That monster is the star, Deneb, familiar in the constellation Cygnus in the northern sky. It is among the brightest stars in the sky, but what cannot be obvious is that it far outstrips its visual rivals in intrinsic brightness, so that it is the bully in our neighborhood. It is a magnitude 1.25 star at a distance of 800 parsecs.

Now, the Sun has an absolute magnitude of 4.83, which is by definition its ideal visual magnitude at a distance of 10 parsecs. So to have a visual magnitude of 1.25, the Sun must appear 10^(.4*(4.83-1.25)) = 27 times brighter than it does at 10 parsecs. That means it must be 1/sqrt(27) times as distant , or about 3.7 parsecs away, or 216 times closer than Deneb, to show the same visual magnitude.

So, conversely, the earth would have to be orbiting Deneb at 216 AU if Deneb were to have the same apparent brightness as the Sun. That’s over five times the distance of Pluto from the Sun, which averages about 40 AU.

The period of an orbiting body is given by M/r^2 ~ r/T^2 or T ~ sqrt( r^3/M ), which embodies Kepler’s third law for a central body of a given mass, but allows us to account for a variable mass. Deneb has about 19 solar masses, so the “year” of this Deneb-earth would be sqrt( 216^3/19 ) = 728 earth years. Well, at least that’s comprehensible!

I was led into these contemplations after having taken the following photograph of Deneb, with its star background in the Milky Way, using my Nikon D5100 mounted on my telescope purely for its use as a “clocked” mount. The exposure is about 20 seconds, so the lack of guidance is not so bad, as long as I get reasonable alignment, for which purpose I have a well practiced system! … ( click to enlarge )