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Not close to being able to read by. Proxima Centauri is about 13,000 AU from the two binary Centari stars. Together they have about twice the luminosity of the sun but at 13,000 AU, that's roughly 2/169,000,000 the visible light that the Earth gets from the Sun. The brightness variation of the full moon to the Sun is about 1 to 440,000, so, some rough ...

11

You possibly are confused by these two entries in Wikipedia (click on the quotations to go to the original distinct entries: Canis Minor contains only two stars brighter than the fourth magnitude, Procyon (Alpha Canis Minoris), with a magnitude of 0.34, and Gomeisa (Beta Canis Minoris), with a magnitude of 2.9. and This is the list of  notable stars ...

9

"ugriz" is short for U-band, G-band, R-band, I-band, and Z-band, detailed on the Wikipedia article on photometric systems: $$\begin{array}{|c|c|} \hline \text{Band}&\text{Effective Wavelength Midpoint }(\lambda_{\text{eff}})\\ \hline \text{U} & \text{365 nm}\\ \hline \text{G} & \sim\text{475 nm}\\ \hline \text{R} & \text{658 nm}\\ \hline \... 9 Take a look at the HR-diagram. You have forgotten to take into account that stars come in vastly different luminosities. For example, we can barely see all red dwarves out to a distance of 10pc, while we know that they vastly outnumber the more luminous stars. The luminous red giants however we'll be able to see on side of the edge of the disc, but ... 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 When quoted like this they would usually be the most likely value (peak of the probability distribution) or the median value (where half the distribution is above or below), and then the superscripts and subscripts would be increments that would contain \sim 34% of the probability distribution. In the M87 papers, the numbers are the 50th percentile (median)... 7 Aabaakawad's answer highlights an important aspect of any constellation: They are defined by a surface area on the sky, behind which many, many stars hide. In order to answer just how many stars, you'll have to ask "above which brightness threshold?". I think you should accept his answer, but I have two comments: "…if one were to map every star visible to ... 6 As @RobJeffries mentions, you can calculate the difference from the equations which define magnitude and flux. But this particular case is even covered in the Wikipedia article on apparent magnitude: For objects within the Milky Way with a given absolute magnitude, 5 is added to the apparent magnitude for every tenfold increase in the distance to the ... 6 The apparent magnitude classification was done arbitrarily by Ptolemy. His thought was to set the 20 brightest stars to the first position, the less bright stars to the second position and so on up to the faintest stars which were given the sixth position. After the use of Pogson's law we were able to give to stars not only natural numbers, but also numbers ... 6 The magnitudes don't sum and subtract that way: consider two 0.00 magnitude stars closely together. Is their combined magnitude 0.00? Actually, it is -0.75. This paper might help you with the derivation of the formula for addition of magnitudes, but you are interested in subtraction. You just need to rearrange the formula in the paper to$$m_a=-2.5\log(10^{-...

5

Yes, $B$ and $I$ are the same things in this context. The $\Delta m$ you derived goes by the name "distance modulus" when one of the distances is set to $10\operatorname{pc}$. The thing to be careful of is that $I$ is also sometimes used for spectral radiance (or surface brightness) when working with radiative transfer. Spectral radiance doesn't obey the ...

5

Mu Sagittarii is a star system, not a single star. If that can be included, then Eta Carinae should be included, and it has an absolute magnitude of -12.0. It's a star system about 7,500 light-years from Earth. It looks like the brightest (absolute magnitude) single star visible to the unaided eye is WR 24 (in Carina Nebula). Its absolute magnitude is −11.1 ...

5

The R in that equation is the distance from the star to observer, not the star radius. The light emitted from the star is distributed uniformly on a sphere of radius R, and when the light arrives to the Earth, that sphere will have a radius equal to the distance Earth-star. Therefore, the second relation for the two fluxes is about the apparent magnitudes (...

4

Caveat this with the fact that I am not familiar with this code. The input asks you for the flux through the filter, but a filter has a bandpass, so if you have a 10 nm bandpass filter and you measure $1 W/m^2$, then you measured $0.1 W/m^2/nm$ (sorry, I'm a physicist so I never got used to measuring wavelengths in Hz). As for your second question, the ...

4

The "Schechter-like" Equation 3 in that paper is meant to give the value of black-hole mass density in units of "number of black holes of a particular mass per volume per log black hole mass", with the volume in units of megaparsecs and the "log black hole mass" in units of 0.1 log solar-mass units -- i.e., 0.1 dex. Since the integral involves multiplying ...

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Albedo is a measurement of reflected light. It varies between 0 and 1, with 0 being completely black (reflecting no light) and 1 being completely white (reflecting all light). The albedo of coal is about 0.04, which is actually very similar to the albedo of our moon. The Albedo of ice is about 0.7. Only planets and other small objects are described using ...

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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 ...

4

To calculate a "luminosity light curve" from a time series of V-band photoetry, you need two things. You need to know the distance. The distance to Betelgeuse is uncertain and that means the absolute value of the luminosity you get will also be shifted systematically up or down by whatever distance you adopt. The V-band only contains a small ...

4

It's hard to say because the situation is very similar to this question: How much of the Sun's disk must be covered for a visible shadow to be cast? The major problem lies in the sensitivity of the human eye, which is able to detect illumination changes sooner or later. Basically, there are people, who might not know about the ongoing solar eclipse until ...

4

The absolute magnitude of the Earth is about $H = -3.99$. This means, from $1$ AU away from the Sun and the Earth, the Earth has an apparent magnitude of $-3.99$. Voyager 1 is $152.26$ AU from the Earth, and $151.85$ AU from the Sun. Using the formula for apparent magnitude, $$m=H+5\log _{10}{\left({\frac {d_{BS}d_{BO}}{d_{0}^{2}}}\right)}-2.5\... 3 Consider for example these bright stars (V and B-V values from Yale Bright Star Catalog). Near V=0, B/V is ill-behaved, and B*V is near 0. Name B V B-V B/V B*V Arcturus 1.19 -0.04 +1.23 -29.7 -0.05 Vega 0.03 0.03 0.00 1.00 0.00 Capella 0.88 0.08 +0.80 11.0 0.07 Rigel 0.09 0.12 -0.03 0.75 0.01 Now imagine ... 3 The unaided eye can typically see mag 6 objects. With your telescope, you can see an additional 8 magnitudes. This requires a factor of 100^{8/5} of additional light gathering power (since 5 magnitudes equals 100 times the brightness). I.e. about 1580 times larger aperture than the pupil of the eye. Assuming the pupil has a diameter of 7mm, then the ... 3 The ugriz values are just measures of the stellar brightness taken in 5 bands from the near ultraviolet u through to the near infrared z. You need the distance to the star to convert the apparent SDSS magnitudes to absolute magnitudes. Then you need to find a relationship (or relationships) between bolometric correction in a particular band and the colour ... 3 I found references which indicate that Venus usually goes dark. Check out the graphs in this paper for one example. The exceptions happen when Venus' orbit, which is slightly out of the ecliptic, brings it to inferior conjunction when projected onto our (Earth's) orbital plane, but Venus remains at a slight angle relative to our viewing lines. From ... 3 As the phase angle of Venus changes, so does its distance to Earth. More about modeling the phase angle dependence of apparent magnitude can be found in this answer to Calculating the apparent magnitude of a satellite. For more on that subject, see What is the difference between albedo, absolute magnitude or apparent magnitude? and also these: 1, 2, 3. When ... 3 TheSkyLive's Mars information page has graphs of its distance and apparent magnitude from Earth. If you set the same year range for each, you can see how the quantities are related. They provide similar pages for all the planets and selected comets and asteroids. 3 The Earth has an absolute magnitude of -4 This means that from 1 AU and at zero phase ("full Earth") it would have a magnitude of -4. Of course the only place that satisfies those two requirements is the location of the sun, but that is not the point; The apparent magnitude depends on the distance from Earth, and the phase angle of the Earth. ... 3 You need to do what is called “propagation of uncertainties”. You can search to get more information on that, but briefly if you have some function f(x) that depends on variable x, then the uncertainty \sigma_x on the quantity x is related to the uncertainty on f by$$ \sigma_f = \sqrt{\left(\frac{\partial f}{\partial x}\right)^2 \sigma_x^2} $$... 3 You are using the formula in an incorrect way. The meaning of the formula is the following: the flux we receive on Earth is$$F = \frac{L}{4\pi d^2} where $L$ is the total luminosity of the star and $d$ is the distance from the star to Earth, not the radius of the star. You may see that this formula is not particularly useful to solve your problem, since ...

3

In addition to @User123 answer, and assuming you are subtracting luminosities and not magnitudes, there is usually one more obstacle: the catastrophic cancelation. In short, if the difference between the total and the source B luminosity is near the accuracy of either of them, you will get no meaningful estimate for the source A. In practical terms, I wonder ...

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