Bram Dorreman wrote:
> > Who can tell me the formula to calculate the stellar
> > magnitude from a self defined absolute magnitude (phase 90°
> > and distance 1000 km), range and sun-target-observer angle?
> >
> > Another question: what might be a reasonable value for
> > Rosetta's standard magnitude?
Ted Molczan replied:
> I have derived the following formula, for use with data provided by the
> ephemeris:
>
> mag = 32.17 + 5 * log10("delta") + 0.01 * ("S-T-O" - 90), +/- 2
>
> where: "delta" = Target apparent range relative to observer. Units: AU
>
> "S-T-O" = Sun-Target-Observer angle; target's apparent PHASE ANGLE
as
> seen at observer's location at print time. Units: DEGREES
>
> The above formula is based on my estimated standard magnitude of 6.3 +/- 2
mag
> (1000 km, phase angle 90 deg), based upon the ESA's description: "a large
> aluminium box with dimensions 2.8 x 2.1 x 2.0 metres"
Thanks Ted and other repliers too.
Before I received these replies I recalculated my topocentric predictions so
that distance unit is km. What would be the formula when "delta" unit is km?
Meanwhile I can start with the original output in which "delta" is in AU.
ESA's statement that Rosetta would be mag +8 or +9 at 10000 km range makes
it mag +3 or +4 at 1000 km (100 times as bright -> mag 5 difference). I now
know what standard magnitude range I can apply in my further calculations.
It may all be in vain because we now have a real winterperiode with clouded
skies, snow and temperatures below zero (Celsius) during the whole days.
This will cause we cannot "hunt happily".
Best regards..Bram Dorreman
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