How do I determine the uncertainty value of the star's absolute magnitude?

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SUMMARY

The discussion focuses on calculating the uncertainty value of a star's absolute magnitude given its apparent magnitude of 13.73 and a distance modulus of 13.9967, resulting in an absolute magnitude of -0.26. The uncertainty in the distance, Δd, was initially assumed to be 1 parsec but was corrected to 500 parsecs, leading to a revised uncertainty in absolute magnitude of 0.21. The calculations utilized the formula Δ(log10(d)) ≈ 0.4343 Δd/d and the combined uncertainty formula ΔM = √(Δx² + Δy²). The final result confirmed that the uncertainty in absolute magnitude is indeed larger than the uncertainty in the apparent magnitude.

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Thomas Smith
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Homework Statement


I have a star that has an apparent magnitude of 13.73 with uncertainty of 0.03303

It's distance Modulus is 13.9967 so it's absolute magnitude is -0.26

The distance is 6300 parsecs

Homework Equations


[/B]
The uncertainty on log10(d) is given by
Δ(log10)≈0.4343 Δd/d

ΔQ) = √(Δx)^2 + (Δy)^2 errors of sums or differences

The Attempt at a Solution



My guess is working out the uncertainty of d first

0.4343 x Δ/6300

d= 0.4343/6300

d=0.00007

and then the uncertainty of the absolute magnitude given by:

∆M= 0.4343 × √(0.03303)^2+(0.00007)^2) =0.0143

I'm not quite sure if this is right.
 
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What is Δd? You seem to assume 1 pc, that is unrealistic in an astronomy context.
Thomas Smith said:
and then the uncertainty of the absolute magnitude given by:

∆M= 0.4343 × √(0.03303)^2+(0.00007)^2) =0.0143
Why did you multiply by 0.4343?
Thomas Smith said:
I'm not quite sure if this is right.
Your uncertainty on the derived absolute magnitude is smaller than the uncertainty on the measured apparent magnitude. Can that be right?
 
mfb said:
What is Δd? You seem to assume 1 pc, that is unrealistic in an astronomy context.
Why did you multiply by 0.4343?Your uncertainty on the derived absolute magnitude is smaller than the uncertainty on the measured apparent magnitude. Can that be right?
∆d is 500 parsec sorry
 
I've did my calculations again and got the absolute magnitude uncertainty of 0.21.

I did 0.4343 x 500/6300
∆log(d) = 0.0345
∆5log(d) = 0.0345 × 5 = 0.1725

∆M = sqaureroot of (0.033030)^2 + (0.1725)^2

∆M = 0.20553 rounded up to 0.21
 
That looks good.

This forum supports LaTeX, by the way: ##\Delta M = \sqrt{0.033030^2+0.1725^2} = 0.2055##

How I wrote it: ##[/color]\Delta M = \sqrt{0.033030^2+0.1725^2} = 0.2055##

Edit: Fixed link
 
Last edited:
mfb said:
This forum supports LaTeX,
Hi mfb:

The LaTeX link does not open. I get the message:
Server error
404 - File or directory not found.
The resource you are looking for might have been removed, had its name changed, or is temporarily unavailable.

I imagine you may want to correct this.

Regards,
Buzz
 
Buzz Bloom said:
Hi mfb:

The LaTeX link does not open. I get the message:
Server error
404 - File or directory not found.
The resource you are looking for might have been removed, had its name changed, or is temporarily unavailable.

I imagine you may want to correct this.

Regards,
Buzz
Missing initial h. Should read https://www.physicsforums.com/help/latexhelp/
 

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