Page:Ueber die Ablenkung eines Lichtstrals von seiner geradlinigen Bewegung.djvu/9

 $$y^2=px+ \frac {p} {2a} x^2$$,

then it transforms into:

$$y^2=\frac {2b^2} {a} x+ \frac {b^2} {a^2} x^2$$.

If we compare this coefficients of x and x² with those in (IX), then we obtain the semi-major axis

$$a=\frac {2g} {v^2- 4g}=AB$$,

and the semi-lateral axis

$$b=\frac {v} {\sqrt{v^2- 4g}}=AD$$.

If we substitute this values for AB and AD into the expression for $$\operatorname{tang}\ \omega$$, then we have:

$$\operatorname{tang}\ \omega=\frac {2g} {v\sqrt{v^2- 4g}}$$.

We now want to give an application of this formula on earth, and investigate, to what extend a light ray is deflected from its straight line, when it passes by at the surface of earth.

Under the presupposition, that light requires 564",8 decimal seconds of time to come from the sun to earth, we find that it traverses 15,562085 earth radii in a decimal second. Thus v = 15,562085. If we take under the geographical latitude its square of the sine ⅓ (that corresponds to a latitude of 35° 16'), the earth radius by 6369514 meters, and the acceleration of gravity by 3,66394 meters (s. Traité de mécanique céleste par Laplace, Tome I, pag. 118): then, expressed in earth radii, g = 0,000000575231. — I use this arrangement, to take the most recent and most reliable specifications of the size of earth's radius and the acceleration of gravity, without specific reduction from the Traité de mécanique céleste. By that, nothing will be changed in the final result, because it is only about the relation of the velocity of light to the velocity of a falling body on earth. The earth radius and the acceleration of gravity must therefore taken under the mentioned