For a lossless line, you have \$Z_0=\sqrt{\frac{L}{C}}\$ and \$VF=\frac{1}{c\sqrt{LC}}\$.
For a line with R, L, G and C parameters, we have \$Z_0=\sqrt{\frac{R+j\omega L}{G+j\omega C}}\$.
But how the formula changes for the VF parameter?
Considering the coaxial cable example, one could say the VF not only depends on \$Z\$, but also in the geometry of the cable, hence a more real expression for VF do not exist?.
Best Answer
For a lossy transmission line
\$ Z_0 = \frac{\sqrt{R+j\omega L}}{\sqrt{G + j\omega C}} \$
Note that for \$ R << \omega L\$ and \$ G << \omega C\$ this reduces to the expression in your question.
Finding the phase velocity of a lossy line is difficult as the propagation constant \$\beta \$ will, in general, be a non linear function of frequency. (Except for a distortionless line that satisfies the Heaviside Condtion). This means that the velocity factor is not a very useful metric, except over a narrow band or small distance.
In fact it means that different frequency components of a signal will travel at different speeds, resulting in distortion of the signal and inter-symbol interference in the case of digital communications.
The geometry of the cable determines the R, L, G and C parameters as well as Z and the phase velocity.
Edit: I'm not sure if a better general expression than this can be found but the complex propagation constant will be:
\$ \gamma = j\omega \sqrt{LC} \sqrt{1 - j(\frac{R}{\omega L} + \frac{G}{\omega C}) - \frac{RG}{\omega^2LC}}\$
Then the phase velocity is
\$ v_p = \frac{\omega}{Imaginary(\gamma)}\$
Thus the velocity factor is
\$ \frac{\omega}{c Imaginary(\gamma)} \$
And I don't think the Gamma expression can be decomposed into real and imaginary parts without making approximations.