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Suppose two stars have the same apparent brightness, but one star is 8 times further away than the other. What is the ratio of their luminosities? Which one is more luminous, the closer or the further star?


marked as duplicate by Rob Jeffries, Reinstate Monica, peterh says reinstate Monica, Jan Doggen, Mike G Feb 2 at 23:52

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    $\begingroup$ Why don't you do your own homework? The answer is readily resolved using the ideas presented to you in the previous homework question you asked. $\endgroup$ – Rob Jeffries Feb 2 at 7:47

Welcome to Astronomy Stack Exchange!
Let the star which is farther away be A, and the fainter one be B.
Since this perceived brightness is something called flux density (sometimes referred to as just flux), which the following formula holds its own for: $F = \frac{L}{4\pi r^2}$ ($L$ is the luminosity, and $r$ is the distance) it would mean the luminosity ratio can be calculated in the following manner:
$F_A = F_B$
$\frac{L_A}{4\pi r_A^2} = \frac{L_B}{4\pi r_B^2}$
Substituting for $r_A = 8r_B$:
$\frac{L_A}{64r_B^2} = \frac{L_B}{r_B^2}$
$L_A = 64L_B$
The farther star is much more luminous than the closer one.
P.S. This is only if there is no extinction, if it is present, I cannot make an estimate without more details.


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