Page:Ueber die Ablenkung eines Lichtstrals von seiner geradlinigen Bewegung.djvu/6

If this and z is substituted into the equation for $d\phi$, the we will have:

$d\phi ={\frac {dz}{\sqrt {D+{\frac {4g^{2}}{v^{2}}}-z^{2}}}}$,

From that the integral is now:

$\phi =\operatorname {Arc\cos } {\frac {z}{\sqrt {D+{\frac {4g}{v^{2}}}}}}+\ \alpha$,

where $\alpha$ is a constant magnitude. By well-known properties it is furthermore:

$\cos(\phi -\alpha )={\frac {z}{\sqrt {D+{\frac {4g}{v^{2}}}}}}$,

and if we also substitute instead of z its value:

$\cos(\phi -\alpha )={\frac {v^{2}-2gr}{r{\sqrt {v^{2}D+4g^{2}}}}}$.

$\phi -\alpha$ would be the angle that r forms with the major axis of the curved line that has to be specified. Since furthermore $\phi$ is the angle which r forms with the line AF (the axis of the coordinates x and y), then $\alpha$ must be the angle that forms the major axis with the line AF. However, since AF goes through the observation place and the center of the attracting body, then by the preceding, AF must be the major axis; also $\alpha$ = 0, and thus:

$\cos \phi ={\frac {v^{2}-2gr}{r{\sqrt {v^{2}D+4g^{2}}}}}$.

For $\phi$ = 0 it must be r = AC = 1, and we obtain from this equation:

${\sqrt {v^{2}D+4g^{2}}}=v^{2}-2g$.

If we substitute this in the previous equation, then the still unknown D and also the square-root sign vanish; and we obtain:

$\cos \phi ={\frac {v^{2}-2gr}{r(v^{2}-2g)}}$;

and furthermore by that