View Full Version : Luminosity and Apparent Magnitude formula for Sun

Launch window
2005-Apr-27, 10:03 AM
I almost recall how I worked this one out before, I think the magnitude of the Sun was about -26.6 and the Mag of our Sun when looking from Mars was about -26.1

But I can't find the right equation for this one again

Does anyone know how to work out the Magnitude of our Sun when viewing from a position in Alpha Centauri system or Barnard's star

Thanks for the help :-?

2005-Apr-27, 10:18 AM
I think it's something like m = -2.5 log I

Launch window
2005-Apr-27, 10:52 AM
Ok thanks Glom,
I just need to be pointed in the right direction

so that's the -2.5 log of 3.9 * 10 power 26

Sunlight takes over 8 mins to reach Earth so 1AU = 8.3 light mins
So for brightness at the Centauri system would that be

Lumonosity / 4 pie distance square
1 AU = 4.8 x 10-6 pc
So how do you submit distance into the equation from Alp Centauri, would you multiply Earth AUs
64000 times Earth's distance is a light yr
= 255000 AU to Alp Cent ( multiply by ) 150 M Kilometres
= 3.9 by 10 power of 16 distance

does this seem right ?

2005-Apr-27, 02:23 PM
Light diminishes with the square of the distance. Taking your figure of 255000 AU, Alpha Centauri receives 1/65025000000 = 1/(255000^2) as much light as earth. Magnitude is a log base 2.5118864315095801110850320677993 (approximately )system , so this is about 27 magnitudes less than earth. Therefore Sol shines at magnitude 1 from Alpha Centauri.

Launch window
2005-Apr-27, 02:39 PM
Ok, thanks for the help Glom and Jfribrg. I had a bit of trouble with this one
and might have estimated some numbers with a bit of error

I think the true answer is about +0.4 magnitude

So it seem you don't have to go to far away in light years before our entire Solar System becomes an insignificant speck,
I wonder how many light years away it takes for our Sun to become invisible to the human eye

2005-Apr-27, 03:10 PM
I wonder how many light years away it takes for our Sun to become invisible to the human eye

This depends on what magnitude you consider to be "invisible". Inverting the same calculations as I did above, the Sun reaches magnitude 7 at a distance of 3,981,071 AU's which is 63 light years. The sun is magnitude 6 at a distance of 2,511,886 AU's or 39.8 light years.

2005-Apr-27, 03:13 PM
The rule is that 5 magnitudes is equivalent to 100 times difference in brightness.