\right) \cos\theta -\left(\frac{1}{12}\frac{1}{r^3} - \frac{1}{4}\frac{1}{r} \right)\cos ^3 \theta   \right]$$ where: $$\label{Meqn}
M_* = \sqrt{\frac{1}{D}}M_{\infty}$$ which can be viewed as the fast-mode Mach number. Using $\mathbf{v}= \nabla \phi$, we get: $$\begin{aligned}
v_r(r,\theta) = & \left(1-\frac{1}{r^2} \right)\cos \theta - {M_*}^2\left[ \left(-\frac{13}{12}\frac{1}{r^2}+\frac{3}{2}\frac{1}{r^4}-\frac{5}{12}\frac{1}{r^6} \right) \cos\theta +\left(\frac{1}{4}\frac{1}{r^4} - \frac{1}{4}\frac{1}{r^2} \right)\cos ^3 \theta   \right] \notag \\
& + O({M_*}^4)\label{vr}\\
v_{\theta}(r,\theta) = & -\left(1+\frac{1}{r^2} \right)\sin \theta + {M_*}^2\left[ \left(\frac{13}{12}\frac{1}{r^2}-\frac{1}{2}\frac{1}{r^4}+\frac{1}{12}\frac{1}{r^6} \right) \sin\theta -\left(\frac{1}{12}\frac{1}{r^4} - \frac{1}{4}\frac{1}{r^2} \right)3 \cos ^2 \theta \sin \theta   \right] \notag \\
& + O({M_*}^4)\label{vthe}\end{aligned}$$

Derivation of density distribution
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It is now necessary to derive an equation for the distribution of the density. We need to convert equation \[reduced\_en\] to cylindrical coordinates: $$\label{derivedeqn}
\rho^{\gamma-1}+\frac{\gamma-1}{2}{M_*}^2(v_r^2+v_{\theta}^2 -1)=1$$ Rewriting Equation \[derivedeqn\], we find: $$\label{dendist}
\rho=\left(1-\frac{\gamma-1}{2}{M_*}^2(v_r^2+v_{\theta}^2 -1)\right)^{1/(\gamma-1)}$$ This equation can now be used to determine the expected density distribution around a plume head.

Figure \[eqnden\] shows the density distribution around a circular cylinder. The high density region, created as a result of the compression of the plasma, is clearly visible at the top of the circular cylinder. This is for the case where the higher order terms were used, i.e. the distribution shown is from Equation \[dendist\].

![The density found for a compressible flow around a circular cylinder. Brighter 