An Astronomical Teaser

For those of you who feel up to a little brain-teaser after last night’s revels, try this little problem which involves the use of everybody’s favourite type of astronomical measurement, the magnitude system. Answers through the comment box please!

A binary star at a distance of 100 pc has such a small separation between its component stars that it is unresolved by a telescope. If the apparent visual magnitude of the combined image of the system is 10.5, and one star is known to have an absolute visual magnitude of 9.0, what is the absolute visual magnitude of the other star?

18 Responses to “An Astronomical Teaser”

  1. I’m getting 5.54, but as I’m about as fond of magnitudes as you are of sigmas, I could easily have made a mistake.

    • telescoper Says:

      This is not the correct answer.

      I sometimes worry that we teach our students about magnitudes for the same reason that people who were abused as children go on to become abusers themselves..

  2. 10.2
    Chris C

  3. Ignore that!. Not only a sign error in the definition of magnitudes, but a failure to notice “absolute”
    Chris |C

  4. I did a study contract once on a space radar where the principal requirement of the Customer was “There shall be no dB in the final Report”. I found dB$ to be an interesting concept in the presentation… but now my brain hurts. “A bit more than 5.5” is good enough for me…
    Chris C

  5. Michael Hubbard Says:

    12.5 ?

  6. Here’s my reasoning. The combined system has an absolute magnitude of 5.5 [ M = m – 5 log(d/10 pc) ]. Luminosity is proportional to 10^(-M/2.5), so we want to solve

    10^(-5.5/2.5) = 10^-(9/2.5)+10^(-x/2.5)

    giving x = 5.544.

    • Bryn Jones Says:

      Yes, that’s right, an absolute visual magnitude of 5.544 mag. Don’t listen to the others.

    • telescoper Says:

      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.

  7. Mark McCaughrean Says:

    Hmmm. Interesting.

    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.


    • telescoper Says:

      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..

    • Bryn Jones Says:

      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.

  8. I almost quit astronomy yesterday in a furious rage because I could not get any other answer except the supposedly wrong 5.54. My neighbours probably wondered why someone was shouting “magnitudes are bullshit!!” every now and then. I’m happy that it was actually right.

    • Mark McCaughrean Says:

      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 đŸ™‚

  9. […] test your understanding you could try these little problems. To warm up you might look at I posted the first of them a while ago. Anyway, here we […]

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