Doppler effect equation derivation I am after to model the Doppler shift of an LEO satellite. At [1] (pp 183) I have found the following formula:
$$\frac{f_r}{f_s}=\frac{1-\frac{u}{c}\cos\theta}{\sqrt{1-\frac{u^2}{c^2}}}$$
$f_s$ stable frequency transmitted from the satellite
$f_r$ received (Doppler shifted) frequency at the observation point.
$u$ satellite velocity 
$c$ velocity of light
$\theta$ angle between the velocity vector of the satellite and the observer-satellite line of sight 
After some further research, I see that falls in one of the theories of estimating the Doppler effect.
Can I please ask how to derive the nominator of the above formula?
I have found several references but am not yet convinced.
[1]Seeber, G. (2003). Satellite Geodesy.New York, Walter de Gruyter
 A: In the following replace $\:^{\prime}$$^{\prime}$Star$\:^{\prime}$$^{\prime}$ by $\:^{\prime}$$^{\prime}$satellite$\:^{\prime}$$^{\prime}$. 
Solution 1

Suppose two light pulses $p'_1$ and $p'_2$ are emitted successively from the Star towards the Earth at time moments $t'_1$ and $t'_2$, apart by an infinitesimal time interval $\mathrm{d}t'=t'_2\!-\!t'_1$. Time $t'$ is the time in the rest frame $\mathrm{S'}$ of the Star. 
These two events happen in the rest frame $\mathrm{S}$ of the Earth at  time moments $t_1$ and $t_2$, apart by the dilated infinitesimal time interval $\mathrm{d}t=t_2\!-\!t_1=\gamma\left(v\right)\mathrm{d}t'$. Time $t$ is the time in the rest frame $\mathrm{S}$ of the Earth. 
Now, let the two light pulses arrive to Earth at  Earth time moments $\hat{t}_{\!1}$ and $\hat{t}_{\!2}$, apart by an infinitesimal time interval  $\mathrm{d}\hat{t}=\hat{t}_{\!2}\!-\!\hat{t}_{\!1}$. If the Star would be at rest relatively to Earth or its motion would be transverse (no radial motion : $v_\mathrm{r}=0$) then $\mathrm{d}\hat{t}=\mathrm{d}t$. But because of the radial motion of the Star relatively to Earth the 2nd pulse, which emitted later, has to run a larger distance than the 1rst pulse if the Star is moving away or has to run  a smaller distance than the 1rst pulse if the Star is approaching. In the first case $\mathrm{d}\hat{t}>\mathrm{d}t$. In the second case, that shown in the Figure, $\mathrm{d}\hat{t}<\mathrm{d}t$.
So, if we could estimate the time interval $\mathrm{d}\hat{t}$ then we would solve the problem since the time intervals are inversely proportional to frequencies that is proportional to wavelengths :
\begin{equation}
\dfrac{\mathrm{d}\hat{t}}{\mathrm{d}t'}=\dfrac{\nu'}{\nu}=\dfrac{\lambda}{\lambda'}=\dfrac{\lambda \text{(observed)}}{\lambda'\text{(emitted)}}
\tag{01}
\end{equation}


As shown in Figure-02 above
\begin{equation}
\mathrm{d}t=t_2\!-\!t_1=\gamma(v)\left(t'_{\!2}\!-\!t'_{\!1}\right) =\gamma(v)\mathrm{d}t'  
\tag{A}
\end{equation}
\begin{equation}
\mathrm{dr}\approx \mathrm{r_2}-\mathrm{r_1}=-v_\mathrm{r}\,\mathrm{d}t=-v \cos\theta\, \gamma(v)\,\mathrm{d}t' 
\tag{B}
\end{equation}
\begin{equation}
\mathrm{d}\hat{t}=\hat{t}_2\!-\!\hat{t}_1=\left(t_2\!+\!\dfrac{ \mathrm{r}_2}{c}\right)\!-\!\left( t_1\!+\!\dfrac{ \mathrm{r}_1}{c} \right)=\mathrm{d}t\!+\!\dfrac{\mathrm{dr}}{c} =\gamma(v)\mathrm{d}t'\!-\!\dfrac{v \cos\theta\, \gamma(v)\,\mathrm{d}t' }{c}   \Longrightarrow 
\nonumber
\end{equation}
\begin{equation}
  \dfrac{\mathrm{d}\hat{t}}{\mathrm{d}t' } = \dfrac{1\!-\!\dfrac{v \cos\theta}{c}}{ \sqrt{1\!-\!\dfrac{v^2}{c^2}}} \stackrel{\left(\beta=\tfrac{v}{c}\right)}{=\!=\!=}\dfrac{1\!-\!\beta \cos\theta}{\sqrt{1\!-\!\beta^2}}=\dfrac{\nu' \text{(emitted)}}{\nu\:\text{(observed)}} =\dfrac{\lambda\:\text{(observed)}}{\lambda' \text{(emitted)}}  
\tag{C}
\end{equation}
QED.
$===================================================$
Solution 2
Link : My answer in About de Broglie relations
For a plane wave the angular frequency 4-vector 
\begin{equation}
\boldsymbol{\Omega} \equiv \left(\omega,c\mathbf{k} \right) 
\tag{2.01}  
\end{equation}
is transformed between frames under the Lorentz transformation. This is proved in the link for a more general configuration of two frames (see the Figure in the end of the link). In (2.01) 
\begin{equation}
\omega= 2\pi\nu
\tag{2.02}  
\end{equation}
is the angular frequency and $\:\nu\:$ the frequency. Also
\begin{equation}
\mathbf{k}= \dfrac{ 2\pi}{\lambda} \;\mathbf{m} , \qquad  \Vert \mathbf{m}\Vert =1
\tag{2.03}  
\end{equation}
is the wave 3-vector and $\:\lambda\:$ the wavelength. The plane wave $^\prime$$^\prime$propagates$^\prime$$^\prime$ with velocity vector
\begin{equation}
\mathbf{w}= \dfrac{ \omega}{\Vert \mathbf{k}\Vert } \;\mathbf{m}=\lambda\nu\;\mathbf{m} = \dfrac{ \omega}{\Vert \mathbf{k}\Vert^{2}}\mathbf{k},  \qquad  \Vert \mathbf{w}\Vert \equiv \mathrm{w} = \dfrac{ \omega}{\Vert \mathbf{k}\Vert }=\lambda\nu
\tag{2.04}  
\end{equation}
From the Lorentz equation (A-14b) in the link we have 
\begin{equation}
\omega^{\boldsymbol{\prime}}  =\gamma\left(\omega\!+\!\dfrac{ \mathbf{v}\boldsymbol{\cdot}c\mathbf{k}}{c}\right)
\tag{2.05}  
\end{equation}
For a light wave $\: \mathbf{k}=(2\pi\nu/c)\mathbf{m}\:$ so
\begin{equation}
\nu^{\boldsymbol{\prime}} =\gamma\left(1\!+\!\dfrac{ \mathbf{v}\boldsymbol{\cdot}\mathbf{m}}{c}\right)\nu
\tag{2.06}  
\end{equation} 
In above equation $\:\mathbf{v}=\!-\boldsymbol{v}\:$ is the velocity vector of the Earth  relatively to the Star , the vector $\:\boldsymbol{v}\:$ shown in Figures-01,-02 and  $\:\mathbf{m}\:$ the unit vector parallel to its radial component $\:\boldsymbol{v}_{\mathrm{r}}\:$ 
\begin{equation}
\mathbf{m}=\dfrac{\boldsymbol{v}_{\mathrm{r}}}{\Vert\boldsymbol{v}_{\mathrm{r}}\Vert}
\tag{2.07}  
\end{equation} 
so that finally
\begin{equation}
\dfrac{\nu' \text{(emitted)}}{\nu\:\text{(observed)}}=\gamma\left(1\!-\!\dfrac{ v \cos \theta}{c}\right)= \dfrac{1\!-\!\dfrac{v \cos\theta}{c}}{ \sqrt{1\!-\!\dfrac{v^2}{c^2}}}
\tag{2.08}  
\end{equation} 
A: Strangely enough, it does not seem the transverse Doppler effect has been demonstrated on Physics SE. This is a trivial consequence of the Lorentz transformation of 4-wave vector. Denoting by $\omega$ the pulsation and by $(k_x, k_y, k_z)$ the wave-vector, $(\omega, k_x, k_y, k_z)$ is a 4-vector and therefore the transformation between the frame $S$ where the satellite is at rest and the frame $R$ where the receiver is at rest is
$$\begin{align}
\omega_R &= \gamma(\omega_S + u\,k_{S,x})\\
k_{R,x} &= \gamma(k_{S,x} + u\, \omega_S)\\
k_{R,y}&=k_{S,y}\\
k_{R,z}&=k_{S,z}\\
\gamma&=(1-u^2)^{-\frac{1}{2}}\\
\end{align}
$$
where the $x$-axis is oriented along the speed of $S$ wrt to $R$. I wrote all of them but we are going to use the first one only. Oh, and I used units where c=1. The next step is to use the relation between the pulsation and the length of the wave vector and introduce an angle for the direction of propagation
$$k_{S,x} = \omega_S\cos\phi,$$
where $\phi$ is the angle between the wave vector and the direction $\overrightarrow{SR}$. Then
$$\frac{\omega_R}{\omega_S}=\gamma(1+u\cos\phi)$$
This is your formula since $\phi = \pi - \theta$ as you consider the angle between $\overrightarrow{RS}$ and the wave vector instead. After re-introducing $c$ of course…
A great reference for undergraduate on that and much more is the classic
Wolfgang Rindler,
Introduction to Special Relativity, 2nd ed.
Oxford University Press, 1991.
