39 votes
Accepted

Is the Sun visible from Proxima Centauri to human eyes?

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 ...
20 votes

Why are nearby stars like Proxima Centauri and Barnard's star not visible to the naked eye?

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 ...
  • 22.9k
18 votes
Accepted

Could Neptune be viewed with the naked eye from Uranus?

According to https://arxiv.org/pdf/1808.01973.pdf, the magnitude of Neptune follows the relationship (formula 17, page 25): $ V = 5 \log_{10} (rd) - 7.00 + 7.944 \times 10^{-3} α + 9.617 \times 10^{-5}...
14 votes

Could Neptune be viewed with the naked eye from Uranus?

Supplementary answer supporting @PierrePaquette thorough and well-source answer: I tried the nice new JPL Horizons interface and fired up Excel which I haven't used in a long time. For years 1800 to ...
  • 30.4k
13 votes

Have there been successful attempts at spotting Neptune with the naked eye?

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 ...
11 votes
Accepted

Determining the extinction by counting stars?

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 ...
  • 16.8k
9 votes
Accepted

As viewed from Mars, what are Jupiter's and Saturn's maximum brightness in apparent magnitude?

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. ...
  • 34.2k
9 votes

How bright is the full Earth during the lunar midnight?

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 $f$ ...
  • 120k
9 votes
Accepted

Brightest Stars (by its Apparent Magnitude) List beyond 300

The Yale Bright Star Catalogue lists stars brighter than about magnitude 6.5. It is available as a compressed text file documented in the readme file I've also made a google sheets version sorted by ...
  • 93.4k
8 votes
Accepted

Snowball Earth brightness

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 ...
  • 370
8 votes

Could Neptune be viewed with the naked eye from Uranus?

The brightness of a Solar System object, seen in reflected light, depends on how far it is from the Sun, $d_s$, and how far away it is from the observer, $d_o$, (and the angles between them). Both ...
  • 120k
7 votes
Accepted

Sun's apparent magnitude at 2.5 AU?

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 ...
  • 14.5k
6 votes
Accepted

Why is there "modulus" word in the "distance modulus" term?

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....
  • 34.2k
6 votes
Accepted

Total apparent magnitude of eclipsing binary system

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 ...
  • 16.8k
6 votes
Accepted

What is the "lunar irradiance" received by the Earth from the full Moon?

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 &...
  • 93.4k
6 votes

How long does lunar opposition surge last? Are there measurements of the full Moon getting suddenly brighter?

The lunar opposition surge has been well studied, likely because we can study it in detail, we have surface samples, and so it serves as a baseline for other bodies in the solar system (as it does for ...
6 votes

Is subtracting one apparent magnitude from another ever practical?

Two stars in a binary system are at the same distance from Earth. If they have similar spectral types then the difference in their magnitudes tells us the ratio of their luminosities. $$\Delta m = m_1 ...
  • 120k
6 votes

What is the maximum magnitude that the JWST can safely observe?

The Bright object observation by JWST is still being studied. The known problems that arises with Bright object observation is image persistence. Although it is said that James Webb space Telescope ...
5 votes

Why do dark objects look white from a distance? (Moon, Ceres, but not Earth!)

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 ...
  • 120k
5 votes

How bright is the full Earth during the lunar midnight?

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 ...
  • 380
5 votes
Accepted

What's the difference between apparent brightness and apparent magnitude? Are they the same?

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 ...
  • 2,866
5 votes

Is the Sun visible from Proxima Centauri to human eyes?

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 ...
5 votes
Accepted

Asteroid visible during the day?

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 ...
  • 7,218
5 votes

Comparing different magnitudes. Is this statement correct?

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) ...
  • 16.8k
5 votes

How many photons does it take to determine the existence of a distant object?

What matters is how many photons you collect versus how many you would expect to see if the object wasn't there. Photons would be present, without a source, for a variety of astrophysical (e.g. ...
  • 120k
5 votes
Accepted

If Planet Nine indeed exists, how large a impact event there should be to be seen from Earth?

We can compare the brightness of a comet hitting planet 9 with the 1994 impact of Shoemaker-Levy 9 / SL9 on Jupiter. It was seen by Hubble, and Galileo which was orbiting Jupiter at the time. View ...
4 votes

Why are nearby stars like Proxima Centauri and Barnard's star not visible to the naked eye?

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 ...
  • 4,346
4 votes

Sun's apparent magnitude at 2.5 AU?

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). ...
  • 93.4k
4 votes

Why do dark objects look white from a distance? (Moon, Ceres, but not Earth!)

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 ...

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