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I'd consider a **loss-less** coaxial cable. It has inductance and capacitance per unit length that are easily derivable and introduce the terms \$\epsilon\$ and \$\mu\$. Then, I'd derive the characteristic impedance of the cable based on the values of inductance and capacitance per metre: -$$$$

![Image_alt_text](https://electrical.codidact.com/uploads/ah7sp9a51vxc5wxvrr7pcwufn8qi)

$$Z_{IN} = Z_0 = sL + \dfrac{\frac{1}{sC}\cdot Z_0}{\frac{1}{sC} + Z_0}$$

$$Z_0 = sL + \dfrac{Z_0}{1 + sCZ_0}$$

$$Z_0 + sCZ_0^2 = sL + s^2LCZ_0 + Z_0$$

$$sCZ_0^2 = sL + s^2LCZ_0$$

$$$$

When s, L and C are very small values we can say that \$s^2LC\$ becomes zero hence: -

$$sCZ_0^2 = sL$$

And, \$\hspace{7cm}Z_0 = \sqrt{\dfrac{L}{C}}\hspace{1cm}\$ (a well known formula).

$$$$

At this point I would consider analysing the circuit above using standard transfer function analysis. This allows us to derive the phase response of the output compared to the input: -

![Image_alt_text](https://electrical.codidact.com/uploads/f5kgycobtrxtkzmwzmn059gablpj)

I am interested in the phase shift at low to medium frequencies. I'm interested in phase shift because that is equivalent to a time delay. I want the analysis to deliberately stay away from the natural resonant frequency of the circuit because, in this area the phase shift rapidly changes.

$$$$

Here is a simulation of the circuit using typical 50 Ω coaxial cable lumped values for L and C: -

![Image_alt_text](https://electrical.codidact.com/uploads/oi9ql7ohs1uqoumw045cbjoyzyli)

And, here's a close-up of the phase response plotted linearly against frequency from DC to 1 MHz: -

![Image_alt_text](https://electrical.codidact.com/uploads/mdv9xbskrx38tvx7psryjhd5y0j8)

 - At 1 MHz, the output phase lag is 1.8° 
 - As a fraction of the period (1 μs) it's 0.005 hence, it's a time lag of 5 ns.

 - At 100 kHz, the phase lag is 0.18° but, it's still a time lag of 5 ns

Given that I've modelled 1 metre lengths of capacitance and inductance, the equivalent velocity of propagation has to be 200 million metres per second. It's just the inverse of the time lag.

A simulator is great as a demonstration but, a formula for the phase angle is needed so that we know algebraically what the dependencies are. It's a simple 2nd order low pass filter and, if you went through the derivation (omitted to keep the answer shorter) you find that the phase angle response is this: -

$$\text{Phase lag} = \arctan\left(\dfrac{2\zeta\dfrac{\omega}{\omega_n}}{1 - \dfrac{\omega^2}{\omega_n^2}}\right)$$

Then because we are considering situations where \$\omega\$ is a lot smaller than \$\omega_n\$ we can make simplifications:

 - The arctan of a smaller number = the small number
 - The denominator equals 1

$$\text{Phase lag} = {2\zeta\dfrac{\omega}{\omega_n}}$$

But we can also determine \$\zeta\$ for the circuit. [Wikipedia RLC circuits](https://en.wikibooks.org/wiki/Circuit_Theory/RLC_Circuits#Damping_Factor) shows it is this: -

![Image_alt_text](https://electrical.codidact.com/uploads/hm38mlmadwihim4yl5jbrudee5i3)

And, because we know that R is \$Z_0\$ \$\left(\sqrt{\frac{L}{C}}\right)\$, \$\zeta\$ must equal 0.5. The phase lag now becomes: -

$$\text{Phase lag} = {\dfrac{\omega}{\omega_n}}$$

If we divide the phase lag by \$\omega\$ we get the time lag: -

$$\text{Time lag} = \dfrac{1}{\omega_n} = \sqrt{LC}$$

And, the velocity of propagation is the reciprocal of the time lag (recalling that the time lag was for a 1 metre length of the modelled cable): -

$$\text{Velocity of propagation} = \dfrac{1}{\sqrt{LC}}$$

Nearly there! 

$$$$

Right at the start I mentioned coaxial cable and its inductance and capacitance per unit length. The formulas are: -

$$L = \dfrac{\mu}{2\pi}\ln\left(\dfrac{b}{a}\right)$$

Taken from [Inductance of a Coaxial Structure](https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electro-Optics/Book%3A_Electromagnetics_I_(Ellingson)/07%3A_Magnetostatics/7.14%3A_Inductance_of_a_Coaxial_Structure#:~:text=Below%20we%20shall%20find%20the,flux%20to%20the%20source%20current.)

$$C = \dfrac{2\pi\epsilon}{\ln\left(\dfrac{b}{a}\right)}$$

Taken from [Capacitance of a coaxial structure](https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electro-Optics/Book%3A_Electromagnetics_I_(Ellingson)/05%3A_Electrostatics/5.24%3A_Capacitance_of_a_Coaxial_Structure)

$$$$

If we multiply L and C we get  \$\epsilon\cdot\mu\$  hence,

$$\text{Velocity of propagation} = \dfrac{1}{\sqrt{\epsilon\cdot\mu}}$$