On the Induced Circularity of Orbits through Collision

Since the moment of momentum is the velocity into the perpendicular upon its direction, in the time $dt$ it is:—

$vpdt=hdt=r^{2}d\theta$

The whole moment of momentum from perihelion to perihelion is therefore:—

${\begin{aligned}\int _{0}^{360^{\circ }}r^{2}d\theta =\,&{\frac {a^{2}\cdot (1-e^{2})^{2}}{1-e^{2}}}\\&{\underset {0}{\overset {360^{\circ }}{{\Biggl [}\;}}}{\frac {-e\sin \theta }{1+e\cos \theta }}+{\frac {2}{(1-e^{2})^{\frac {1}{2}}}}\tan ^{-1}\left({\sqrt {\frac {1-e}{1+e}}}\cdot \tan {\frac {\theta }{2}}\right){\Biggr ]}\end{aligned}}$
$=2\pi a^{2}\cdot (1-e^{2})^{\frac {1}{2}}$,

which is twice the area of the ellipse.

The energy in the ellipse during an interval $dt$ is

${\frac {1}{2}}mv^{2}dt={\frac {1}{2}}m\mu \left({\frac {2}{r}}-{\frac {1}{a}}\right)dt$

from the well-known equation for the velocity in a focal conic. The integral of this for the whole ellipse is

$\int _{0}^{T}{\frac {1}{2}}mv^{2}dt=\int _{0}^{360^{\circ }}{\frac {1}{2}}{\frac {m\mu }{h}}\left(2r-{\frac {r^{2}}{a}}\right)d\theta =m\mu ^{\frac {1}{2}}\pi a^{\frac {1}{2}}$

Since

$\int rd\theta =\int {\frac {a\cdot 1-e^{2}}{1+e\cos \theta }}d\theta ={\frac {2a\cdot 1-e^{2}}{(1-e^{2})^{\frac {1}{2}}}}\tan ^{-1}\left({\sqrt {\frac {1+e}{1-e}}}\tan {\frac {\theta }{2}}\right)$

and $\int r^{2}d\theta$ is given above.

By collision a part of this energy is lost, being converted into heat. The major axis, a, is, therefore, shortened. But from the expression $\scriptstyle {2\pi a^{2}\cdot (1-e^{2})^{\frac {1}{2}}}$ for the moment of momentum we see that this is greatest when *e* is least. If, therefore, *a* is diminished, *e* must also be diminished, or the moment of momentum would be lessened, which is impossible.