Page:The Solar System - Six Lectures - Lowell.djvu/51

Mercury

33

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$\scriptstyle r={\frac {a\;(1-e^{2})}{1+e\cos v}}\,\!$,

and equate it to that of the circle supposed described about that focus with the length of radius $\scriptstyle {\sqrt {ab}}\,\!$. This geometrically is the point of intersection of the two curves, since the value of r is common to both.

Consequently for the point sought whence, since

$\scriptstyle {\sqrt {ab}}={\frac {a\;(1-e^{2})}{1+e\cos v}}\,\!$,

whence, since $\scriptstyle b=a{\sqrt {1-e^{2}}}\,\!$,

$\scriptstyle (1-e^{2})^{\frac {1}{4}}={\frac {(1-e^{2})}{1+e\cos v}}\,\!$,

and $\scriptstyle \cos v={\frac {(1-e^{2})^{\frac {3}{4}}-1}{e}}\,\!$.

In the case of Mercury, e = .205605; v, the true anomaly of the point of maximum libration, is therefore 98° 55'.13.

But $\scriptstyle {\frac {a-r}{a\;e}}=\cos E\,\!$,

where E is the eccentric anomaly; and $\scriptstyle E-e\sin E=M\,\!$, where M is the mean anomaly; whence $\scriptstyle v-M=\zeta \,\!$, which is the amount of the maximum libration, is 23° 40' 38".

The gain or loss of the rotation over the revolution is the same thing as the equation of the centre.

We have, then, in the libration, a most conclusive and interesting proof of the isochronism of rotation and revolution.

The next point to consider is what caused this