p.s. I agree. Very convenient logarithmic unit.

]]>Very nice idea. And actually perfectly well-suited to Peter, if he shares the same view re: gongs as I do.

That is, the more times you’re awarded the Order of Magnitude, the _lower_ down the rankings you go; to be awarded it zero times would put you on top of the pile.

And perhaps being offered said gong but refusing it would put you into negative numbers, thus making you even more worthy ðŸ™‚

]]>In fact, it’s quite easy to see that the answer =~ 5.5 to 6 mag, without having to do the full calculation.

A combined apparent visual magnitude of 10.5 mag and a distance of 100 pc implies a combined absolute visual magnitude of 5.5 mag (either from the m – M = 5 log d – 5 formula directly, or from 100 pc being 10 times the distance of 10 pc in the definition of an absolute magnitude, so there is 10^2 = 100 times more flux at 10 pc than 100 pc, which from the definition of magnitudes means the absolute magnitude is 5 mag brighter than the apparent magnitude).

Then, if the combined absolute visual magnitude is 5.50 mag, and that of one component is a measly 9.0 mag, almost all the flux must come from the brighter component, which means the absolute visual magnitude of the brighter component is only a little bit fainter than 5.5 mag.

P.S. I actually think magnitudes are quite sensible, actually.

]]>Is it sufficient if the answer is of the correct order of magnitude?

Hey, I know Peter won’t accept honours, but for other astronomers, instead of the Order of the Garter or whatever, why not the Order of Magnitude?

]]>Actually what happened is that I remembered 10.54 for the apparent magnitude of the other star, not the answer to the actual question. Hence 10.2 being close, but not cigarworthy…

I should, of course, have checked..

]]>The answer of 5.544 was easy to arrive at, but Peter’s categorical “This is not the correct answer” to Ted’s first post and then “Close but no cigar” to Chris C.’s 10.2 gave me the nagging feeling that this was one of his archly fiendish trick questions.

But it wasn’t.

Perhaps there’s something to be learned here about the intellectual bluff and bluster of cosmological theoreticians, but as a lowly common-or-garden observer condemned to carry the heavy millstone of the magnitude system around my neck, I’m clearly incapable of working out what it might be.

ðŸ˜‰

]]>*“I relied on my memory, which one should never do at my age.”*

Right. Make sure you don’t forget that! ðŸ™‚

]]>Yes, 5.54 is the correct answer. Sorry if I said otherwise earlier. I relied on my memory, which one should never do at my age.

The point is that magnitudes of two stars can not be added but the fluxes can. The way I did was to work out the apparent magnitude of the M=9 star at 100 pc (14). Then it’s easy to work out the ratio in flux between the composite star (F) and it (f) using

14-10.5 = 2.5 log (F/f), which gives an answer 25.11 for the ratio. This yields a ratio of 24.11 for the ratio from the two separate stars, from which it can easily be deduced that the apparent magnitude of the brighter star would be 10.54, so its absolute magnitude is 5.54.

Note that this is only slightly larger than the combined system; the fainter star contributes very little flux.

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