How bright is the Milky Way at night?

[Dr. Doolittle]

How bright is the Milky Way at night?

That silvery expanse of nebulosity, rich in magnificent and colourful
stars so closely packed together as to appear as grains of silicates
glinting in a bed of oceanside sand across the vast firmament of the
black sky on a moonless night.

But if you had to use math to give an analytical estimate of that
total Milky Way brightness, what value would you assign it? How would
you work the sums?

AA Institute formulated an original methodology using natural
logarithmic equations, and published his results online thus:

The Milky Way galaxy's absolute magnitude, accepted in most official
journals as an astronomical constant, is -20.5. That figure is based
on the assumption that the *entire* galaxy is viewed face on, as one
integrated object, from a standard distance of 10 parsecs (32.6 light
years). Now, since we are located in one of the spiral arms of the
Milky Way not far from the galactic plane and only get an "edge-on
view" looking inwards towards the centre of the Milky Way, we see only
50% of the galaxy's total brightness stretching across the night sky
(since the remaining 50% is on the *other side* of the dense galactic
core, and not directly in view to us).

Now, the standard formula for evaluating the brightness ratio, R,
between any two objects of magnitudes M1 and M2 is given by:-

R = 10^[0.4*(M1-M2)]

Hence, this formula can be used to "reduce" the Milky Way galaxy's
total absolute magnitude of -20.5 by 50% to give a figure of -19.7,
representing the "portion" that we see stretching across our night
sky. Since we are located at a distance, d, of about 8,200 parsecs
from the galactic centre [Source: Handbook of the British Astronomical
Association], the apparent magnitude, m, of the bulk of this "portion"
can be calculated from:-

m = M - [5 - 5 * log10(d)] = -19.7 - [5 - 5 * log10 (8200)] = - 5.1

Hence, the net integrated magnitude of the "visible" Milky Way
stretching across our night skies ought to be about -5.1.

However, there are various dark, intervening clouds of interstellar
gas and dust, such as the "Cygnus Rift", the "Coal Sack" near Crux,
many dark clouds in Sagittarius looking towards the centre of the
galaxy, etc. which contribute to dimming the overall light reaching
Earth from the broader Milky Way. Hence, if one makes a 10% (0.1-
magnitude) allowance for light extinctions owing to such obscuring
interstellar media, one will arrive at a net magnitude of -5.0.

This would be one way that I would *analytically* estimate the Milky
Way's total integrated brightness as -5.0 magnitudes.

This figure of course relates to a full 360-degree view of the whole
Milky Way. In actual practice, from a particular location on the
Earth's surface, only a fraction of this total brightness will be
experienced by an observer depending on various factors such as how
low the horizons are, which particular quadrant of the galaxy is on
view (e.g. the Cygnus region is much brighter than the Auriga region),
airglow and light extinctions due to the Earth's own atmosphere which
depends on the observer's elevation above mean sea level, etc...

AA Institute
http://uk.geocities.com/aa_spaceagent/

[Sourced from: Astronomy Physics forum, November 7, 2004 ]


Above from a myspace blog evidently run by an abdul ahad fan:

http://blog.myspace.com/index.cfm?fu...ogID=300459132

[John]

Dr. Doolittle wrote:

The same as it is during the day...

John.

[Beeblebrox]

"John" <me@zen.co.uk> wrote in message
news:46c5d3f1$0$11434$db0fefd9@news.zen.co.uk...

Works for me.....

Beeb