# RE: Standard magnitude

From: Matson, Robert (ROBERT.D.MATSON@saic.com)
Date: Thu Dec 21 2000 - 11:24:21 PST

• Next message: Tristan Cools: "ISS"

```Jonathan wrote:

> According to the formula
> A - 15.75 + 2.5 x log (BxB/C),

> where A = the Standard magnitude,
> where B = the range in kilometers, and
> where C = the satellite illumination,

> If I input 5.94 for A, 1000km for B, and 75% for C, I get +5.50.

> Conclusions?

The equation is wrong, or at best incomplete.  "75% illumination"
is the problem.  What does that mean?  There is no easy way for
a satellite observer to know when an idealized satellite has
75% the brightness of "full phase".  Most people seem to think
that if you have a spherical satellite and you illuminate it
from the side (so that the satellite appears like a half-circle),
that it is exactly 1/2 has bright as if you illuminate it from
the front (full circle).  Seems logical, but it's wrong.  Side
illumination (90 degree phase) leads to only 31.83% the brightness
of front (0 degree phase) illumination.

This is why I want to know how standard magnitudes are computed.
By defining the standard magnitude to correspond to 90 degree
phase, I'm not at all optimistic that the correct phase formula
was used to derive it.  --Rob

-----------------------------------------------------------------
Unsubscribe from SeeSat-L by sending a message with 'unsubscribe'
in the SUBJECT to SeeSat-L-request@lists.satellite.eu.org
http://www2.satellite.eu.org/seesat/seesatindex.html
```

This archive was generated by hypermail 2b29 : Thu Dec 21 2000 - 11:28:31 PST