Satellite latency (fwd)

Wed Feb 27 14:29:07 UTC 2002

> >The radius of the earth is about 6400km.  Geostationary orbit is, as
> >you note, 36000km above the equator.  The path from the satellite to
> >the North Pole is the hypotenuse of a right triangle with legs of
> >6400km and (6400+36000)km.  That gives a distance from the North Pole
> >to the satellite of 43000km.  It's reasonable to conclude that the
> >distance from either New York or San Diego is less than that.
> 
> Is that 36000 km above the equator measures from the surface of the earth,

The surface of the Earth.  (Or, more precisely, the surface of the
Earth that is directly below the satellite.  It's further to other
parts of the surface of the earth.  So, geostationary orbit is 36000km
above the equator.  Or (36000+6400)km from the center of the earth.)

> Looking at a spam I just received from satcast.com, it looks like they are 
> considering the distance from the dish to the satellite to be roughly 44000 
> miles.  That's roughly  70000 km.  Assuming that they are stating that from 
> the dish to the bird, that would account for the larger RTT times that 
> people see.

I'm not sure why spam should be considered a reliable source of
information, but, anyway, there's nowhere on earth where the distance
from the dish to a geosynch satellite is 44000 miles.  I would guess
that they are referring to the round-trip distance -- 22K miles up,
22K miles down.  That's not surprising, since geostationary orbit is
22K miles above the equator.  (They are presumably ignoring the
slightly increased distance to the satellite from anywhere not directly
under the satellite.)

     -- Brett