Page:The Solar System - Six Lectures - Lowell.djvu/59



but $$-\frac$$ is the diff. coefficient of $$-\frac$$ with regard to $$\zeta$$ that is the diff. coefficient of $$-\fraccos$$

The potential necessary to bring the Earth to rest is then $$-\fraccos$$

The potential of M with regard to the particle is $$\frac$$, while the potential of m upon the particle is $$\frac$$ plus a constant. This constant we determine by the condition that the potential at the planet's centre shall be zero, since we are seeking the motion of the particle relative to this centre, and it becomes $$\frac-\frac$$

Since $$r$$ is very large compared with $$p$$, we may advantageously expand the last in powers of $$\frac$$, which gives:—

$$ \frac\lbrack \fraccos+\frac +$$ $$ \frac(\frac\cos^-\frac\cos)+$$ etc.$$\rbrack.$$.

The first term cancels with the potential for bringing the Earth to rest, and we have for the whole potential urging the particle,—

$$ \frac{M}{\rho}+\frac{m\rho^{2}}{r^{3}} (\frac\cos^2-\frac)+ $$ $$\frac{m\rho^{3}}{r^{4}} (\frac{5}{2}\cos^{3}{z}-\frac{3}{2}\cos{z}) $$.