This last must be resolved into two, one in MS, the other perpendicular to it.

Therefore, the whole central force on the moon in MS

and the proposition will be proved if we shew that this force

cos M-1, and is then

E

m'

SM EM

It is, therefore, always positive, or the path always concave to the sun.

At new moon the force with which the moon tends to the sun is, therefore, greater than that with which she tends to the earth: the earth being itself in motion in the same direction, and, at that instant, with greater velocity, will easily explain how, notwithstanding this, the moon still revolves about it.

Central and Tangential Disturbing Forces.

93. We have hitherto considered the effects of the central and tangential disturbing forces in combination; but it will be interesting to determine to which of them the several inequalities principally owe their existence.

(1) To determine the effect of the central disturbing force.

1—3 k2 — { m2 + 3⁄4 m2e cos(c0-a)—3m2 cos{(2—2m)0—2ß}

[ 1 - 3 k2 - 1m2 + e cos(c0− a) +1⁄2m2 cos {(2—2m)0—2ẞ} + me cos{(2−2m − c) 0 − 2B+ a}

— 4k2 cos2 (g0—y)—§m2e' cos(m0+B−5). If we compare this with the value of u found Art. (48), we see that the elliptic inequality, the reduction, and the annual equation are due to the central or radial force, as also one half of the variation and about a third of the evection.

It would perhaps be proper to separate the absolute central force from the central disturbing force; the terms due to the latter are those which contain m; therefore, the elliptic inequality and the reduction are the effects of the former, except that in the elliptic inequality the introduction of c, or the motion of the apse, is due to the disturbing force.

(2) To determine the effect of the tangential disturbing force. Let the central disturbing force be zero;

м

h2

(1-3)= a neglecting the inclination,

- §m2 sin {(2 — 2m) 0 – 2ß}

+3m2e sin {(2 – 2m − c) 0 − 2B+a}

I

3m2ae cos {(2 - 2m − c) 0 − 2ẞ + a},

do=3m2 cos ((2-2m) 0-2,3}

d2u

d02

-3m2e cos((2-2m-c) 0-2B+a},

+u= a to the first order.

We have here the remaining half of the variation and rather more than two-thirds of the evection as the effects of the tangential disturbance. Also c=1, or, to the second order, the tangential force has no effect on the motion of the The inequalities in the longitude could be easily obtained from the relation

apse.

but they would lead to the very same conclusions as the

discussion of the values of u.

To calculate the value of c to the third order.

94. We must here make use of the results which the approximations to the second order have furnished; but as the value of c is determined by that term of the differential equation whose argument is c✪ — a, we need only consider those terms which by their combinations will lead to it without rising to a higher order than the fourth.

We shall simplify the arguments by omitting e, a, ß, which can easily be supplied by remarking that cơ – a and me+B always enter as one symbol, c and m will therefore be sufficient to distinguish them. This only applies to the arguments.

We have, Arts. (48), (23),

u = a {1+e cos (c) + m2 cos (2 − 2m) + 15 me cos (2 − 2m — c)},

P

h2u2

T

hu

T du

h3u3 do

= a − ¿m2a {1 +3 cos (2 – 2m)} {1 – 3e cos (c)

+......- 45me cos(2 — 2m - c)}

= a+ &m3ae cos(c) + 13,5m3ae cosc),

— §m2 sin (2 – 2m) {1 – 4e cos(c) — 4m2 cos (2 — 2m)

=

=

− §m2 sin(2−2m)+3m3e sin(2− 2m−c) + 45m3e sin(c),

= {— §m2 sin (2 — 2m)} {− 15mae sin (2 — 2m — c)}

+ {3m2e sin (2 − 2m — c)} {— 2m3a sin (2 – 2m)}

= 11⁄2m3ae cos(c),

the other term is of the fifth order,

[128 d0 = 4m2 cos (2 – 2m) — 3m3e cos(2—2m—c) — 45m3e cos (c),

(dru

+u)=2a{1+...—3m2 cos(2—2m) + 45 m2e cos(2—2m—c)};

2

de

= a {1+(3m2e + 135m3e — §§m3e + 45m3e) cos(c)+.....}

= a {1+(3m2e + 225 m3e) cos (c) +...}.

Assume ua {1+e cos(c) +.....} ;

95. This is to be obtained in a very similar manner from

8 =

+8=&c. We shall, in the argument, write

· k {sin (g) + 3m sin (2 — 2m — g)},

Ps-S 3m2a+s

2u1

{1+cos(2 − 2m)},

3m2 a1

sin (2-2m).