Find period of circular-orbiting source based on max observed freq

[cyberpixel44]

TL;DR Summary

Deriving an equation to calculate the time period of a circular-orbiting source from an observer at a distance away from the circle based on max frequency observed

This question refers to Doppler Effects observed in circular motion (at non relativistic speeds, so $v\ll c$, ignoring transversal Doppler shifts).

Suppose there is a source emitting a frequency, $f_s$. An observer at the center will experience no shift in observed frequency ($f_r$). As they move towards the circle of motion, $f_r$ starts to shift up and down "sinusoidally". I use quotes, as it starts go skew as the observer approaches the source's orbit, because $f_r$ gradually increases as the source approaches, and then instantly drop to a minimum after it passes the observer.

My goal here is to derive an equation that, given the parameters of this environment:

$$\begin{array}{l}{{X:Distance\;of\;observer\;from\;center}}\\ {{R:Radius\;of\;orbit\;of\;source}}\\ {{C:Speed\;of\;wave\;in\;medium}}\\ {{f_{\mathrm{max}}\;:Maximum\;observed\;frequency}}\\ {{f_{\mathrm{s}}\;:Source\;frequency}}\end{array}$$

I've tried my hand at deriving something, but it is *almost* correct. It just breaks down when $X$ approaches $R$ and the sinusoidal-look starts to break down and the actual Time period, $T_s$, starts to deviate from the actual value. Here's my take:

The source moves in a circle of radius $R$, with its position as a function of time:
$$\mathbf{r}_{s}(t)=(R\cos(\omega t),R\sin(\omega t)),$$
where $ω=2π/T_s$ is the angular frequency of the source, and $T_s=1/f_s$ is its time period.
And for the distance,
$$d(t)=\sqrt{(R\cos(\omega t)-X)^{2}+(R\sin(\omega t))^{2}}$$
$$d(t)=\sqrt{R^{2}-2R X\cos(\omega t)+X^{2}}$$
From the Doppler Effect equation,
$$f_{r}(t)=f_{s}\left(1-\frac{v_{\mathrm{rel}}(t)}{c}\right)$$
Substituting $v_{rel}$,
$$f_{r}(t)=f_{s}\left(1+\frac{R\omega X\sin(\omega t)}{c\sqrt{R^{2}-2R X\cos(\omega t)+X^{2}}}\right)$$
Since the maximum $f_r$ occurs when $sin(\omega t)=1$,
$$f_{\mathrm{max}}=f_{s}\left(1+{\frac{R\omega X}{c\sqrt{R^{2}+X^{2}-2R X\cos(\pi)}}}\right)$$
$$f_{\mathrm{max}}=f_{s}\left(1+\frac{R\omega X}{c(X+R)}\right)$$
Since $\omega = 2\pi/T_s$,
$$f_{\mathrm{max}}=f_{s}\left(1+{\frac{2\pi R X}{c T_{s}(X+R)}}\right)$$
$$T_{s}=\frac{2\pi R X}{c(X+R)\left(\frac{f_{\mathrm{max}}}{f_{s}}-1\right)}$$
This equation is almost right. I do not know where to proceed from here. My go at it might also be completely incorrect. Additionally, here is the equation I used to check the "actual" values of T should be:
$$d\left(t\right)=\sqrt{\left(R^{2}-2RX\cos\left(\omega t\right)+X^{2}\right)}$$
$${\frac{f_{r}}{f_{s}}}={\frac{1-{\frac{v}{c}}\cos\theta}{\sqrt{1-{\frac{v^{2}}{c^{2}}}}}}$$
$$r_{rel}=-v\cos{\theta}=\frac{d}{dt}\left(d\left(t\right)\right)$$
$$\boxed{f_{r}=f_{s}\left(1-\frac{r_{rel}}{c}\right)}$$
With this form of the equation, $\omega$ cannot be separated to only one side of the equation, right? I don't suppose the above equation can be used to rearrange it to get what I want?

 

[thomsj4]

We analyze the Doppler shift by tracking individual wave crests emitted by a source orbiting with angular frequency $\Omega$ at a radius $R$. The source's position at emission time $t$ is $\bigl(R\cos(\Omega t),\,R\sin(\Omega t)\bigr)$. The emitted wave has an angular frequency $\omega_s=2\pi f_s$. We index the crests by integers $n\in\mathbb{Z}$, with the nth crest emitted at

$$t_n=\frac{2\pi n}{\omega_s}.$$

The phase of the emitted wave at $t=t_n$ is thus $2\pi n$. An observer is fixed at $(X,0)$. The distance traveled by the nth crest is

$$d_n = \sqrt{(X-R\cos(\Omega t_n))^2 + (R\sin(\Omega t_n))^2}.$$

This crest arrives at the observer at time

$$\tau_n = t_n + \frac{d_n}{c}.$$

Therefore, consecutive crests arrive at times

$$\tau_n = \frac{2\pi n}{\omega_s} + \frac{1}{c}\sqrt{\,X^2-2XR\cos\left(\Omega\,\tfrac{2\pi n}{\omega_s}\right)+R^2}.$$

For large $n$, we approximate the observed frequency by

$$f_r(n) \approx \frac{1}{\,\tau_{n+1}-\tau_n\,}.$$

Using the expression for $\tau_n$, we find

$$\tau_{n+1}-\tau_n = \left(\frac{2\pi}{\omega_s}\right) + \frac{1}{c}\left(\sqrt{\,X^2-2XR\cos\left(\Omega\,\tfrac{2\pi(n+1)}{\omega_s}\right)+R^2} - \sqrt{\,X^2-2XR\cos\left(\Omega\,\tfrac{2\pi n}{\omega_s}\right)+R^2}\right),$$

which leads to

$$f_r(n) = \frac{1}{\displaystyle \frac{2\pi}{\omega_s} + \frac{1}{c}\left(d_{n+1}-d_n\right)}.$$

To determine the maximum of $f_r(n)$, we treat $n$ as continuous and set the derivative with respect to $n$ equal to zero:

$$0 = \frac{d}{dn}\bigl[f_r(n)\bigr] = \frac{d}{dn}\left[\,\bigl(\tau_{n+1}-\tau_n\bigr)^{-1}\right].$$

Applying the chain rule gives

$$\frac{d}{dn}\bigl(\tau_{n+1}-\tau_n\bigr) = \frac{d}{dn}\left(\frac{2\pi}{\omega_s} + \frac{1}{c}\bigl(d_{n+1}-d_n\bigr)\right),$$

where

$$d_{n+1} = \sqrt{\,X^2-2XR\cos\left(\Omega\,\tfrac{2\pi(n+1)}{\omega_s}\right)+R^2}$$

and

$$d_n = \sqrt{\,X^2-2XR\cos\left(\Omega\,\tfrac{2\pi n}{\omega_s}\right)+R^2}.$$

Differentiating $d_{n+1}-d_n$ with respect to $n$ involves applying the chain rule to each square root. This gives a transcendental equation in $\tfrac{2\pi n}{\omega_s}$. Let $n_{\ast}$ represent a real solution satisfying

$$\frac{d}{dn}\bigl(\tau_{n+1}-\tau_n\bigr)\Bigl|_{n=n_{\ast}}=0.$$

Thus, $f_r(n)$ reaches an extremum at $n=n_{\ast}$. The peak observed frequency is then

$$f_{\max} = f_r(n_{\ast}) = \frac{1}{\,\tau_{n_{\ast}+1}-\tau_{n_{\ast}}\,}.$$

Note that the orbital phase $\Omega\,\tfrac{2\pi n_{\ast}}{\omega_s}$ is entangled within a trigonometric function inside a square root, preventing us from isolating $\Omega$ (or $T_s=2\pi/\Omega$) in a closed form using elementary functions. We must solve for $\Omega$ numerically by iteratively adjusting $\Omega$ until the peak of $\frac{1}{\,\tau_{n+1}-\tau_n\,}$ matches the measured $f_{\max}$. In the limit $X\gg R$, $d_{n+1}-d_n$ becomes a small perturbation, simplifying the derivative test and leading to a far-field Doppler approximation where a closed-form relation between $T_s$ and $\tfrac{f_{\max}}{f_s}-1$ exists. However, when $X$ is on the order of $R$, the transcendental nature of the exact discrete-crest condition prevents a purely algebraic solution, which means that numerical methods are needed.