |
Absolutemagnitude |
Distance(light years) |

Sun | -26.7 | - | - |

Venus | -4.4 | - | - |

Jupiter | -2.2 | - | - |

Sirius | -1.46 | +1.4 | 8.7 |

Rigel | 0.1 | -7.0 | 880 |

Arcturus | -0.1 | -0.2 | 35.9 |

Proxima Centauri | 10.7 | +15.1 | 4.2 |

Vega | 0.0 | +0.5 | 26.4 |

Betelguese | -0.4 | -5.9 | 586 |

Deneb (a Cygni) | 1.3 | -7.2 | 1630 |

Andromeda galaxy | 5 | -17.9 | 2 200 000 |

Our Galaxy | - | -18.0 | - |

1. Calculate the absolute magnitude of a star of apparent magnitude +2.5 which is at a distance of 25 pc from the Earth.

M = m + 5 – 5lg(d) = 2.5 + 5 – 5lg(25) = 7.5 – 6.99 = +0.51

2. Calculate the distance of a star with an apparent magnitude of +6.0 and an absolute magnitude of +4.0.

6 – 4 + 5 = 5lg(d) Therefore: lg(d) = 0.6 d = 3.98 pc

3. Calculate the apparent magnitude of a star whose absolute magnitude is -1.8 if the star is at a distance of 35.5 parsecs from the Earth.

m = M – 5 + 5lg(d) = -1.8 – 5 + 5x1.55 = + 0.95

4. Calculate the apparent magnitude of a star that has an absolute magnitude of +2.0 if the star is 25 parsecs from the Earth.

m = 5lg(d/10) + M = 5lg(25/10) + 2.0 = 1.98 + 2.0 = 3.989 = 4