Page:Ueber die Ablenkung eines Lichtstrals von seiner geradlinigen Bewegung.djvu/4

${\frac {ddx\ \cos \phi +ddy\ \sin \phi }{dt^{2}}}=-{\frac {2g}{r^{2}}}$

(IV)

To reduce in these equations the number of variable quantities, we want to express x and y by r and $\phi$. We easily see that

$x=r\ \cos \phi$ and $y=r\ \sin \phi$.

If we differentiate, then we will obtain:

$dx=\cos \phi \ dr-r\sin \phi \ d\phi$ , und $dy=\sin \phi \ dr+r\cos \phi \ d\phi$.

And if we differentiate again,

$ddx=\cos \phi \ ddr-2\sin \phi \ d\phi \ dr-r\ \sin \phi \ dd\phi -r\ \cos \phi \ d\phi ^{2}$,

and

$ddy=\sin \phi \ ddr+2\cos \phi \ d\phi \ dr+r\ \cos \phi \ dd\phi -r\ \sin \phi \ d\phi ^{2}$,

If we substitute these values for ddx and ddy in the previous equations, the we obtain from (III):

${\frac {ddy\ \cos \phi -ddx\ \sin \phi }{dt^{2}}}={\frac {2d\phi \ dr+r\ dd\phi }{dt^{2}}}$.

Thus we have

${\frac {2d\phi \ dr+r\ dd\phi }{dt^{2}}}=0$

(V)

And furthermore by (IV),

${\frac {ddr-r\ d\phi ^{2}}{dt^{2}}}=-{\frac {2g}{r^{2}}}$

(VI)

To make equation (V) a true differential quantity, we multiply it by rdt, thus:

${\frac {2r\ d\phi \ dr+r^{2}\ dd\phi }{dt}}=0$,

and if we again integrate, we will obtain:

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

where C is an arbitrary constant magnitude. To specify C, we note that $r^{2}d\phi (=r\ rd\phi )$ is equal to: the double area of the small triangle which described the radius vector r in the time dt. The double area of the triangle that is described in the first second of time, is however: = AC · v; thus we have C = AC · v. And if we assume the radius AC of the attracting body as unity, what we will always do in the following,