Page:The Solar System - Six Lectures - Lowell.djvu/55

Mercury

37

---

If M be far away compared with CP,

$\textstyle BN=2CB=2p$, say;

for since

$\textstyle PM=BM=D$

and

$\textstyle CB=p$,

$\textstyle {\frac {(D+P)^{2}}{D^{2}}}={\frac {D}{NM}}$;

whence

$\textstyle NM=D-2p$,

whence

$\textstyle BM-NM=BN=2p$.

.

The tide-raising force PN may be resolved into a normal disturbing force PL and a tangential disturbing force LN. From the fact that BN is always twice CB, we find for the vanishing points of the normal force a and b, those where the angle BCP = 54° 44’. The whole disturbing force is there tangential.

Now consider the action of the two components; first, that of the tangential factor. At F, the whole force is normal and acting inward. From its minimum here the tangential force rises to a maximum at a, where it comprises the whole force. It then subsides to zero at A. During this quadrant it has been urging the particle onward in its own direction of movement FA. At A, it changes sign and becomes a retarding force, which attains its maximum at b, and then sinks to zero again at E.

In consequence, the velocity of the particle due