Why is my reasoning to derive the bistatic radar equation wrong? The ideal effective antenna aperture is calculated from $A_{eff}=G\lambda^2/4\pi$, G being the antenna gain.
Why is it that in a bistatic radar setup with a transmitter and receiver fixed on the Earth and a moving aircraft, the bistatic radar equation only contains the transmitter wavelength, and nowhere the two Doppler-shifted ones?
If I wanted to determine the received power, I would consider the transmitter's wavelength $\lambda_0$ to calculate its $A_{eff, 0}$, the Doppler-shifted wavelength $\lambda_1$ on the reflecting aircraft as an intermediate receiver and transmitter antenna to calculate its $A_{eff, 1}$, and finally the Doppler-shifted wavelength $\lambda_2$ at the receiver antenna on the ground to calculate its $A_{eff,2}$.
What is wrong with this reasoning?
 A: Your reasoning is logically right even for monostatic radar when detecting a moving target. As for your question, you need to note that the radar equation is merely used to calculate the power received from the target to the receiver antenna, in order to calculate the maximum detection range of the radar given its minimum acceptable SNR: (The $A_{eff}$ formula is incorporated in this equation)
$$P_r = {{P_t G_t G_r \sigma \lambda^2}\over{{(4\pi)}^3 R_t^2R_r^2}}$$
you can see that power drops with $1/R^4$ and signals at the receiver are usually exceedingly attenuated. 
Now the doppler frequency of the reflected signal is:
$$f_D \approx \frac{2V}{c}f_0\rightarrow \lambda'=\frac{c}{f_0+f_D}\approx \lambda_0(1-\frac{2V}{c})$$
The maximum possible speed that you can think of when designing a radar is much lower than 20 Mach (around 7 km/s) even for ballistic missile detection. A speed of 15 km/s gives $\dfrac{2V}{c}=10^{-4}$
Thus, insterting this modified $\lambda$ in the radar equation doesn't affect anything at all and is always neglected compared to other losses and sources of noise in the SNR equation.
