View Full Version : How to use intensity Ratio Formula for Magnitudes?
2004-Jan-22, 01:18 AM
Well finally get into some adult college (Night School) for Astronomy class (about the Solar System) ><...yea the first day was cool (Watching TV), but the 2nd day... ;) ok never mind here's the question :
when Mars made its closet approach to Earth, it shone very brightly with a magnitude of -2.9. Use the intensity ratio formula for magnitudes to compare this brightness to the star Fomalhaut (aphla Piscis Austrini), which ranks among the brightest stars in the sky with a magnitude of +1.1 and which appears to the South of Mars inAugust of 2003. Which appeared brighter, Mars or FomalHaut? How Many times brighter?
>< big problem I haven't learn about using the intensity ratio formula before ><...
2004-Jan-22, 03:37 AM
Ok I have read about the formula from other website, and see if I do it right (Correct me if i am wrong><)
Mars's Magnitude is -2.9 and Fomalhaut is 2.0, therefore Famalhaut is the brighter star, but due to the distance from Earth to Mars and Mars to the Sun, so Mars is the brightest object...
Each Magnitude = 2.512 IR (Intensity Ratio)
Mars has -2.9 Magnitude, so its 1 magnitude, therefore its 2.512 IR
FomalHaut has 2 Magnitude, so its (2.512)^2 = 6.31 IR
So Mars its like 3.8 times brighter than Fomalhaut...
2004-Jan-22, 07:27 AM
Your mistake was that you should be using the log (base 10) of the intensity ratio. Magnitude and intensity have a relationship governed by the inverse square law you see. These a diagram on this page that's a nifty explanation of this law:
As for the calulation, it sounds as though you should be using the figure of +1.1 that was quoted to you for Fomalhaut.
Here it is:
(m1 - m2) = 2.5 x log(I2/I1)
(1.1 - -2.9) = 2.5 log(I2/I1)
4 = 2.5 log(I2/I1)
1.6 = log(I2/I1)
I2/I1 = 10^1.6
Mars is roughly 40 times brighter.
A more intuitive approach would be:
You know that for every magnitude point, a star is 2.5 times brighter. Since Mars' magnitude is 4 higher than Fomalhaut, the difference in brightness will just be:
this also equals 39.8
Hope this helps
2004-Jan-22, 06:48 PM
I don't get it @@, where did u get 12/11? Is this the brightest object?
I did a little research last night and I found this equation :
Ia/Ib = (2.512)^(Mb-Ma) and Ma-Mb = 2.512xlog(Ia/Ib)
Mb = 1.1
Ma = -2.9
Ia/Ib is the intensity Ratio
Ia/Ib = (2.512)^(1.1--2.9)
Ia/Ib = (2.512)^4
Ia/Ib = 39.81
So its about 40 times brighter
2004-Jan-23, 01:50 AM
Your a and b is my 1 and 2. As you said correctly, that fraction is the intensity ratio.
The equation you found is exactly the same as mine.
2004-Jan-23, 01:55 AM
:lol: thanks for helping me out
2004-Jan-23, 02:01 AM
I'd suggest studying some more algebra, and logarithmic/exponential functions.
2004-Jan-23, 02:10 AM
I will while attending to college this year ^^ (after June)
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