Astronomy Stack Exchange is a question and answer site for astronomers and astrophysicists. It only takes a minute to sign up.
Sign up to join this community
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?
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...
