then C = v. If we substitute this value for C into the previous equations, then:
$r^{2}\ d\phi =v\ dt$,
Thus we have
$d\phi ={\frac {v\ dt}{r^{2}}}$

VII

If this value for $d\phi$ is substituted into equations (VI), we obtain:
${\frac {ddr}{dt^{2}}}{\frac {v^{2}}{r^{3}}}={\frac {2g}{r^{2}}}$.
If we multiply this equations by 2dr, then:
${\frac {2dr\ ddr}{dt^{2}}}{\frac {2v^{2}\ dr}{r^{3}}}={\frac {4g\ dr}{r^{2}}}$,
and if we integrate again,
${\frac {dr^{2}}{dt^{2}}}+{\frac {v^{2}}{r^{2}}}={\frac {4g}{r}}+D$,
where D is a constant magnitude, that depends on the constant magnitudes which are contained in the equation. From this equation that is found now, the time can be eliminated, hence:
$dt={\frac {dr}{\sqrt {D+{\frac {4g}{r}}{\frac {v^{2}}{r}}}}}$,
If we substitute this value for dt into equation (VII), then we obtain:
$d\phi ={\frac {v\ dr}{r^{2}{\sqrt {D+{\frac {4g}{r}}{\frac {v^{2}}{r^{2}}}}}}}$,
To integrate this equations, we bring it into the form:
$d\phi ={\frac {v\ dr}{r^{2}{\sqrt {D+{\frac {4g^{2}}{v^{2}}}\left({\frac {v}{r}}{\frac {2g}{v}}\right)^{2}}}}}$,
Now we put
${\frac {v}{r}}{\frac {2g}{v}}=z$,
then we have
${\frac {v\ dr}{r^{2}}}=dz$.