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This website explains:

The apparent brightness of a star is proportional to 1 divided by its distance squared. That is, if you took a star and moved it twice as far away, it would appear 1/4 as bright; if you moved it four times the distance, it would appear 1/16 as bright.

It then goes about explaining how this works.

The problem I have is that I don't understand the reasoning. Thus: in simpler terms than this website used, how does this work?

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    $\begingroup$ They don't. Your title is an incorrect statement. Perhaps you are talking about brightness? $\endgroup$
    – ProfRob
    Nov 5, 2021 at 8:46
  • $\begingroup$ @ProfRob. Aha, there is a difference between apparent magnitude and apparent brightness. But if so, what exactly is the difference between the two? $\endgroup$ Nov 5, 2021 at 9:16
  • $\begingroup$ Although this website implies that the two are the same: astronomy.swin.edu.au/cosmos/a/Apparent+Magnitude $\endgroup$ Nov 5, 2021 at 9:20

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The statement of the title is not correct, however I think just a few terms get incorrectly used.

In general it's simply geometry or simple physics: the inverse square law:

You have a star with a certain luminosity (thus overall energy output), let's call it $L$. At a distance $r$ this gives an illumination or irradiance of x W/m² as the total light created by the source is distributed over a sphere of radius $r$. The sphere has an area of $A(r)=4\pi r^2$. Now, if you double the distance $r$, the area to illuminate the whole sphere is now $A(2r) = 4\pi (2r)^2 = 4\pi \cdot 4 \cdot r^2$, thus four times the area, thus you get only 1/4 of the light per unit area. The distance doubled, the brightness decreased by a factor of four.

If you now talk about (apparent) magnitude, it's different as magnitude is the negative logarithm of brightness. So you double distance, you simply add a certain fixed amount. Due to how the apparent magnitude is defined (5 magnitudes represent a factor of 100 in brightness, $m - m_{ref} = -2.5\log_{10}(I/I_{ref})$, a doubling of distance of a star increases its magnitude linearily by about 1.50 magnitudes.

It's very hard to define "ease of visibility". So anything which has a magnitude less than 6 can be visible, if you have normal eye sight and a dark night and sky. As our eyes (and all our senses) also work logarithmically, if the apparent magnitude of an object decreases by 1 (or 1.5), it gets quite much better visible already. Whether that subjectively qualifies as four times as bright ... I cannot say and probably depends on subjective perception and time and circumstance as well: sensitivity of human eyes also depends on the surrounding brightness, adaption time, etc...

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  • $\begingroup$ Thx for the "inverse square law" link. It was really interesting and I especially liked the chart. // Actually I don't bother too much about labels, what I want to know is if a star is four times easier to see in the night sky if it just half the distance away. Is it? $\endgroup$ Nov 5, 2021 at 10:10
  • $\begingroup$ That's quite a bit different question. Yet I amended a closing paragraph $\endgroup$ Nov 5, 2021 at 10:20
  • $\begingroup$ How do you know that the star "appears 72% closer than before"? What observation is that based on? I have no perception of distance of stars at all - or light sources. I can only judge them, when I see their motion with respect to anything I can relate to. $\endgroup$ Nov 5, 2021 at 10:26
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    $\begingroup$ I don't understand the hypothetical situation. Elaborate. What circumstance would put me in the situation that it would appear closer? Or are you asking what circumstance would make me assume it is 72% closer (thus at 1/4 distance)? Hint: I wouldn't. I would first assume a brightness change and not a distance change. $\endgroup$ Nov 5, 2021 at 10:31
  • $\begingroup$ Let's rephrase this: (Astronomical) distance is never visible. It is always an inferred property - from brightness, from knowledge about the type of object, from spectral analysis about its movement in line-of-sight, or from positional comparison over an extended period of time. But there is no-way I can see its distance change. I can only derive a distance change from analysing data I (or other people) gathered. And brightness alone NEVER does the trick - at least I need to know what I look at AND know that this type of object has a fixed absolute brightness. Few objects do. $\endgroup$ Nov 5, 2021 at 15:26

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