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Re: [OM] Get your umbrellas ready....

Subject: Re: [OM] Get your umbrellas ready....
From: clintonr@xxxxxxxxxxxxxxxx
Date: Tue, 05 Aug 2003 13:46:51 -0500
Gawd, I love this list!   ;^)

"John A. Lind" wrote:

> At 11:49 PM 8/4/03, you wrote:
> >OK you did your cut and paste but what the hell is
> >magnitude of -2.9 and will appear 25.11 arc seconds wide?
> >Daniel
>
> Magnitude is the brightness of a celestial object.  It is based on star
> brightness.  Counter-intuitive, the lower the magnitude number the brighter
> the object.  The brightest stars are magnitude -1.  It is 2.512 time
> brighter than a magnitude 0 star.  Thus, a magnitude of -2.9 means it will
> be roughly six times brighter than the brightest star (-1 magnitude).
>
> The "magnitude" system dates to Hipparcus and Ptolemy when they assigned
> magnitude numbers of 1 through 6 to the stars they could see with the
> unaided eye.  They assigned first magnitude to a set of about 20 of the
> brightest stars they could see from their location.  Sixth magnitude stars
> were just barely visible to the unaided eye under the most favorable
> conditions (very dark sky).  Today the magnitude system assigns -1 to the
> brightest star(s).
>
> An arc-second is an angular measure.  There are 360 degrees to a circle, 60
> minutes to a degree, and 60 seconds to a minute.  To distinguish the use of
> minutes and seconds of angular measure from time measure, they are referred
> to as arc-seconds and arc-minutes.  25.11 arc-seconds is 0.006975
> degrees.  Angular measure is used to describe the diameter of a celestial
> object (typically planet or planetary moon) as viewed from Earth.  That and
> its distance can be used to determine its absolute diameter using simple
> trigonometry.  The utility of describing size by angular measure is in
> informing someone who wishes to observe the object how much magnification
> will be required (if any) to make it visible, and how much magnification
> will be required to make it appear a specific size.  If only provided with
> absolute diameter (e.g. in miles), one would have to know its distance to
> calculate how big it will appear.  It's much easier with direct information
> about how big it appears to an Earth observer without regard to
> distance.  Furthermore, in very early astronomy it was much easier and more
> immediate to measure size through a calibrated telescope using a reticle in
> terms of angular measure (and still is).  Measuring distance of an object
> requires several measures of object location against the most distant
> starfield over time as the Earth orbits the sun (e.g. one at the Vernal
> Equinox and another at the Autumnal).
>
> -- John
>
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