RE: OT: Luminosity question

From: Derek C Breit (breit_ideas@poyntsource.com)
Date: Thu Nov 28 2013 - 19:09:08 UTC

• Next message: Brad Young: "BY C (LEOs) 112813"

```See "Absolute Magnitude"..

Derek

-----Original Message-----
From: seesat-l-bounces+breit_ideas=poyntsource.com@satobs.org
[mailto:seesat-l-bounces+breit_ideas=poyntsource.com@satobs.org] On Behalf
Of Björn Gimle
Sent: Thursday, November 28, 2013 1:19 AM
To: Bob King
Cc: SeeSat
Subject: Re: OT: Luminosity question

I might find the answer from the W figure with some research.

But the original question is easy, since the Sun's magnitude is given as
-26.7 here, and diminishes by a factor of distance^2.

Using the Brigg (10-) logarithm for a distance of 10 AU (=1) where the Sun
is 100 times fainter = 5 magnitudes,
you have the formula -26.7 + 5 * log AU
With parsec you would subtract the constant 5*log(206265)

/Björn

2013/11/28 Bob King <nightsky55@gmail.com>

> Hi everyone,
> I realize this is off topic, but how do you calculate the sun's apparent
> magnitude at any given distance? Or, to turn the question around, how do
> you find out at what distance the sun appears a particular apparent
> magnitude?
> I know that if anyone could tell me, this would be the place to ask since
> satellite magnitudes depend on size, distance. If this helps, the sun's
> luminosity is 3.9x10 to the 26th W (watts).
> Thank you for your help.
>
> Best regards,
> Bob
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Björn Gimle, COSPAR 5919
59.2617 N, 18.6169 E, 51 m
Phone: +46 (0)8 571 43 312
Mobile: +46 (0) 704 385 486
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