we use the constraint \[9082013g\] in this approximation, we will obtain $$r_2 { \put(2,0){$>$} \put(2,-5){$\sim$}\quad \ }c\sqrt{\frac{M}{g}},  \label{15082013f}$$ Writing $r_2$ in the form $r_2=10^AM$ and substituting it into Eq. (\[15082013f\]), one finds that $A{ \put(2,0){$>$} \put(2,-5){$\sim$}\quad \ }\log(c/\sqrt{Mg} )$. On the other hand, if we impose the conditions \[9082013e\] and \[9082013f\], we will have $10^{-A} \ll 1$. Based on the latter inequality, it is reasonable to take $A{ \put(2,0){$>$} \put(2,-5){$\sim$}\quad \ }6$. It can be shown that $\log(c/\sqrt{Mg} )$ is larger than $6$ only for $M< 9171\ m$, which allows us to take $A{ \put(2,0){$>$} \put(2,-5){$\sim$}\quad \ }6$ whenever $M{ \put(2,0){$>$} \put(2,-5){$\sim$}\quad \ }9171\ m$.

With respect to $b/R\ll 1$ evaluated at $w=-w_1$, we have to be very careful because in this case the function $b/R$ does not decrease as $w$ goes to $-\infty$ ($r\to r_0$). In fact, it diverges there. Nonetheless, this function vanishes at $r_1$. That is the reason why we are using $w_1=-w(r_1)$ with $r_1$ given by Eq. (\[29012014b\]).

Now we show that the condition \[9082013f\] is not satisfied. From Eqs. (\[29012014b\]) and (\[10082013g\]), we get $$\Phi(r_1)=S\ln\left(\frac{S-1}{S+1} \right). \label{27122013e}$$ The minimum value of $|\Phi(r_1)|$ is $2$ and occurs when $S$ goes to infinity. So we cannot have 