What Causes the Discrepancy in Doppler Shift for Satellites in Orbit?

[pan Hole]

At page http://www.qsl.net/vk3jed/doppler.html" is discussed Doppler shift at various ham radio frequencies. I calculated for this Doppler shift relative velocity of their satellite as 6720 m/s. Their satellite in height of 800 km above ground and at circular orbit should have velocity of 7453 m/s (orbital speed). I thought that difference between these two values should be max 465 m/s (rotation of Earth at equator), but it is almost twice as much... Any ideas?

 

[log0]

The 6720m/s will be the maximum velocity of the satellite relative to the receiver on ground. You have to consider the angle between satellite and receiver direction.

For a satellite at horizon the relative velocity(neglecting Earth rotation) would be 7453*cos(pi/2-asin(6378/(6378+800)))=6622.

 

[berkeman]

The 6720m/s will be the maximum velocity of the satellite relative to the receiver on ground. You have to consider the angle between satellite and receiver direction.

For a satellite at horizon the relative velocity(neglecting Earth rotation) would be 7453*cos(pi/2-asin(6378/(6378+800)))=6622.

He said their answer is bigger, not smaller, so the effect you mention is probably not what is causing the difference.

 

[berkeman]

At page http://www.qsl.net/vk3jed/doppler.html" is discussed Doppler shift at various ham radio frequencies. I calculated for this Doppler shift relative velocity of their satellite as 6720 m/s. Their satellite in height of 800 km above ground and at circular orbit should have velocity of 7453 m/s (orbital speed). I thought that difference between these two values should be max 465 m/s (rotation of Earth at equator), but it is almost twice as much... Any ideas?

I get numbers close to yours. So if you add the two velocities, you get close to 7900m/s. Which band number are you using to say that they get twice the shift?

 

[pan Hole]

I get numbers close to yours. So if you add the two velocities, you get close to 7900m/s. Which band number are you using to say that they get twice the shift?

For any frequency in Table 1 I got same relative velocity -6718 m/s caluclated this way:

v = - \frac{\Delta f c}{f_0}

For orbital speed:

v = \sqrt{ \frac{G M}{ R + h }}

at altitude of 800 km i got 7466 m/s. Difference is 748 m/s. That means, I didn't get twice the shift, from shift which I suppose is correct I got twice the velocity.

 

[log0]

OK. Let's try it again. The speed the OP has calculated(6718 m/s) is the maximum relative speed between the satellite and the receiver, not the orbital speed of the satellite(7453 m/s).

The satellite is never moving directly towards the receiver on ground. There is a height difference of 800km between both. You have to consider receiver position and velocity relative to the orbit.

The example calculation I have offered has been for the most simple case I can think of. Having a fixed receiver placed in the orbit plane and the satellite passing the horizon. A satellite passing the zenith would have a relative speed of 0 m/s.

 

[pan Hole]

OK. Let's try it again. The speed the OP has calculated(6718 m/s) is the maximum relative speed between the satellite and the receiver, not the orbital speed of the satellite(7453 m/s).

The satellite is never moving directly towards the receiver on ground. There is a height difference of 800km between both. You have to consider receiver position and velocity relative to the orbit.

The example calculation I have offered has been for the most simple case I can think of. Having a fixed receiver placed in the orbit plane and the satellite passing the horizon. A satellite passing the zenith would have a relative speed of 0 m/s.

I am starting to understand the problem... Orbital velocity is calculated fot Earths centre and receiver is 6378 km from that centre, right? But I still don't see the way to calculation you posted...

 

[log0]

Here is a sketch. :)

 

[pan Hole]

Thank you very much, now I understand it. I plotted Doppler shifts calculated from your formula and I got exactly same graphs as showed on that webpage.