Inverse Law Force

Since the universe is 1-D, we start off with Newton's law in the r-axis: $m \frac{{d}^{2}r}{{dt}^{2}} = \frac{-GMm}{r} .$

Let $\frac{dr}{dt} = v$ , thus $\frac{{d}^{2}r}{{dt}^{2}} = \frac{dv}{dt} = \frac{dv}{dr} \frac{dr}{dt} = v \frac{dv}{dr}$ .

$mv\frac{dv}{dr} = \frac{-GMm}{r}$ $\frac{1}{2} {v}^{2}= -GMln\left| r \right| +C$

Since $v=0$ and ${r}_{0} = 1.5 \times {10}^{11} m$ at $t=0$ , we find $C = GMln({r}_{0})$ .

Hence $v = \sqrt{2GMln\left|{r}_{0}/r \right|}$ $\frac{dr}{dt} = \sqrt{2GMln\left|{r}_{0}/r \right|}$ $dt = \sqrt { \frac { 1 }{ 2GM } } \int _{ 0 }^{ a }{ \frac { dr }{ \sqrt { ln\left| { r }_{ 0 }/r \right| } } }$

We let $ln\left| {r}_{0}/r \right| = u$ and $\frac{-1}{r} dr =du$ . With some algebra we arrive at $t=a\sqrt{\frac{1}{2GM}}\int _{ 0 }^{ \infty }{ { u }^{ -1/2 } } { e }^{ -u }du$ which by the Gamma function yields ${r}_{0}\sqrt{\frac{\pi}{2GM}}.$