be used to obtain:

$$V_{\theta} = \left( \frac{A \bar{\rho}_{\theta}}{C_{r}} \right) \left( \frac{3cos^{2}(\theta)-1}{g(\theta)}  \right)$$

$$U_{\theta}=- \left( \frac{6 A \bar{\rho}_{\theta}}{C_{\theta}} \right) \left( \frac{cos(\theta)sin(\theta)}{g(\theta)}  \right)$$

The case of the radial system is more cumbersome, as the linear operator which appears is of third order. A good approximation can be obtained by discarding the term on the right hand side of eq.(29), which can then be readily integrated to give:

$$V_{r}=\left( \frac{B C_{r}}{A} \right) r \rho_{r}.$$

In the above we have taken the integration constant which appears as being equal to zero, from requiring $V_{r}  \rightarrow 0$ for $\rho_{r} \rightarrow 0$. Substituting the above relation for $V_{r}$ into eq.(27) leads to:

$$\frac{d \rho_{r}}{d r} \left[ \left(\frac{B}{A} \right)r -r^{2} \right] +  \left(\frac{B}{A} \right)\frac{\rho_{r}}{2} =0.$$

The second term in the above equation can be dismissed for $r<2GM/c^{2}$, which is in any case the validity regime defined by having previously neglected the right hand side of eq.(29), as it is easy to check from the final complete solution. In this regime we now obtain:

$$\rho_{r}=\bar{\rho}_{r} \left( \frac{\bar{r}_{\rho}}{r}\right)^{1/2},$$

$$V_{r}=\left( \frac{B C_{r}}{A} \right) \bar{\rho}_{r} \bar{r}_{\rho}^{1/2} r^{1/2},$$

Where $\bar{r}_{\rho}$ is a characteristic radius at which $\rho_{r}=\bar{\rho}_{r}$. Now from eq.(28),

$$U_{r} = \left( \frac{2 C_{\theta} \bar{\rho}_{r} \bar{r}_{\rho}^{1/2} }{5} \right) r^{3/2}.$$

In 