38

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 ...


25

One way to answer would be to consider the brightest star in our sky (other than the Sun), which is Sirius. Then determine how far you would have to be from our Sun for it to be as bright as Sirius is from here. That turns out to be 1.8 light years. That's not even halfway to the nearest star, so if you're in any other star system, then our Sun is just ...


20

John's answer is correct. For a few more details: Stars brightness works out to roughly the 4th power of their relative mass. This falls off a bit for very large stars, but for smaller stars it's in the ballpark. (Red-giant phases not included.) Proxima Centauri is about 12% the mass of our sun, and about 14% the sun's diameter. At 12% its mass, its ...


17

The easiest way to determine the magnitude of a given star is probably to use the Pogson relation. The idea is to determine the magnitude of a star knowing the magnitude of a reference star; it is thus quite easy, using a well-known reference as Vega or Sirius. The Pogson relation is given by: $$m_1-m_2=-2.5\ log\ \left({\frac{E_1}{E_2}}\right)$$ where $...


15

As Mark Adler mentioned, the best way is to compare the brightness to other nearby stars. I'm going to assume that you have instantaneous travel time, and also take into account that you are actually getting closer to stars depending on the direction you go. I'm using this table from Wikipedia. I'm going to go no further on the list than Sirius, and assume ...


13

Well, having seen Neptune and identifying Neptune are two totally different things. Let's tackle this one step at a time. Your link says: From these observations it would appear that, if seen upon a perfectly black background, a star of magnitude approximately $8.5$ would be at the limit of unusually good vision. The problem with this is that there are ...


11

The general idea is that the more extinction there is, the less you'll be able to see stars. So there is a direct relation between the decrease in flux observed for any individual star and the number of stars you can see in that general direction. So the surface distribution of star counts is an excellent tracer of optical extinction, and $\Delta m = 2.5 \...


9

Let's take an average albedo for the Earth of 0.3 (it depends, which hemisphere is visible, how much cloud cover etc.). That means the Earth reflects 30% of the light incident upon it. The flux $f$ falling on the Earth is given by $$ f_{\odot} = \frac{L_{\odot}}{4\pi d^2} = 1.369\times10^{3}\ Wm^{-2}$$ where $L_{\odot}=3.85\times10^{26}\ W$ from the Sun and ...


8

The inverse-fourth-power law you're referring to is valid for light emitted from a source, reflected non-specularly — i.e. in all directions — from a reflector, and detected by the original emitter. If the reflector is a mirror, the observed flux just follows the normal inverse-square law with the nominator equal to $(2d)^2$ instead of $d^2$, since the ...


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

What is the apparent magnitude of the Sun at that distance? Fortunately, there's a formula known as the Distance Modulus which is defined to calculate precisely this information. The equation is given by $$m-M=5\log_{10}(d_{pc}) - 5$$ where $m$ is the apparent magnitude of your star, $M$ is the absolute magnitude, and $d$ is the distance from the star, in ...


6

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 ...


6

According to Oxford Dictionary of English, the word "modulus" is the diminutive of the Latin modus, meaning measure (modus, in turn, comes from Proto-Indo-European mod-os, Nocentini & Parenti 2010. Hence, the distance modulus is a "measure of distance", just like the modulus of a vector is a way of measuring its size, and Young's modulus is a way of ...


6

You can work in magnitudes, but the magnitude scale is logarithmic. Instead you can use luminosity. The moon has a luminosity that is 400,000 times less than the sun, so the maximum theoretical "lunar irradiance" is about 0.0034 watts per square metre. In practice, the luminosity at the surface is less than this, with Wikipedia quoting "how ...


5

A device called a photomultiplier is attached to a telescope. The telescope is pointed at the target star and an aperture is closed down to avoid as much spurious light as possible, and a reading is taken. The photomultiplier literally count the photons hitting it. Next the telescope is pointed at a well-known reference star in the same part of the sky, ...


5

Here's my take on it. I'm not sure what the mystery is - there appear to be two contributing factors. Both the Moon and the Earth simply reflect/partially absorb the light that is incident upon them from the Sun. The overall of albedo is of little consequence, since the reflection of any light will result in an object that appears bright against the night ...


5

The key to answering your question is to know the albedo of the asteroid compared to the Moon. The Moon is rather dark (reflecting 7% to 12% of the light if I remember correctly, about as black as worn asphalt). I think, but do not know with certainty, that the asteroid would not be much darker. For the dimensions that you give, the asteroid would be ...


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, ...


5

Textbooks typically express the flux vs. magnitude relation something like this: $$m_2 - m_1 = -2.5 \log_{10} \frac{f_2}{f_1}$$ which we can transform into this: $$\frac{f_2}{f_1} = 10^{(m_1 - m_2) / 2.5}$$ so the original version of your formula was correct. If m2 - m1 = 0.5, then f2 / f1 = 0.63. Some readers find comparisons like "x% fainter" ...


5

Without information about stellar radii, I think it's reasonable to assume $R_A \approx R_B$. Then your equation becomes $$ m_p - m_s = -2.5 \log \frac{F_A}{F_B} $$ and you can compute $k$ and the non-eclipsed total magnitude.


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

It's too bright to look at. Your astronaut has permanent retinal damage The apparent magnitude of the sun from Earth is -26.7. From the asteroid belt it is -24.7 (http://www.1728.org/magntudj.htm). So it is dimmer, but it is still far too bright for her to look at without eye protection. Since she is standing in the vacuum of space, there is no atmosphere ...


4

Ask Ethan: How bright is the Earth as seen from the Moon? has some detailed explanations, including discussions of lunar eclipses as seen from the moon. It doesn't include magnitude calculations, but it concludes that a “full Earth” as seen from the Moon is about 43 times brighter than the full Moon is as seen from Earth. When the icecaps are larger and ...


4

Stars vary a lot in both size and brightness. The nearby stars you mentioned are less bright than many stars that are much further away. We see the bright stars but not the less bright ones. Its like comparing the effectiveness of the old 2-battery flashlight to the 5-battery one. Both are flashlights but one is much brighter than the other.


4

The thing about detecting exoplanets by eclipses is that the eclipses are repeatable. You have the same star, whose light curve dims in the same way with each eclipse, which gives you information about the planet's orbital period and its apparent size relative to the star. The Kepler mission observed many fluctuations in stellar brightness that were not ...


4

First you must estimate the luminosity. That may be difficult unless the magnitude you have is a bolometric magnitude. Then you assume some mass to light ratio (actually mass to luminosity ratio, where in solar units the M/L of the Sun is 1). The M/L ratios of galaxies depend on galaxy type (and age, metallicity, star forming history) and vary from about ...


4

Yes, brightness is inversely proportional to the square of the distance, and the square of 632 is about 400,000. It's probably appropriate to round the value to 1s.f. to give "600AU". Of course, at that distance the sun appears point-like. The distance to the Oort cloud is rather hard to define. It's not like you reach 1000 AU and see a Star Wars like "...


3

The photo in your question is -- well, not exactly fake, but a composite. The biggest clue is that Earth is too close to the horizon; it would have had to be taken from within a few degrees of the boundary between the near and far sides of the Moon, and none of the Apollo missions landed there. Furthermore, take a close look at the cloud patterns. The view ...


3

If your sources are within 100pc then the best thing to do is assume the extinction is zero, unless you are looking for absolute magnitudes that are a lot more precise than say +/- 0.1 mag. I think you would be very unfortunate if the extinction was any more than about ~0.1 mag in the V-band (and less than that at R and I). Beyond that you are really ...


Only top voted, non community-wiki answers of a minimum length are eligible