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Meteor detection using Graves beacon


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Could someone check my maths?

Looking at the display the frequency went down by about 5 khz which is about 0.0035%

300000 x 0.0035% = 10.5 km per second. if the meteor was an early Geminid then the entry speed (from the net) is 35 km per second which means that it went from 35 km per second to 24.5 km per second in about a quarter of a second.

Kind of makes sense?

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It may not be as simple! See https://www.britastro.org/radio/projects/meteorproj.html, where there are a a couple of interesting papers by Morgan, especially "Detection of Meteors by Radar".

Note that he uses the frequency shift equation Δf/F = V/c, where F is the "transmitted" frequency. I'm not convinced that this is correct here. It is fine where the object is transmitting, but here the object is reflecting, and the reflected frequency will also be frequency shifted by virtue of its movement relative to the Graves transmitter.

Ian

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Wouldn't you need to know the meteor track relative to the transmitter and receiver to draw any conclusions about the speed of the meteor? 

The basic doppler shift equation is correct but this is just the component  of the relative velocity between a transmitter and receiver in the direction of  a line joining them (the radial velocity) so the observed shift depends on the relative positions of the transmitter and receiver and their direction of travel. 

Then for a scattered signal you  have to add two doppler shift components.  

Consider first riding on the meteor. You would observe a doppler shift dependent on the component of the velocity in the direction of the transmitter. (Note that unless the meteor happens to be travelling directly towards the transmitter, this will not be the same as the speed of the meteor.)

This doppler shifted signal is then effectively "retransmitted" as a scattered signal.  We receive this signal on the ground, doppler shifted again according to the component of the velocity of the meteor in our direction. The total shift will be the sum of these two shifts.  

For example, a radar gun measures a doppler shift equal to 2x your speed when you are far away compared to how far the gun is from the midddle of the road. The measured doppler shift then drops as you approach the gun and becomes zero as you pass it.  

 

Robin

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"For example, a radar gun measures a doppler shift equal to 2x your speed when you are far away compared to how far the gun is from the midddle of the road."

Exactly, Robin. That's how I see it. This doubling of frequency shift is certainly very apparent when detecting the ISS as it moves towards or away from both Graves and your receiver (i.e. when on the horizons).

Ian

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Indeed.  I saw the same effect when measuring the rotation of Saturn using reflected sunlight

http://www.threehillsobservatory.co.uk/astro/spectra_28.htm

(bottom of the page)

Looking at the fig 3.4 and the equation below it in the paper you mentioned and simplifying it so the meteor is moving horizontally at constant velocity and making the distance between the transmitter and receiver large compared with the meteor altitude, I would expect delta f to be +2V at the far left, falling to ~V when overhead of the receiver, 0 when at the mid point between transmitter and receiver, ~-V when overhead of the transmitter and -2V at the far right.

Robin

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