- #1

Thomas Smith

- 12

- 0

## 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

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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.