[a] Show that the phase shift of a radiation field over any path can be written:
where n(s) is the phase refractive index along the path.
Note then, that if the path changes continuously with time (due say to the source
of the radiation moving), there will be a doppler shift:
imposed on the signal which depends on the properties of the medium. Measurement
of this doppler shift could, in principle, be used to measure properties of the
medium.
SOLUTION:
Since the phase shift must be 
for every wavelength:
where 
is the vacuum wavelength.
Under the circumstances we discuss here:
The frequency shift will then be:
The first term in the integral gives the vacuum Doppler shift:
Where 
is the satellite radial velocity toward the observer.
The ionospheric plasma `doppler shift' is much smaller:
Since the maximum value of X is about 
.
(We can get at 
from 
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[b] An artificial Earth satellite is in circular orbit 1000km above the Earth's
surface. It transmits a 200MHz radio signal. At some instant, the line of sight
makes an angle of 
to the vertical for an observer on Earth who is in
the plane of the satellite orbit.
At such a high frequency we can show that the effects of the ionospheric plasma on the radio wave are very weak (i.e. the deviation of the refractive index from unity is very small). Consequently, the effect of the plasma on the phase of the signal can be computed using the simplest theoretical expression, neglecting the magnetic field:
Show that the doppler shift seen by this observer consists of:
SOLUTION:
Using a flat Earth approximation, if the zenith angle of the satellite is ,
the satellite radial velocity is 
where v is
the speed of the satellite in its circular path around the Earth. We can calculate v
using Newtonian gravity. If R is the radius of the Earth and h is the height of the
satellite above theEarth and M is the Earth mass:
Using h=1000km, 
, 
, 
:
The vacuum doppler shift would be:
Since the vacuum wavelength at 200MHz is 1.5m.
The ionospheric doppler shift:
A more precise estimate of 
:
From above the magnitude of the ionospheric phase shift is:
Putting 
(where 
is the zenith angle of the satellite):
Assuming the ionosphere is horizontally stratified so 
does not change
with time:
Using 
:
would be a typical daytime value for 
so:
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[c] Clearly the ionospheric doppler shift would be hard to measure in the presence
of the much larger vacuum doppler shift.
Show that the ionospheric doppler shift can be separated out if the satellite
simultaneously coherently emits a harmonic of the 200MHz.
Show that this could lead to a measurement of the integrated electron concentration
along the path from the satellite to the observer.
SOLUTION:
In general:
So for the fundamental frequency (200MHz in this case):
For the nth harmonic, 
and 
.
So to subtract out the vacuum doppler contributions and leave the ionospheric doppler shift we need to divide the nth harmonic by n and subtract (in a mixer circuit).
Note that the process of subtracting out the vacuum doppler shift does not completely
subtract out the ionospheric doppler shift because it has a different frequency dependence
from the vacuum effect.