First I shall prove the formula:

${\displaystyle a=v\frac{\mathrm{dv}}{\mathrm{dx}}\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(1)}$

I shall write the formula for acceleration $\frac{\mathrm{dv}}{\mathrm{dt}}$ as

${\displaystyle a=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{dv}}{\mathrm{dx}}.\frac{\mathrm{dx}}{\mathrm{dt}}=v.\frac{\mathrm{dv}}{\mathrm{dx}}.}$

From the problem, I know that $a=-c/x^2$, as vectors $a$ and $x$ have opposite signs. Using this and relation (1), I obtain: In this example

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

and hence

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

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

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

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

which gives $x=\frac{4\times {10}^5}{2.5}=160,000$. So the space craft moves at 5 km per second when it is 160,000 km from the Earth.

I solved the problem in two ways, as above using the relation just proved, and using conservation of energy. The methods are essentially the same and the constant given as $c$ combines the gravitational constant $G$ and the mass of the Earth.