Astronomy Answers
AstronomyAnswerBook: Magnitudes

[Astronomy Answers] [Dictionary] [AnswerBook] [Universe Family Tree] [Science] [Starry Sky] [Planet Positions] [Calculate] [Colophon]

1. Star Numbers ... 2. Different Magnitudes ... 3. Limiting Magnitude versus Lens Size and Instrument ... 4. How Bright Is It?

This page answers questions about magnitudes. The questions are:

Astronomers use the magnitude scale to express the brightness of celestial objects. The scale's history goes back to observations by ancient Greeks (notably Hipparchos and Plato) who divided the visible stars into six brightness classes, with the brightest stars in class 1 and the faintest stars in class 6. In modern times, astronomers noted that stars in each next brighter class were about 2.5 times brighter than stars in the previous class. Modern astronomers standardized the brightness measure and called it "magnitude", such that a star which has a magnitude that is one less than another star is about 2.512 times as bright as that other star. A difference of five magnitudes corresponds to a brightness ratio of exactly 100: a star of magnitude 0 is 100 times as bright as a star of magnitude 5, and 10,000 (= 100 times 100) times as bright as a star of magnitude 10.

The brighter the star, the lower its magnitude. The brightest stars have magnitudes of 1 or less; there are only 20 such stars. The faintest stars that one can see on a very dark night with a healthy, dark-adapted eye without visual aids such as binoculars or telescopes have a magnitude of about 6; there are about 5000 such stars. In big cities, where there is a lot of light pollution, the limiting magnitude (the highest magnitude that one can see with the unaided eye) may be as low as 2 or 3, limiting the number of stars that one can see to about 170. The number of stars that one can see from a particular place on Earth depends not just on the limiting magnitude at that place and time, but also on the latitude of the place. At the poles, one can see about half of the stars that one can see at the equator, and for places between the poles and the equator one can see intermediate numbers of stars.

If objects are bright enough, then their magnitude is negative. For instance, the star Vega in the constellation of the Lyre (Lyra) is one of the brightest stars and has magnitude 0. The star Sirius in the constellation of the Large Dog (Canis Majoris) is almost four times as bright as Vega, and has a magnitude of −1.5. The full Moon is about 70,000 times as bright as Sirius at a magnitude of about −13.6, and the Sun is about 177,000 times as bright as the full Moon at a magnitude of about −26.7.

1. Star Numbers

There are many more faint stars than there are bright stars. For instance, there is only one star (other than the Sun) with a visual magnitude equal to or less than −1.46. There are ten stars of magnitude 0.50 or brighter. There are a hundred stars with a magnitude of 2.59 or brighter, and there are a thousand stars that are of magnitude 4.60 or less (according to the Bright Star Catalogue). For each magnitude increase, there are about three times more stars with a magnitude less than that.

The following table lists estimates for the number of stars up to particular magnitudes (on the left), and for the magnitudes at which particular numbers of brighter stars are attained (on the right). The numbers are based on the Bright Star Catalogue for magnitudes up to 5, and for larger magnitudes the numbers are estimates based on "Astrophysical Quantities" by Allen. A number with "k" behind is counted in thousands, and a number with "M" behind it in millions. The left-hand table has three columns. For instance, there are 513 stars with a magnitude of up to 4.00, and 15.5 million stars with a magnitude of up to 14.0. The one hundred millionth brightest star has a magnitude of about 16.2.

Table 1: Star Counts

Magnitude Count
−1.46 1
−1 1
0 4
0.50 10
1 15
2 48
2.59 100
3 171
4 513
4.60 1000
5 1602
6 4.8 k
6.67 10 k
7 14 k
8 42 k
8.82 100 k
9 121 k
10 340 k
11 927 k
11.1 1 M
12 2.46 M
13 6.29 M
13.5 10 M
14 15.5 M
15 36.9 M
16 83.7 M
16.2 100 M
17 182 M
18 374 M
19 733 M
19.5 1000 M

Here is a 3rd-degree least-squares fit to the star numbers:

\begin{equation} N = 10^{0.754 + 0.4896 V + 0.001159 V^2 - 0.000235 V^3} \end{equation}

where \(N\) is the number of stars with a magnitude up to \(V\).

2. Different Magnitudes

There are many different magnitude scales:

Some of these scales can be combined. For example, you can speak about an absolute visual magnitude or an apparent photographic magnitude.

3. Limiting Magnitude versus Lens Size and Instrument

If you use an optical aid such as a pair of binoculars or a telescope, then you can see fainter objects. Generally, the larger the light-collecting surface of the instrument (i.e., the first lens or mirror that the light encounters), the more magnitudes you gain. For perfect instruments, the light gain in magnitudes for point-like objects is about

\begin{equation} G = 5 \log\left( \frac{d}{d_\text{eye}} \right) \end{equation}

where \(d\) is the diameter of the main lens or mirror, \(d_\text{eye}\) is the diameter of the pupil of your eye, and \(\log(.)\) is the decimal logarithm which for nice round numbers consisting of a one followed by a number of zeros counts the number of zeros. For example, the light gain of a perfect 7x50 pair of binoculars relative to a healthy dark-adapted eye with a pupil diameter of 7 mm (about 0.28 inch) is about 4.3 magnitudes. If without the binoculars you can see stars down to a magnitude of 5.0, then with the binoculars you can see stars down to a magnitude of about 9.3, of which there are about 100 times more than of stars down to magnitude 5.0.

Since optical instruments are never perfect, they do not yield as much magnitude gain as a perfect instrument would. On a recent reasonably dark night, using common 7x50 binoculars, I could easily see stars down to about magnitude 8.8. This leads to the following formula for the limiting magnitude of optical instruments with comparable quality to my binoculars on a dark night:

\begin{align} V_\text{lim} & = −0.4 + 5 \log(d/[\text{mm}]) \\ V_\text{lim} & = 6.6 + 5 \log(d/[\text{in}]) \end{align}

where \(d\) in the first formula is measured in millimeters, and in the second formula in inches. If the instrument yields an image for each eye (for instance, binoculars), then add 0.7 to the limiting magnitude. Some limiting magnitudes for viewing by eye derived from this formula are listed in the following table.

Table 2: Limiting Magnitudes

Diameter \(V_\text{lim}\) Candle Stars Remark
mm in km
7 0.3 4.5* 1.4 900 eye
50 2.0 8.8* 9.8 98,000 7x50
100 3.9 9.6 14 226,000
150 5.9 10.4 20 509,000
500 19.7 13.0 68 6.3 M
1000 39.4 14.6 140 26 M
2400 94.5 16.5 340 124 M Hubble Space Telescope
10000 393.7 19.5 1300 1004 M Keck telescope

The first two columns list the diameter of the instrument (i.e., of the main lens or mirror) in millimeters (first column) or inches (second column). \(V_\text{lim}\) is the limiting magnitude obtained on a reasonably dark night with a dark-adapted eye and an instrument of comparable optical quality to my own set of binoculars. If a * follows the number, then two-eye observation is assumed, otherwise one-eye observation. The column headed "Candle" lists how far away a typical candle has to be to appear at the limiting magnitude. Multiply by 0.6 to get miles. The column headed "Stars" lists the approximate number of stars in the whole sky that could be seen with such an instrument, with "M" meaning million. The "Remarks" column lists some optical instruments of that size. The limiting magnitudes are as seen from the Earth: the Hubble Space Telescope in space might gain another 1.5 magnitudes because it does not have to look through an atmosphere that blurs and scatters light.

The magnitudes listed in the table above are for looking through a telescope with your eye. If you use the telescope to take pictures, then you can get very long exposure times and this allows you to record much fainter objects. For instance, a few years ago, the Hubble Space Telescope's WPC2 recorded galaxies with a magnitude down to about 30 after exposing the pictures for about 5.9 days in total (1.4 days per passband). How long one has to expose a picture to record a very faint object depends on the magnitude of the object, on the optical quality of the instrument, on the efficiency of the recording device, and on whether the object is point-like or extended. From the quote about the recent HST observations, I find the following estimate of the limiting magnitude of a HST WPC2-like instrument for observing point-like objects (with a CCD camera):

\begin{equation} V_\text{HST} = 2.5 \log\left( \frac{t/[\text{sec}]}{(d/[\text{mm}])^2} \right) - 1.2 \end{equation}

where \(t\) is the exposure time and \(d\) the diameter of the main mirror or lens of the instrument. In comparison, for exposure of 35-mm film, and assuming a 3-magnitude difference between the limiting magnitude (i.e., the magnitude of a barely visible object) and the magnitude of a well-exposed object, the limiting magnitude is about

\begin{equation} V_\text{35mm} = 2.5 \log\left( I \frac{t/[\text{sec}]}{(d/[\text{mm}])^2} \right) - 7.5 \end{equation}

where \(I\) is the film speed in ISO. The HST WPC2 camera is about as sensitive as 330-ISO film, and the human eye, with a fixed effective exposure time of about 0.1 seconds, is about as sensitive as 500,000-ISO film.

4. How Bright Is It?

How bright an object appears in the night sky is indicated by its (apparent visual) magnitude, as described above. How bright an object appears depends on quite a few things.

If the object shines all by itself (for instance, a star or a galaxy), then its brightness depends on its distance to you, on how much absorption there is between you and the object, whether the object appears extended or point-like, and on how much light it emits (its intrinsic brightness, indicated by its absolute magnitude). The formula for the apparent magnitude \(V_\text{app}\) is:

\begin{equation} V_\text{app} = V_\text{abs} + 5 \log(d/[\text{pc}]) - 5 \end{equation}

where \(d\) is the distance in parsecs, and \(V_\text{abs}\) the absolute magnitude.


The absolute magnitude is the magnitude the object would have if it were at a distance of 10 parsec (32.6 lightyears). So, a star that has a visual magnitude equal to its absolute visual magnitude has to be at a distance of 10 parsec, regardless of what the value of the magnitude actually is. The star Arcturus in the northern constellation Bootes has a visual magnitude of 0 and is at a distance of 34 lightyears, so its absolute visual magnitude is also almost 0.

Here is a table with absolute magnitudes for a number of objects.

Table 3: Absolute Magnitudes

Object Vabs[d]/pc Vtext{app}
Jupiter +26.5 0.00002 −2.0
Sun +4.83 0.000005 −26.72
Vega +0.5 8.1 0.03
Rigel −7.1 280 0.12
supernova −19
galaxy −20
quasar −25
quasar APM 08279+5255 −32.2

If the limiting magnitude of your instrument is \(V_\text{lim}\), then you can see a point-like object with absolute magnitude \(V_\text{abs}\) out to a distance

\begin{equation} d_\text{max} = 10^{1 + 0.2×(V_\text{lim} - V_\text{abs})} \end{equation}

if none of the light is absorbed by material along the way. For instance, with an unaided eye (assuming \(V_\text{lim}\) = +4.5) you can see an unobscured supernova out to about 500,000 pc.

If instead the object shines only because it reflects light, then its brightness depends on its distance to you, its distance to the light source, and the relative positions of it, you, and the light source (because not all of the object is illuminated). The formula is

\begin{equation} V_\text{app} = V_{1,0} + 5 \log\left( \frac{d}{[\text{AU}]}\frac{D}{[\text{AU}]} \right) + f(φ) \end{equation}

where \(d\) is the distance of the object to you, in AU, \(D\) is the distance of the object to the light source (e.g., the Sun) in AU, \(f(φ)\) is the influence of the phase angle \(φ\), and \(V_{1,0}\) is the brightness of the object when it is at 1 AU from both you and the light source and at zero phase angle. When the phase angle is zero, then the object appears full (with no unilluminated parts visible), and then \(f(φ)\) is equal to zero.

In our Solar System, the planets have the following values for \(V_{1,0}\), the equatorial radius \(R_\text{eq}\), and the albedo \(A\):

Table 4: Planet Magnitudes and Albedos

\(V_{1,0}\) \(R_\text{eq}\)/km \(A\)
Mercury −0.42 2439 0.106
Venus −4.40 6052 0.65
Earth −3.86 6378 0.367
Mars +0.21 1738 0.12
Moon −1.52 3397 0.150
Jupiter −9.40 71492 0.52
Saturn −8.88 60268 0.47
Uranus −7.19 25559 0.51
Neptune −6.87 25269 0.41
Pluto −1.0 1700 0.38

The albedo can have any value between 0 (object reflects no light at all, is perfectly black) and 1 (object reflect all light). If you know the radius and albedo of an object of interest, then you can calculate its \(V_{1,0}\) yourself according to this formula:

\begin{equation} V_{1,0} = 14.10 - 2.5 \log\left( \left( \frac{R_\text{eq}}{[\text{km}]} \right)^2 A \right) \end{equation}

For instance, an asteroid with a radius of 10 km and an albedo of 0.1 has a \(V_{1,0}\) of about +11.6. If such an asteroid lies 0.01 AU from the Earth (5 times further than the Moon) in the direction away from the Sun, then its apparent magnitude is approximately +1.6: easily visible with the unaided eye. Such an asteroid may be visible to the unaided eye if it is less than about 7 million km from our planet.


[../../pic/docsleft.gif vorige][../../pic/docsrigh.gif volgende]

languages: [en] [nl];
Last updated: 2016−02−07