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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