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

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then C = v. If we substitute this value for C into the previous equations, then:

${\displaystyle r^{2}\ d\phi =v\ dt}$,

Thus we have

 ${\displaystyle d\phi ={\frac {v\ dt}{r^{2}}}}$ VII

If this value for ${\displaystyle d\phi }$ is substituted into equations (VI), we obtain:

${\displaystyle {\frac {ddr}{dt^{2}}}-{\frac {v^{2}}{r^{3}}}=-{\frac {2g}{r^{2}}}}$.

If we multiply this equations by 2dr, then:

${\displaystyle {\frac {2dr\ ddr}{dt^{2}}}-{\frac {2v^{2}\ dr}{r^{3}}}=-{\frac {4g\ dr}{r^{2}}}}$,

and if we integrate again,

${\displaystyle {\frac {dr^{2}}{dt^{2}}}+{\frac {v^{2}}{r^{2}}}={\frac {4g}{r}}+D}$,

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:

${\displaystyle dt={\frac {dr}{\sqrt {D+{\frac {4g}{r}}-{\frac {v^{2}}{r}}}}}}$,

If we substitute this value for dt into equation (VII), then we obtain:

${\displaystyle d\phi ={\frac {v\ dr}{r^{2}{\sqrt {D+{\frac {4g}{r}}-{\frac {v^{2}}{r^{2}}}}}}}}$,

To integrate this equations, we bring it into the form:

${\displaystyle d\phi ={\frac {v\ dr}{r^{2}{\sqrt {D+{\frac {4g^{2}}{v^{2}}}-\left({\frac {v}{r}}-{\frac {2g}{v}}\right)^{2}}}}}}$,

Now we put

${\displaystyle {\frac {v}{r}}-{\frac {2g}{v}}=z}$,

then we have

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