Page:The evolution of worlds - Lowell.djvu/284

with that of the Earth. Then those approaching it at any angle $\theta$ less than that which makes $\theta _{1}=90^{\circ }$ will be visible at sunset ; those at a greater angle, at sunrise. The angle $\theta _{1}$is given by the relation,

$\cos {\theta _{1}}=+{\frac {a}{x}}$,

in which $a$ is the Earth's velocity, $x$ the meteor's, and $\theta _{1}$ is reckoned from the Earth's quit.

The portion of the celestial dome covered at sunset is, therefore,

$\int _{0}^{\theta _{1}}\int _{0}^{360^{\circ }}\sin {\theta }\cdot d\theta \cdot d\phi$,

where $\phi$ is the azimuth,

that at sunrise, $\int _{0}^{180^{\circ }}\int _{0}^{360^{\circ }}\sin {\theta }\cdot d\theta \cdot d\phi$.

If the meteors have direct motion only, $\theta$ can never exceed 90°, and the limits become,

for sunset, $\int _{0}^{\theta _{1}}\int _{0}^{360^{\circ }}\sin {\theta }\cdot d\theta \cdot d\phi$,

and for sunrise, $\int _{0}^{90^{\circ }}\int _{0}^{360^{\circ }}\sin {\theta }\cdot d\theta \cdot d\phi$.

The mean inclination at sunset is

${\frac {\int _{0}^{\theta _{1}}\int _{0}^{360^{\circ }}\theta _{1}\cdot \sin {\theta }\cdot d\theta \cdot d\phi }{\int _{0}^{\theta _{1}}\int _{0}^{360^{\circ }}\sin {\theta }\cdot d\theta \cdot d\phi }}$,

in which $\theta _{1}$ must be expressed in terms of $\theta$, etc.

From this it appears that the relative number of bodies, travelling in all directions and at parabolic speed, which the Earth would encounter at sunrise and sunset respectively would be:—

sunrise ⁠. ⁠. ⁠. ⁠. ⁠. ⁠. ⁠. ⁠. ⁠5.8
sunset ⁠. ⁠. ⁠. ⁠. ⁠. ⁠. ⁠. ⁠. ⁠1.0

and with the speed of the short-period comets,