Absolute Magnitude of Lightbulb


by tomeatworld
Tags: absolute, lightbulb, magnitude
tomeatworld
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#1
Jun12-10, 08:13 AM
P: 51
1. The problem statement, all variables and given/known data
By comparing with the luminosity of the Sun, calculate the absolute magnitude of a 60W lightbulb. [THe absolute magnitude of the Sun is +4.75; you may ignore colour differences]

2. Relevant equations
m-M=5log(d/10pc)
[tex] f = \frac{L}{4\pi d^{2}}[/tex]
m1 - m2 = -2.5long(f1 / f2)

3. The attempt at a solution
I'm not really sure where to start. I know the luminsoty of the Sun to be 3.8x1026W but I'm not sure how it helps.
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George Jones
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#2
Jun13-10, 05:02 AM
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If d = 10 pc for both objects, what does the second equation become?
tomeatworld
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#3
Jun13-10, 09:07 AM
P: 51
I'm not sure what you mean? Surely, it just has 10pc put into it.

George Jones
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Jun13-10, 09:29 AM
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Absolute Magnitude of Lightbulb


What is the definition of absolute magnitude? What are the quantities on the right side of the second equation when d = 10 pc?
George Jones
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#5
Jun13-10, 11:15 AM
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Quote Quote by George Jones View Post
What is the definition of absolute magnitude? What are the quantities on the right side of the second equation when d = 10 pc?
Oops, I meant "on the left side of the second equation'".
astrozilla
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#6
Jun14-10, 01:01 PM
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First, The Greek Astronomer Hiparchos divided the stars we see on the sky in 6 magnitudes.The brightest were of first magnitude ,the next fainter stars were second and the stars just visible with naked eye were 6th magnitude.
British astronomer Pogson(1854) ,taking into account the law of Weber and Fechner ,according to which the intesnity of feelings is proportional to the logarithm of the stimulus
stated the relationship: lm/ln=c(exp)(n-m) (1)
Where lm and ln are the brightnesses and m ,n are the corresponding magnitudes.Here the subjective feeling is the apparent magnitude of the star and the stimulus is the brightness.
Since the first magnitude star is 100 times brighter than a 6th magnitude star from (1) we have:

l1/l6=100=c (exp)(6-1) therefore c=2.512 ,that is the ratio of two successive magnitudes is 2.512 ,
Setting the value of c into the pogson equation we get : lm/ln=2.512(exp)(n-m) ,from which if you log both sides you get the third relevant equation ,you show us.

Now: According to the principle of physics that brightnesses are inversly proportional to the squares of the distances we have:

l/L=(10pc)^2/r^2 , where L is the abolute brightness at distance of 10 parcecs and l is the brighness at distnce r

NOW IT MUST BE CLEAR THAT f1=lm and f2=ln , i.e BRIGHTNESS (l) =FLUX (f)

USING THE THIRD RELEVANT EQUATION we have:

M-m= 2.5log(l/L)=2.5log(100/r^2) ,{warning :NOTE THE - SIGHN AND THE ORDER OF FLUXES !}

Solving the last equation whith respect to M( abolute magnitude)
we get M=m+ 5-5 logr

Which is absolutely equivalent to your first relevant equation.

now use LaTeX Code: f = \\frac{L}{4\\pi d^{2}} you want to find the absolute brightness/flux (f) therefore you put in this equation d=100 (YOU DONT NEED TO CONVERT PARSECS INTO METERS BECAUSE YOU WILL DIVIDE THE FLUXES) and put L= 3(10)^23

This is the ABSOLUTE luminosity of the sun.

Divide the ABSOLUTE luminosity of the sun with the ABSOLUTE luminosity of the bulb.

THE KEY IS THAT THE ABSOLUTE LUMINOSITY OF AN OBJECT IS GIVEN WHEN YOU USE THE FORMULA LaTeX Code: f = \\frac{L}{4\\pi d^{2}} AND YOU PUT d=10 pc.

Finally when you have the ratio of the fluxes put it in your third relevant equation and put m1=Msun=4.75 ,
The only uknown is the absolute magnitude of the bulb Mbulb.
Now its easy to do the calculations to find the answer.

hehe


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