The brighter the star, the lower its magnitude. The brightest stars in our sky have magnitudes of 1 or less; there are only 20 such stars, according to the Bright Star Catalogue. 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 5026 such stars, according to the Bright Star Catalogue. In big cities, where there is a lot of light pollution, the limiting magnitude (the highest magnitude that one can see with the naked 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. If the limiting magnitudes are equal, one can see roughly the same number of stars at any given time of the night from any place on Earth, but your latitude determines which stars you can see at all at any time of the year. At the poles, you can see about half the total number of stars (in a year) that one can see at the equator, but the stars that are visible from the poles are always above the horizon there, while the stars that are visible from the equator are above the horizon only half of the time. For places between the poles and the equator one can see intermediate numbers of stars: some are always above the horizon, and others are above the horizon only part of the time.
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.
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. In the left one, 10 must be subtracted from the magnitudes V. In the right one, 10 must be added. 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.
A 3rd-degree linear least-squares fit to the star numbers is the following:
log(N) = 0.754 + 0.4896 V + 0.001159 V^2 - 0.000235 V^3where log(N) is the decimal logarithm of the number of stars with a magnitude up to V.
N = 10^(log(N))
G = 5 log(d/d_eye)where d is the diameter of the main lens or mirror, d_eye is the diameter 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 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:
V_lim = -0.4 + 5 log (d/mm)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.
V_lim = 6.6 + 5 log (d/in)
|50||2.0||8.8*||9.8||98,000||pair of 7x50 binoculars|
|150||5.9||10.4||20||509,000||6-inch telescope (e.g. Newtonian)|
|2400||94.5||16.5||340||124 M||e.g., Hubble Space Telescope|
|10,000||393.7||19.5||1300||1,004 M||e.g., Keck telescope|
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, the Hubble Space Telescope's WPC2 recently 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 length of the exposure, 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):
V_hst = 2.5 log([t/sec][d/mm]^2) - 1.2where 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
V_35mm = 2.5 log(ISO [t/sec][d/mm]^2) - 7.5where ISO 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.
[LS 14-23 October 1996]
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). If there is no absorption of light between the object and you, then the formula is:
V_app = V_abs + 5 - 5 log([d/pc])where d is the distance in parsecs. The absolute magnitude is the magnitude the object would have if it were at a distance of 10 parsec (32.6 lightyears). Below is a table with absolute magnitudes for a number of objects.
d_max = 10^(1 + 0.2*(V_lim - V_abs))if none of the light is absorbed by material along the way. For instance, with an unaided eye (assuming V_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, its reflective properties, the relative positions of it, you, and the light source (because not all of the object is illuminated), and the absorption (if any) of light along the path of the light.
Assuming that there is no absorption, and that the reflection is diffuse (not like a mirror, but reflecting parts of each ray in many directions), we have
V_app = V_(1,0) + 5 log([d/AU][D/AU]) - 2.5 log(f(phi))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(phi) is the influence of the phase angle phi, 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(phi) is equal to zero. For diffuse reflection where all illuminated parts of the object have the same brightness (per unit spherical angle) as seen from any direction, we have
f(phi) = (1 + cos phi)/2.
In our solar system, the planets have the following values for V_(1,0), the equatorial radius R_eq, and the albedo (diffuse light reflection fraction) A:
V_(1,0) = 14.10 - 2.5 log([R_eq/km]^2 A)The constant 14.10 is equal to V_sun + 5 log(AU/km). For a derivation of these equations, see Explanation Page 9.
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.
If the object reflects like a mirror, then you'll only see it if its orientation is exactly right to reflect light from the light source (the Sun) to you. For a plane mirror surface the formula is
V = -37.1 - 2.5 log(S_app/d^2)where S_app is how big the surface area of the mirror looks to the observer (which varies between the actual surface area if the mirror is facing the observer, and zero if the mirror is seen edge-on), and d is the distance of the mirror to the observer. This formula is correct as long as the mirror seems smaller in the sky than the Sun and the mirror is oriented so that it reflects the sunlight to the observer.
For instance, a plane mirror in the sky with an apparent surface area of 1 square foot (0.093 m^2) at a distance of 200 miles (320 km) would appear to have a magnitude of -7.0 if it was oriented just right: this would be much brighter than the brightest star, but much fainter than the full moon. The same mirror in a geostationary orbit at 22,000 miles distance would appear to have a magnitude of +3.2: still visible to the unaided eye, but no longer very bright.
[LS 4 November 1997]