# Re: Satellite visual magnitude equations

From: Jonathan T Wojack (tlj18@juno.com)
Date: Thu Apr 26 2001 - 08:05:26 PDT

• Next message: Matson, Robert: "RE: Satellite visual magnitude equations"

```> Alas, these formulae have no practical application to
> the computation of satellite brightness.  "Illumination
> fraction" is a term that should never be applied to
> visual satellite observing -- it's irrelevant,
> unobservable, and not directly useful in predicting
> satellite visual magnitude.
>
> > P.S.:  How well do these formulas work for non-spherical objects?
>
> They don't even work for spherical ones -- that's the
> whole point.  Many people have been erroneously applying
> illumination fraction to satellite observing (and the
> prediction of satellite brightness), and it has no
> place there.

So ... these are only theoretical equations, and don't truly reflect
reality?

> Jonathan's other question had to do with how another
> list member computed a predicted magnitude of 6.5
> from a standard magnitude of 5.5 for an object at
> 1600-km range and ~90-degree phase.  This case is
> straightforward (due to the 90-degree phase) since
> this is the phase at which standard magnitude is
> defined.  So it's a simple case of adjusting the
> magnitude for the different range.
>
> Standard magnitude is defined for 1000 km range;
> brightness is inversely proportional to range
> squared.  So at 1600 km, the satellite is dimmer
> by a factor of (1600/1000)^2 or 2.56 (which is
> close to one visual magnitude).  The exact value
> would be given by:

So, just confirming - that is just the application of the inverse square
law?

> Mag = Std. mag + 2.5*LOG((1600/1000)^2)   -or-
> Mag = Std. mag + 5*LOG(1600/1000)
>     = 5.5 + 1.02 = 6.52

I'm sorry, but the above equations seem to be ignoring the phase angle.
It must be possible for an object to be at 1600km distance in more than
one spot.  Or perhaps the phase angle is irrelevant?

> A simplified general equation for object at 90-degree
> phase angle is:
>
> Mag = Std. mag - 15 + 5*LOG(Range)
>
> where Range is in km.  --Rob

I'm just trying to understand the math involved - where does the -15 come
from?

It's certainly the correct equation, though - they both provide the same
solution - to the 10th decimal place, no less.

Thanks a lot for the responce, Rob!

------------------------------
Jonathan T. Wojack                 tlj18@juno.com
39.706d N   75.683d W

4 hours behind UT (-4)

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