In this example

${\displaystyle -\frac{\mathrm{dv}}{\mathrm{dt}}=-\frac{\mathrm{dv}}{\mathrm{dx}}.v=\frac{c}{x^2}}$

and hence

${\displaystyle \int -v\mathrm{dv}=\int \frac{c}{x^2}\mathrm{dx}}$

${\displaystyle \frac{-c}{x}=\frac{-v^2}{2}+k}$

and we are given $c=4\times {10}^5$, and $v=10$ when $x={10}^4$ so

${\displaystyle k=-40+50=10.}$

Hence when $v=5$ we have

${\displaystyle \frac{-c}{x}=-12.5+10=-2.5}$

and hence $x=\frac{4\times {10}^5}{2.5}=160,000$. So the space craft moves at 5 km s ${}^{-1}$ when it is 160,000 km from the Earth.