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Well, there's two things we'll need for this: apparent magnitude (the brightness that an object appears to have) and absolute magnitude (the actual brightness an object has). Both of these scales are logarithmic, with brighter objects being lower and dimmer objects being higher. Astronomers have determined that the Sun's absolute magnitude is 4.83. Knowing ...

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Yes. You derive the absolute bolometric magnitude by reference to the Sun. The Sun has a bolometric magnitude (i.e. integrated over all wavelengths) of about 4.75. Thus the difference in bolometric magnitudes is given by the log of the ratio of luminosities like so $$M_{\rm bol} = 4.75 -2.5 \log_{10} \left(\frac{L_*}{L_{\odot}}\right)\ ,$$ where $L_{\odot} =... 8 There isn't a one-to-one relationship between spectral type and absolute magnitude. Instead, there is a mean relationship with a fair bit of scatter around it. The reason is that the luminosity of a star of a given effective temperature depends on its composition/metallicity and how far along in its main sequence lifetime it is. Basically, late B-type main ... 8 Earth is already pretty bright due to cloud cover, with a typical albedo of .3-.35 -- that is, it reflects about a third of the visible light that hits it. That means it couldn't get more than about three times as bright even if it were perfectly reflective (albedo 1.0), which means about 1.2 magnitudes brighter. Spotting Earth from a great distance with ... 7 You just use the Luminosity:$M = -2.5 \log_{10}(L / L_0)$Where$L_0$is the luminosity of a magnitude 0 star$L_0=3\times10^{28}\mathrm{W}=79L_\odot$You don't need the other variables. You would need the distance if you wanted to calculate the apparent magnitude. You could also use the radius and mass to estimate the luminosity, if you weren't given it. ... 6 A star with magnitude 0 would be 85 times brighter than the sun (since Magnitude=-2.5 log(Luminiosity)) Referring to the H-R diagram on Wikipedia shows that there is quite a range of spectral types possible with this luminosity: from B type main sequence stars, and A type sub-giants, such as 4 Sco There are also G and K and M type Giant stars with this ... 5 Alpha Centauri A and B happen to be rather similar to Sol, and their absolute magnitudes are 4.38 and 5.71 respectively (Wikipedia). Add them together and you get absolute magnitude 4.10 (the scale is logarithmic, and backward). Sol, with absolute magnitude 4.83, should look 0.73 magnitude dimmer than αCen at the same distance, so magnitude +0.46, ... 4 The key to this is the so called Absolute Magnitude, which represents the visual magnitude from a distance of 10 parsecs (about 32 light years). The sun is much brighter than Proxima Centauri. It has an absolute magnitude of 4.8, and at a distance of 4 light years (the distance of Proxima), it would be be somwhat brighter than 1st mag, and so very easily ... 4 There are a couple of standard papers containing the table you want. Kenyon & Hartmann (1995) http://adsabs.harvard.edu/abs/1995ApJS..101..117K Table A5 contains many colours for stars as a function of spectral type. You need to combine this with something that gives absolute V magnitude along the main sequence, like that of Schmidt-Kaler (1982). An ... 4 Yes, you are correct. The luminosities can be added. Luminosity is the amount of electromagnetic energy emitted per unit of time (measured in$ \textrm{J} \cdot \textrm{s}^{-1} $or$\textrm{W}$). So if you have a multiple star system, the total amount of energy emitted (i.e. the total luminosity) is simply the sum of the energy emitted by each of its ... 4 There is a relationship between$E(B-V)$and the reddening in any other colour. The exact value depends on the type of dust, the extinction value itself and the intrinsic spectrum of the star (as does the 3.0 coefficient mentioned in your question). However, for the purposes of estimation, the canonical relationship is$E(U-B) = 0.72 E(B-V)$(e.g. Pandey et ... 3 Neither. You use the Luminosity Distance. $$M = m -5\log D_L + 5$$ This assumes bolometric magnitudes. If you are trying to estimate it in some photometric band then you must also calculate and apply a K-correction that will depend upon the intrinsic spectrum of the source. For$z=11.09$this cosmology calculator gives a model-dependent luminosity-distance ... 3 The absolute magnitude of an object is defined as the brightness of the object observed at a distance of$d = 10\,\mathrm{pc}$. With this distance, you can convert the luminosity density$L_\nu$in$\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{Hz}^{-1}$to a flux density$f_\nu$in$\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}\,\mathrm{Hz}^{-1}$: $$f_\nu = \... 3 "Accurate star size" is a problem. Obviously you do not have the resolution on your screen for accurate angular resolution of stars (they would require an absurdly fine resolution), and the dynamic range of brightness is also too low -- stars range over many orders of magnitude in brightness, far more than a screen can show. Deeply annoying. What one might ... 3 This is all absolutely true (pun intended). Finding absolute magnitudes is hard. For many types of star, we don't really know their absolute magnitude and so we don't really know their distance. For close-by stars we can get the distance by measuring the parallax (how far the star appears to move over a year due to the orbit of the Earth. Nearer stars ... 3 Your maths looks ok, bar the fact that 1/parallax is a biased estimate of the distance (but that can be forgiven so long as you are using data where the parallax uncertainties are much smaller than the parallax). Your main problem is that stars do indeed have a vast range of sizes. Thus if you really do want to show the relative sizes of the stars you ... 3 Here is my attempt to reconcile your calculations. If the AB g-band apparent magnitude is 22.5, then the flux density in the g-band is given by$$f_{\nu} = 10^{(-48.6-22.5)/2.5} = 3.63 \times 10^{-29}\ {\rm erg\ cm}^{-2} {\rm s}^{-1} {\rm Hz}^{-1}$$If the distance is 1991 Mpc, then the absolute g magnitude is$$ M_g = m_g - 5\log d + 5 = -19.0$$The ... 3 The page you quote is guilty of a gross simplification (or error). The luminosity ratio cannot be estimated from the difference in visual magnitudes for stars of different spectral type. You need to use the difference in bolometric magnitude, which indicates the luminosity integrated over all wavelengths. This will involve making bolometric corrections to ... 3 This graph is not the light curve of a cepheid. It is used to find the average brightness of the star, and from that, estimate its distance. The rate at which cepheids pulse is related to their average luminosity (and so their average absolute magnitude). This graph relates the period to their average absolute magnitude. To use the graph, observe a cepheid ... 3 But you demonstrate that you know the answer to this question! Fainter objects have larger magnitudes. So as the white dwarf becomes cooler and redder, its blue magnitude (G_{BP}) grows by more than its red magnitude (G_{RP}) and the colour-index (G_{BP} - G_{RP}) becomes larger. This is very similar to the commonly used B-V colour, which is larger ... 3 Color is a difference, not an absolute value Being "blue" doesn't (necessarily) mean that a light source has a large flux in the B band. A color is not an absolute value; it is the ratio between two fluxes or, equivalently, the difference between two magnitudes. Being blue means "More flux in some short wavelength band (e.g. B) than in a ... 2 The JPL CNEOS asteroid size estimator and various asteroid albedo papers cite Harris and Harris 1997. That paper is behind a paywall, but Stuart and Binzel 2004 attribute this formula to it:$$ H = C - 5 \log_{10} D - 2.5 \log_{10} p_V$$where$H$is absolute magnitude,$p_V$is albedo, and$C$= 15.618. 2 A star with absolute magnitude of -7 is big and powerful, It would be among the most powerful stars in the galaxy. The star would be near the end of its life, so comparing with other aging stars: Aldebaran is slightly more massive than the sun. It is a red giant (or orange giant according to some sources) It has an absolute magnitude of about -0.6. ... 2 The SDSS exercise shows how to estimate a star's actual radius. If you use this radius, you should also use different model materials for different color index values, since luminosity per unit area is a function of temperature. If you prefer to avoid that complexity, base your model stars' radii on visual magnitude alone. If you put the model stars at a ... 2 Stars born together in clusters have more-or-less the same age. As a rule of thumb, any spread in age, measured in millions of years, is smaller than the extent of the cluster in parsecs. For most stellar clusters, smaller than a few pc, the only chance of measuring age differences occurs in the first 10 million years of life. There is no evidence for age ... 1 The absolute magnitude quantifies the luminosity of an object at a standard distance of$10\ \mathrm{pc}$from Earth. For example, in the case you mentioned, Vega becomes dimmer than at its actual distance (about$7\ \mathrm{pc}\$). To answer your question, I don't think there is an actual star with exactly 0 absolute magnitude. If there is then, following ...

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Spica (α Vir) is a binary star. The Wikipedia article cites Herbison-Evans et al. 1971, whose Table III says: Absolute magnitude of primary (Mv1)   -3.5±0.1 Absolute magnitude of secondary (Mv2)   -1.5±0.2 So the confusing "-3.55 (-3.5 / -1.5)" can be interpreted as "combined (primary / secondary)."

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Luminosities cannot be calculated from magnitudes without knowing the distance to the source (and perhaps something about extinction). I assume therefore that you are talking about absolute magnitudes in those filter band passes. Luminosity can also not be arrived at from a single absolute magnitude measurement in the way you suggest in your post, since ...

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