# Re: A doppler 2.4GHz question

• Subject: [amsat-bb] Re: A doppler 2.4GHz question
• From: James R Miller <g3ruh@xxxxxxxxxxxxxxxxxxxx>
• Date: Wed, 22 Dec 2004 00:41:02 +0000

```To Graham, G3VZV

--------------------------------------------------
Assuming the satellite is in a circular orbit, and passes overhead then, with
a little bit of (easy) manipulation of Pythagorus' theorem, you find:

RangeRate_Max       = Re W           km/s    at the horizon      (1)

Re Rs
RateOfRangeRate_Max = ----- W^2      km/s^2  at overhead         (2)
Rs-Re

where:
Re  is Earth radius 6378.140        km
Rs  is satellite orbital radius     km
W   is satellite mean motion        rad/s

And               Re^2
W^2 =  g  ------   where g = 9.81 m/s^2                   (3)
Rs^3

Plugging in some numbers, taking care with units:

Norad 2-line elements for the ISS give mean motion 15.716 rev/day which
equates to W = 1.143E-3 rad/s.  Using (3) this leads to a value for
Rs = 6735 km. Substituting into (1) and into (2) we find:

RangeRate_Max       = 7.290 km/s   at the horizon

RateOfRangeRate_Max = 157   m/s/s  at overhead

To convert these to frequency shift, divide by the wavelength, which at
S-band is 0.125 m.  Thus:

RangeRate_Max       = +/-58.3 kHz     at the horizons

RateOfRangeRate_Max = -1257 Hz/s      at overhead

The minus sign is because doppler shift is -ve for a +ve velocity.

[The above also assumes that the Earth is non-rotating].

73 de James G3RUH

--
==========================================================================
James R Miller      WWW/PGP:     http://www.jrmiller.demon.co.uk/
Cambridge, England   Stardate:        2004 Dec 21 [Tue] 1732 utc
==========================================================================
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