# How to calculate absolute magnitude of a stellar body?

1. May 23, 2014

### AotrsCommander

I'm trying to find the appropriate formula (and abstractions to plug into it as necessary) to work out the absolute magnitude (from which I can dervive an apparent magnitude) of a solar body.

I've found a lot of formula which relative absolute and apparent magnitude, but trying to find how to calculate the former is proving really hard - especially since planetary bodies use a slightly different system.

Wiki - ironically, considering I google searched for ages - has been most helpful so far (which is not very), especially with the moon example.

With just one problem - the Hmoon value. From what I'm understanding, that's the moon's absolute magnitude (so that formula is for the apparent magnitude). But it doesn't appear to say anyway where that's dervived from. It says something about subtracting 31.57 from something to get the planetary as opposed to stellar magnitude... Trouble is, the wiki page on the moon doesn't give you that information, so I can't back-work the numbers to get where Hmoon is coming from.

What I am looking for is a first order approximation of absolute magnitude for a stellar body, given I know the luminosity of the star, the size, distance (or the stellar flux, since I've already got that calculated) and the body's albedo. (I am assuming maximum, so the phase angle wouldbe 0º if I understand right.)

Can someone point me in the right direction?

2. May 23, 2014

### Bandersnatch

Hey, AC.
That's for your "aesthetic moon", isn't it?

How about you drop the absolute magnitude concept altogether, and focus on luminosity and apparent magnitude?

Luminosity follows the inverse square law. What you need to do, is calculate the total flux per square metre at the distance of the planet's orbit(i.e., where the moon is as well):
$F_0=\frac{L_0}{4∏R_0^2}$
then multiply that by the cross-sectional area of the satellite and its albedo to get the total wattage reflected by it:
$L_1=F_0*∏r^2*α$
(r - moon radius; α - albedo)

then treat the satellite as the source of light with the just-calculated luminosity, and use inverse square law again to find the flux incident on the planet from the moon:
$F_1=2\frac{L_1}{4∏R_1^2}$
(R1 - moon-planet distance; the factor of 2 comes from the fact that a moon reflects light in only half of the directions a star shines, so the total wattage is that much less spread out)

Then use $log_{2.512}\frac{F_0}{F}=m-m_0$ with F0 and m0 being either our Sun's, or the star's reference luminosity and magnitude, you can find the apparent magnitude of the moon.

Plug in values for our Sun, Earth and Moon to see how it works. The results are not perfect but close enough.

(All these assume phase angle 0°, i.e., full moon)

3. May 23, 2014

### AotrsCommander

Yeah, it is! (I figured this was a more straight-foward physics question, hence posting it here.) (Still slowly working on the rest of it...!)

In that last equation, should it be F0/F1 or L0/F1? You said luminosity, but used the F1 value? I'm having trouble getting the right numbers out...

Plugging in the numbers for Earth and the moon:

$F_0 =\frac{L_0}{4∏149597870.7^2}$(km => m, so /1000000) $= 1367.6 W/m²$

where L0 = 1 * 3.846E+26W, R0 = 149597870.7 km

$L_1 = 1367.6 * ∏ * 1737.1^2 * 0.136 = 2199.1$

where a= 0.136, r = 1737.1

$F_1 = 2*\frac{2199.1}{4*∏*384399^2}$(km => m, so /1000000) $= 1.46E+12$

where R1 = 384399km

Which gives you either

$m = m_0 + log_2.512 \frac{F_0}{F_1}$

where m0 = 4.83, F0 = 1367.6 W/m², F1 = 1.46E+12 W

or

$m = m_0 + log_2.512 \frac{L_0}{F_1}$

where L0 = 1 * 3.846E+26W

depending which is right! But neither of which are coming out close to -12.74 (using the bottom end one, it's coming out as about 30). I must be going wrong somewhere - probably on that last equation or the F1, or fracking my units up somewhere.

4. May 23, 2014

### Bandersnatch

It should be F0. My mistake.

First of all, units! Remember to use metres everywhere. r=1737100 metres.
Additionally, I've no idea how you get that number out of it. It comes out to about 1.76*10^15 (10^9 if you incorrectly use kilometres).

Everthing afterwards is screwed up by the above. Although I must say that there's definitelly something iffy with your calculations in general. I get a completely different result for F1 even when using the numbers you plugged in.

5. May 23, 2014

### AotrsCommander

Aha. As my chemistry teacher always used to say, first write the equation that gives you the answer.

Aside from the units mistake, the 2199 figure was the L1 for the wrong moon, and for a kick off and then there was an extraneus 2* in there, which I think had crept in from the F1 equation somehow (perils of the spreadsheet, and why a calculator is not a bad idea as a back-up...) That (and making sure I converted everything to metres...!) gave the answer as 1.76E+15 for L1.

So then:

$F_1 = 2 * \frac{1.76E+15}{4*∏*384399000^2} = \frac{1.76E+15}{1.8568E+18} = 0.001899$

So that gives

$m_0 + log_2.512\frac{1367.6}{0.001899} = 4.83 + 14.61 = 9.812$

Which is closer, but is positive.

That said, -26.74 + 14.61 is about -12, which would appear to be correct; so m0 must be the apparent magnitude, yes? (On the basis that it appears to be right - as with all else, try both and the one that's right probably is, if everything else is correct!)

[STRIKE]Oddly enough, if all the above is right, plugging the numbers in for said "aethetic moon" comes out at ≈7 apparent magnitude, despite being the same angular diameter, a third the distance from the planet and getting nearly twice the stellar flux with an albedo of 0.8. That doesn't sound right...[/STRIKE]

And if frackwit over here actually remembers to put in the apparent magnitude of the sun, it comes out to -13, which actually sounds like it's right, as I'd have expected it to be brighter in those conditions. (And that was sort of the idea.)

6. May 23, 2014

### Bandersnatch

Yes, m0 is the apparent magnitude. Should've made it clearer.

7. May 23, 2014

### AotrsCommander

Righto! Like everything else, once you know what the answer should be, it's easy to work out where you went wrong.

Much appreciated - I'm actually getting there slowly. I might just have chance to tackle the next bit before I go away on holiday for the week (where hopefully I can actually start thinking about the planet ecology...!)

If nothing else, all this is a grand learning exercise!