Two stars—A and B, of absolute magnitudes three and eight, respectively—are obse
ID: 1551225 • Letter: T
Question
Two stars—A and B, of absolute magnitudes three and eight, respectively—are observed to have the same apparent magnitude. Which one is more distant, and how much farther away is it than the other? Hint: We know that a dierence in 5 absolute magnitudes amounts to a lumi-nosity ratio of 100. This means that star A is 100 times more luminous than B (as a smaller absolute magnitude means larger luminosity). Then, apply brightness = luminosity/(distance)^2 (inverse square law).
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Explanation / Answer
According to the given problem,
Star A is more distant.
The absolute magnitude of a star is what its apparent brightness would be at 10 parsecs.
The higher the apparent brightness the dimmer the star.
Therefore A would appear brighter than B if they were both at 10 parsecs.
However, they appear the same brightness therefore A must be further away.
How much farther away is it than the other?
A,B = Absolute magnitudes of star A & B respectively
a,b = Distances of stars A & B respectively
From the equation in the source, since the apparent magnitudes are the same:
A + 5 ( log a - 1 ) = B + 5 ( log b - 1 )
This can be rearranged to:
log a - log b = ( B - A ) / 5
Hence:
a / b = 10 ^ [ ( B - A ) / 5 ] = 10
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