Page:Ueber die Ablenkung eines Lichtstrals von seiner geradlinigen Bewegung.djvu/5

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

${\frac {v\ dr}{r^{2}}}=-dz$.