simple calculation shows that $r_+$ is in the interval $[r_1,r_2]$, but $r_-$ is not. Thus, the maximum value of $f_1$ can occur only at $|f_1(r_1)|$, $|f_1(r_+)|$ or $|f_1(r_2)|$.

Calculating $|f(r_1)|$, one finds that $$|f_1(r_1)|=\frac{16}{M^2}\frac{S^6 (S^2-1)}{(1+S)^8} \left(  \frac{S-1}{S+1} \right)^{2S-4}. \label{08012014f}$$ It is easy to check that $f_1(r_2)\approx 2\times 10^{-3A}/M^2 < |f_1(r_1)|$. By comparing $|f_1(r_1)|$ with $|f_1(r_+)|$, we also find that $|f(r_1)|$ is always bigger (see Figs. \[3022014a\]). Hence, $|f(r_1)|$ corresponds to the largest value of the left-hand side of the inequality (\[9082013l\]) during the trip.

![In this figure the black curve represents $|f_1(r_+)|$ as a function of $S$, while the grey one is the plot of $|f_1(r_1)|$. The interval between $0$ and $1$ has been suppressed so that the qualitative behavior of these curves in the interval $(1,\infty)$ be better visualized. []{data-label="3022014a"}](f6.jpg)

From the inequality (\[9082013l\]) we see that $|f_1(r_1)|{ \put(2,0){$<$} \put(2,-5){$\sim$}\quad \ }10^{-16}/m^2$. As shown in Fig. \[3022014a\], the minimum value of $|f_1(r_1)|$ occurs when $S\to \infty$. Therefore, this limit is our best choice for $S$. By taking this limit, we get $$\frac{16}{M^2e^4} { \put(2,0){$<$} \put(2,-5){$\sim$}\quad \ }10^{-16}/m^2,  \label{08012014h}$$ which yields $M{ \put(2,0){$>$} \put(2,-5){$\sim$}\quad \ }5\times 10^7\ $m.

Condition (\[9082013m\]) {#08032014b}
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Treating the left-hand side of the inequality (\[9082013m\]) 