Take a look at this image:
Airplane Doppler traces recorded at 55.275 MHz from a 353km distant TV transmitter
Those curved lines are the Doppler traces left by airplanes over the main carrier ( the horizontal line ). It's easy to see that all Doppler signatures are within +/- 60 Hz from the carrier, and the asymptotes are further away. Using the bistatic Doppler formula, we can see their asymptotes are at ±2V/λ, so we can deduce the speed of these airplanes is larger than about 162 m/s, or 583 km/h
Airplanes have another advantage: Sometimes you can know their positions. And because you know the transmitter position, everything is easier to study. See carefully this video:
Airplane Doppler traces recorded at 112.7 MHz from a 66km distant VOR transmitter (VTB)
You can see how the airplanes produces a trace whenever they approach the TX-RX line, and you can see how this Doppler trace cross the carrier when the airplane cross the TX-RX line. You can observe also how faster planes seem to produces traces with higher slope than slower ones.
For a target crossing the TX-RX line flying at some altitude, the zero Doppler point is almost over the TX-RX line. But targets can produce zero Doppler shifts without crossing it*: Pay attention to the airplane appearing at 1:30 from the top of the image, very near the RX point. Observe it produces zero Doppler shift without crossing the TX-RX line. In bistatic radars, the zero Doppler point occurs when target's velocity vector is tangential to an ellipsoid with focii at TX and RX places:
In this case, because the point of closest approach (PCA) respect to the receiver is so far away than the PCA from the transmitter, the Doppler signature is "distorted".
*Actually none of the targets (airplanes or meteors) cross the TX-RX line. In a real 3D world with an spherical planet, the TX-RX line goes underground and no airplane can cross it. The zero Doppler point still occurs whenever the airplane velocity vector is tangential to an ellipsoid with focii at TX and RX places: