23 November 2001Little Discoveries Using Stellar Magnitudes
by Art Winfree
Part 2Today is 23 November, a good time (if not cloudy in the afternoon) to notice the Moon lit up on its upper right, but a Sun that lies lower in the west. Use a taut string or a long straight pipe to extend that up-tilting direction and see where it goes. Next week I will finish the topic of Curved Lines in the Sky, at least as far as I could take it in the spirit of unsophisticated thinking about the obvious.While waiting on the Moon, last week we tried some estimations using the astronomer's magnitude scale, as presented in Alan MacRobert’s piece 9 November. We rationalized the apparent magnitude of Venus as remembered from late last summer and as expected again early this coming summer.
Jupiter, though farther away on its orbit of 5.2 AU, is additionally so much bigger that at its closest to Earth it subtends about 2/3 min of arc, twice the angle of Venus. And we almost always see its full sunny face because we are so much nearer the Sun that we look basically outward to see Jupiter no matter where it is. How brightly would we expect a mirror of that size to reflect sunlight? Well, if it were a mirror the view would be as though we see the Sun though a peep-hole, thus cutting its total intensity by factor ( 2/3 arc min / 1/2 60 arc min)2= 0.0005 , and further cutting it in proportion to total distance squared, thus by an additional factor of ( 1 AU/ (5.2 AU + 5.2-1.0 AU))2= 0.01 . In terms of powers of the unit of magnitude, 1001/5, these are diminutions by 8 magnitude units and by 5 more, thus from -27 to -14. This is like full moonlight, way brighter than Jupiter’s observed maximum of about -2.5 . Same result as we encountered at Venus.
Then, just as at Venus, suppose Jupiter's atmosphere scatters the incident sunlight. Jupiter's atmosphere is illumined by a Sun 5.22 = 27 = 3.6 magnitudes less bright than ours, and then its radiation is attenuated by the angle factor twice that of Venus, so 10-4 radians, so 108 fold, or 20 magnitude factors more. The Sun's -27 + 23.6 = -4.4. Subtract a magnitude for losing half the light to heating in orange-red clouds, and we end at -3.4, a bit brighter than Jupiter's maximum brightness when we are as close as we ever get. I don't know whether this reveals that the whole method is bonkers, and comes close only by coincidence, or maybe the clouds absorb or degrade more than a magnitude's worth of incident light. If you find out, please tell me.
Why is ballpark estimation (ideally followed by a more serious effort to quantify) indispensable for Discovery? So long as we don't bother we are implicitly supposing that magnitudes are more or less arbitrary facts that could turn out any way at all, and that their cataloging is boring exercise, sometimes called "butterfly collecting". If you observed Jupiter at magnitude -6 you might just record it and never guess what such an observation would be trying to tell you: that Jupiter has some flat mirrors of substantial collective area always pointed our way (each too small to see individually in a telescope, even as scattered points of light) or that it is emitting more visible light than it gets from the Sun, perhaps because in its core the stellar fires are flickering on.
Sometimes such crude estimations suffice to rule out an idea that might otherwise seem plausible. Sometimes they seem too crude to serve any useful purpose at all. When it comes to stars, this might seem inevitable, due to (one would presume) huge differences in nuclear mechanism, star size, and so on. But if we restrict attention to Sun-like G stars, recognized by their color, then the observed magnitude provides a first estimate of distance. Take Alpha Centauri for example. This summer while lecturing in Buenos Aires I got to see it for the first time. It seemed nearly Sun-color with magnitude near 0, which is 27 magnitudes = 10(27 x 2 /5) times dimmer than Sol looks at our distance of 8 light minutes. So it may be 10(27/5) times more distant. That many times 8 light minutes is 4 light years. Look it up, now: astronomers say Alpha Centauri is indeed a G-type star and they reckon its distance by geometric parallax measurement (almost an arc second from one end to the other of Earth's orbit: check this for yourself, using 16 light minutes) at 4 light years.
Another answerable question: How many stars are visible at each magnitude? We might think this cannot be answered without counting them. (And indeed, any answer we might come to cannot be checked without counting.) But part of the answer can be discovered by thinking. Start with Sun-like stars of comparable intrinsic Magnitude (remember the big M from Alan MacRobert’s contribution of 9 November: intrinsic visual Magnitude, not apparent, distance-affected magnitude.) Their apparent magnitudes differ according to distance. Those sqrt(1001/5) times further away are 1 magnitude dimmer and if such standard unit stars are indiscriminately scattered within 1000 light years or so around us (thus staying inside the galaxy), there are 1001/5 times as many of them, because the volume of a thin spherical shell grows as the square of its radius. So each magnitude class 2.5 times fainter contains 2.5 times more stars, and in general, looking at things P-fold fainter, I can expect to see P times as many. Strange: the total intensity received from any such class seems to be a fixed constant, just broken into smaller individual pieces! The same argument applies to each other kind of star. So if we switch from a 5 mm "telescope" aperture (the pupil of my naked eye) to 50 mm binoculars, thus with 10x the aperture and so 100x the photon catch, we can expect to see 100x as many stars of the faintest class visible, which will be those that were 100 times dimmer because mostly 10 times farther away.
This simple-minded inference does not seem to agree with counts available, e.g., I saw somewhere a count of 500 stars to magnitude 4, and saw a catalog recently boasting 2862 stars to magnitude 5.5 and another with 250,000 to magnitude 9, all advertised as essentially complete, Another with 2.5 million to magnitude 11.5 might under-sample the dimmest ones. How should you plot these data to see if we are getting about 2.5 times more per magnitude? The result will lead you to a small puzzle helpful in "Discovering" the full concept packaged in the foregoing words and examples: How is this 100-fold increase compatible with the expectation that, at 10 times the range, I am surveying a 1000 times the volume? It is cheating and no fun to click the link before working this through explicitly, on paper, to your own satisfaction. With that done, you can click then correct my inadequate explanation or else learn something unexpected.
Here is another: given that dark rocks like those found on the Moon reflect back only about 1/15 of the incident sunlight, how bright would you expect the night to be under a full moon, compared to daytime brightness under the Sun? Of course we already know, having seen plenty of such nights, and having Alan MacRobert’s estimate of -12.5. But can you also rationalize it in such terms as we used for Venus and Jupiter? This problem is a little simpler because the Moon is about the same distance as Earth from the Sun. See how close your considerations bring you to the observed full moon brightness.
There is a stunningly obvious observation to make here, which I never thought of before this week, and it will lead directly to a surprising Discovery (or at least conjecture) about the surface of the Moon in next week's column, timed to coincide with Full Moon.
As a last Discovery exercise, consider Iridium satellites. You have probably seen them flashing in the early evening; if not, type your exact latitude and longitude and time zone into http://www.heavens-above.com and get a prediction of the next time to watch for one near your house. How bright could you expect the flash to be? Here you need to know we are not dealing with mere scattering of sunlight, as off Moon dust or Venusian or Jovian cloud tops. The flash is actually a reflection off two 1x2 meter solar panels. It is not a perfect reflection, as the panels are not optically flat, but let's suppose they were so perfect, so when we look at the satellite we are basically looking at the Sun in a mirror. What is the brightest flash you could ever expect to see? Suppose the shiny parts of panels taken together are maybe as big as 25 square feet, and we see this 5-foot mirror from 500 miles slant distance, so it subtends 5/(500x5000) = 2 10-6 radians. This is not much compared to the Sun's 1/2 deg = 10-2 radian. The area of Sun we see reflected over the horizon is thus ( 1/2 104)2= 1/4 108 less, or 18.5 magnitudes less. So we might expect flashes as bright as magnitude -27+18.5 = -8.5. In fact, brightest reported have been about magnitude -8. (Problem: the panels are in fact way imperfect as mirrors, so the reflected light is dispersed over a larger area than a mirror reflection would illuminate: how come its maximum observed intensity is not less than this?)
How bright is that compared to moonlight? If you attempted the exercise at the start of this follow-up you may have come to the idea that the full Moon has integrated intensity near -12 magnitudes. But that is spread over its disk-like image in the sky. For purposes of naked-eye viewing this is a certain number of least-focusable areas, like pixels, on an imperfect retina. 20/20 vision can resolve points about a minute of arc apart, about 1/30 Moon diameter. So each of the 700 little disks of my pixilated image carries only its share of the illumination and accordingly looks about 7 magnitudes dimmer: about -5. This leads me to believe the very best Iridium flash, being from my point of view an unresolvable "point", may look lots brighter than the surface of the Moon.
And how long would you expect this display to last? Well, the sunbeam diverges at 1/2 degree or 10-2 radians from a mirror (we guessed) 500 miles distant, so it might be 5 miles wide where we see it (or more if the mirror is imperfect). The mirror is moving in low Earth orbit about 5 miles per second, so something like a second (or more if the beam is more diffuse and correspondingly dimmer because reflected off a very imperfect "mirror".)
Note also that were the mirror perfect it would be presenting a 5-mile wide image of the Sun (on the ground, at a tilt, elliptically elongated to more than 5 miles in one direction) in which you could walk around in the detail of a sunspot projected, essentially through the 5-foot pinhole of this astronomical pinhole camera, to an area as big as a football field. That could be a great visual astronomy project … but the mirror is not that good, and the image zooms by too fast.
How "long" in terms of degrees of arc across the sky? If we are looking at the Sun in a tiny remote mirror, we expect to see only part of the Sun, and to have seen it all when the mirror has moved 1/2 degree across the sky. Is that the same thing as moving 5 miles? Nope. Depends on direction of viewing the mirror and its direction of motion, right? So there is a wee Discovery exercise awaiting your attention here: how long would you expect the flash to last, and how long does it last?
Have a look at one and see what you notice. Videotape it if you can: it is plenty bright enough, and it happens so suddenly, and usually not where you were initially looking, that the whole experience can finish before you get fully oriented, so it is nice to have a quantitative record to review. If you are lucky some identifiable bright star might be in the field of view for a magnitude comparison.
So much for temporizing. Next week we return to the Mystery of the Curved Straight Lines. Write down your own inferences from your own observations before the 30th.
Copyright 2001 by Art T. Winfree. All rights reserved. Used by permission.