$v$ is constant. In this case, the condition (\[9082013h\]) can be rewritten as $$\Delta \tau=\frac{\sqrt{1-v^2/c^2}}{v}10^AM { \put(2,0){$<$} \put(2,-5){$\sim$}\quad \ }1\ yr, \label{15122013d}$$ where we have used $w_1+w_2\approx w_2\approx r_2=10^AM$. Notice that, here, we are using $w$ rather than $l$.

With respect to the condition (\[9082013i\]), the constancy of $v$ leads to $$\Delta t=\frac{1}{v}\int_{-w_1}^{w_2}W^{-S/2}dw { \put(2,0){$<$} \put(2,-5){$\sim$}\quad \ }1\ yr. \label{23122013b}$$ Note that, due to the similarity between $l$ and $w$, we can simply exchange $l$ for $w$ in Eq. (\[9082013i\]) to write this condition in terms of $w$. Using \[see Eq. (\[18122013a\])\] $$dw=W^{-S/2}dr, \label{31012014b}$$ we find that $$\Delta t=\frac{1}{v}\int_{r_1}^{r_2}W^{-S}dr. \label{23122013c}$$ The integrand of this expression can be expanded in the form $$(1-r_0/r)^{-S}=1+\frac{Sr_0}{r}+\sum_{n=2}^{\infty} \frac{r_0^n}{n!}r^{-n}\prod_{j=0}^{n-1}(S+j). \label{16092013h}$$ Using this expansion in Eq. (\[23122013c\]), we arrive at $$\Delta t=\frac{1}{v}\left[ r_2-r_1+Sr_0\ln(r_2/r_1)+\sum_{n=2}^{\infty} \frac{r_0^n}{n!(1-n)}\left(r_2^{1-n}-r_1^{1-n}\right)\prod_{j=0}^{n-1}(S+j)  \right].   \label{23122013d}$$ Since the largest value of $r_1$ is $2M$, while $r_2=10^AM$ with $A{ \put(2,0){$>$} \put(2,-5){$\sim$}\quad \ }6$, we can approximate the above expression to $$\Delta t\approx \frac{r_2}{v}=\frac{10^AM}{v}.\label{23122013g}$$ The smallest value for $M$ that is allowed by the condition \[9082013g\] is $M=5\times 10^{15}\ m$. Using this value in Eq. (\[23122013g\]) with $A=6$ and taking $v$ as the speed of light, one gets $\Delta t \approx 5\times 10^5\ yr$. Since this is the best-case 