Bjoern wrote:
> > If the satellite is spherical, the illuminated fraction of the total
> > visible disk is (1 + cos v)/2
>
Jonathan Wojack replied:
> Yes, I was close to deriving that formula...
> To solve for the angle (if you know the illumination fraction, and want
> to find the angle), using the formula:
> Theta = cos ^ -1 ((2 Frac) - 1)
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.
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:
Mag = Std. mag + 2.5*LOG((1600/1000)^2) -or-
Mag = Std. mag + 5*LOG(1600/1000)
= 5.5 + 1.02 = 6.52
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
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This archive was generated by hypermail 2b29 : Wed Apr 25 2001 - 14:48:14 PDT