Doppler radar measurement by an aircraft

[LCSphysicist]

Homework Statement

All below

Relevant Equations

All below

u is the aircraft speed.
c is light speed
f is the initial frequency
λ is the initial wavelength
λ' is the apparent wavelength

λƒ = u +λ'ƒ
λƒ = u + (c/ƒ')*ƒ
c = u + (c/ƒ')*ƒ
u = c(1-(ƒ/ƒ'))

u = 1500m/s

The answer is half of it, where is my error?

 

[berkeman]

The answer is half of it, where is my error?

Remember that this is the reflected signal. So if it were measured at a stationary detector at the reflector, you would get one value. If it is measured at the aircraft which is moving toward the reflector, you get twice the value...

 

[LCSphysicist]

Remember that this is the reflected signal. So if it were measured at a stationary detector at the reflector, you would get one value. If it is measured at the aircraft which is moving toward the reflector, you get twice the value...

Sorry, I am not sure if i get... So my math equations is wrong? The speed i measured is what would be measured in the aircraft?

 

[berkeman]

So the Doppler shift equation you used is for the shift observed by a stationary observer and a moving aircraft. But what if the observer is moving with the aircraft too? That makes the closing speed with the reflected signal how much bigger?

 

[etotheipi]

The aircraft will see $f= \frac{c+v}{c-v}f_0 = (1 + \frac{2v}{c-v})f_0$, which means that the speed of the plane is, using $f = f_0 + \Delta f$,$$v = \frac{\frac{\Delta f}{f_0}c}{2 + \frac{\Delta f}{f_0}} \approx \frac{c\Delta f}{2f_0} \approx 749.999 \text{ms}^{-1}$$i.e. very close to 750$\text{ms}^{-1}$, given $\frac{\Delta f}{f_0}$ is small.