View Full Version : (Apparent magnitude) What have I done wrong?

nubeoort

2014-Nov-03, 11:05 PM

Hello, I was using these equations

20087

to calculate the apparent magnitude of the Solar System planets.

But, using the second one, the results are not very accurate.

According to Stellarium, the phase of Mars today is 0.91, and is at a distance of 1.7 AU, (0.7 from Earth). According to NASA, Mars' albedo is 0.17, and the radius is 0.533 * Earth.

So, I do the maths. The result of the second equation is 2.512 log (32.22) - 5.159, that equals to -1.37. But the apparent magnitude of Mars today is +1.1. I've tried to do it for another dates, and the error has gone from 1.3 to 2.6 magnitudes.

What have I done wrong?

I had the same problem in Venus: assuming r = 0.95 earth, distance from Sun = 0.72 and distance from Earth = 0.28, albedo = 0.65 and phase = 1, the result I obtained was a -8.1 magnitude, when the max. magnitude of Venus is -4.9 at best. I know something is wrong, but don't know what.

Can you help me, please? Thank you very much! :)

ngc3314

2014-Nov-04, 12:25 AM

Not a big problem, but in the magnitude definitions the multiplier is 2.5 (exactly) rather than 2.512 (which is 10^0.4, the intensity ratio corresponding to a difference of 1.0 magnitude).

nubeoort

2014-Nov-04, 08:56 AM

If I try with 2.5, then the difference is bigger: -1,4. So far from the real 1.1. But I think all the data is well put. And the equation is supposed to work.

thank you, ngc3314.

Hornblower

2014-Nov-04, 01:47 PM

I cannot help you because I cannot tell what is what in those equations. Please tell me what each letter variable stands for. Only then can I plug in some numbers for simple cases like Venus at superior conjunction or Mars at opposition, as test cases involving a full phase of the planet in question.

nubeoort

2014-Nov-04, 01:57 PM

In the second equation, from left to right m(p) is the apparent magnitude of the planet, d(p) is the distance between the planet and its star, d(o) is the distance between the Earth and the planet. At the lower part, P is the phase of the planet (from 0 to 1), A is the geometric albedo of the planet (from 0 to 1 again), L(s) is the luminosity of the star in solar units, and r(p) is the radius of the planet in Earth radii units.

I found the equation (and it's explained too there) in this webpage: http://phl.upr.edu/library/notes/theapparentbrightnessandsizeofexoplanetsandtheirst ars.

By the way, all the distances are given in AU.

Hornblower

2014-Nov-04, 05:13 PM

I agree with ngc3314 that the multiplier that gives magnitude from the logarithm of the luminosity should be 2.5 exactly, not 2.512, though the difference is really slight considering the uncertainties of some of our parameters here. The magnitude difference between two objects that differ by a factor of 100 is by definition exactly 5. The log of 100 is 2, thus the factor 2.5 exactly. The number 2.512 is the 5th root of 100 expressed to three decimal places. Its log is 0.4 for a magnitude difference of 1.

According to Stellarium, the phase of Mars today is 0.91, and is at a distance of 1.7 AU, (0.7 from Earth). According to NASA, Mars' albedo is 0.17, and the radius is 0.533 * Earth.

So, I do the maths. The result of the second equation is 2.512 log (32.22) - 5.159, that equals to -1.37. But the apparent magnitude of Mars today is +1.1. I've tried to do it for another dates, and the error has gone from 1.3 to 2.6 magnitudes.

What have I done wrong?

I had the same problem in Venus: assuming r = 0.95 earth, distance from Sun = 0.72 and distance from Earth = 0.28, albedo = 0.65 and phase = 1, the result I obtained was a -8.1 magnitude, when the max. magnitude of Venus is -4.9 at best. I know something is wrong, but don't know what.

Can you help me, please? Thank you very much! It appears that you have misinterpreted the given distance figures and inferred that the planets are much closer to Earth than they really are. On November 1, according to Sky and Telescope, Mars was about 1.687 AU from Earth and receding, which would make about 1.7 yesterday. I don't know why you thought it was 0.7. Venus is at superior conjunction, placing it about 1.7 AU away, not 0.28 as you somehow inferred. That shorter distance is about what we would get if Venus is at opposition as seen from Mercury at the latter's aphelion, and your magnitude is in pretty good agreement with what I calculated while daydreaming as an impressionable 12-year-old many decades ago. Using the correct numbers for distance in the given equation, I got about -4, in good agreement with published values.

The given equation is not even close for small phases. It would have the Moon fading by only a factor of 2 at the half phase, but it actually fades by a factor of approximately 10, because of the grazing illumination near the terminator and the scattering characteristics of the rough surface. The folks at PHL should hang their heads in shame.

nubeoort

2014-Nov-04, 06:06 PM

Yes, true, it works very well, but has a little error. I mixed some distance values, so the results were all wrong.

With a d(p) of 1.7129 AU and a d(s) of 0.72391 AU, albedo of 0.65, L of 1, Phase of 0.999 (almost 1) and earth radii = 0.949, the final result is a -4.1 magnitude. For JPL, today's magnitude for Venus is -3.9X, so the difference between -4.1 and -3.9x is not so big.

For Mars, taking a d(p) of 1.70123 AU, a d(s) of 1.39170 AU, albedo of 0.17, L of 1, Phase of 0.906 and earth radii = 0.532, the final result is a 0.1-0.2 magnitude. But JPL gives a magnitude of 0.9. Here the gap is much bigger. And I don't know why the difference with Venus (the distances are not so different) is only 0.2 magnitudes and for Mars is from 0.7 to 0.8. Anyway, for doing a quick calculation is not that bad.

Hornblower

2014-Nov-04, 06:13 PM

Yes, true, it works very well, but has a little error. I mixed some distance values, so the results were all wrong.

With a d(p) of 1.7129 AU and a d(s) of 0.72391 AU, albedo of 0.65, L of 1, Phase of 0.999 (almost 1) and earth radii = 0.949, the final result is a -4.1 magnitude. For JPL, today's magnitude for Venus is -3.9X, so the difference between -4.1 and -3.9x is not so big.

For Mars, taking a d(p) of 1.70123 AU, a d(s) of 1.39170 AU, albedo of 0.17, L of 1, Phase of 0.906 and earth radii = 0.532, the final result is a 0.1-0.2 magnitude. But JPL gives a magnitude of 0.9. Here the gap is much bigger. And I don't know why the difference with Venus (the distances are not so different) is only 0.2 magnitudes and for Mars is from 0.7 to 0.8. Anyway, for doing a quick calculation is not that bad.

I am as sure as I can be is that the fault is with their simplistic assumption about the illumination at phases other than full. Even at 0.9 the Moon fades by a larger amount than that and my educated guess is that Mars does likewise. I will need to search for a graph I have seen for the Moon to check on this.

nubeoort

2014-Nov-04, 06:42 PM

And also, doing the calculation of Uranus for today with r = 3.981, albedo = 0.51, d(p) = 19.13032 AU, d(s) = 20.01125 AU, L = 1 and phase = 1, the result is -aprox.- +4.8. Today's magnitude for Uranus is +5.7. We have again that 0.9 magnitude error, but now with a phase of 1. This makes me lose sleep.

Hornblower

2014-Nov-05, 03:03 PM

And also, doing the calculation of Uranus for today with r = 3.981, albedo = 0.51, d(p) = 19.13032 AU, d(s) = 20.01125 AU, L = 1 and phase = 1, the result is -aprox.- +4.8. Today's magnitude for Uranus is +5.7. We have again that 0.9 magnitude error, but now with a phase of 1. This makes me lose sleep.

Plugging in your numbers for distance and albedo, I got about 5.5, not a major error. You must have made a mistake in your number crunching. Again, I find decent agreement at full phase, but all bets are off at small phases. I stand by my opinion that using a constant number for a given planet's albedo and merely factoring in the percentage of the apparent disk that is illuminated is not going to be close to correct, and that the discrepancy will vary with the type of surface. See the following Wiki page:

http://en.wikipedia.org/wiki/Phase_curve_(astronomy)

Once again, Wiki is not the last word, but in my experience they are very good in routine science topics like this one.

Scroll down to the various planets and click on the graphs to enlarge them, and click on the Back icon to return to the main page. Note the difference between the cloud tops of Venus and the rocky surfaces of the Moon and Mercury. Also bear in mind that "phase", as used here, is the angle through which the lighted hemisphere is rotated from a reference point, not the percentage of illumination of the apparent disk. Those numbers are related by a trig function and can be calibrated to match at the half phase but not elsewhere. The PHL team did not make it clear which they were using in their equation, but the error at the half phase makes it clear that they gave a seriously flawed presentation for anything but the full phase.

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