AND THE FIGURE OF THE EARTH. 
95 
21. The equation of equilibrium will be formed exactly as in (16.), except that 
the expression for the velocity must be found more accurately. By ( 1 7*)? force 
of gravity at the equator is 4 tt % jy (l + s — ; and the centrifugal force is co 2 a ; 
! 9 . M a ( 3 ra 2 \ 
and co- = 4 tt yy \m + m £ «y 
therefore m = 
Ma 
3 a 3 
(i+-x> 
Me/ 3 ra\ 
The equation of equilibrium becomes 
/^ = C = V + 2T f rMl -f- 2 ) 
which at the surface becomes 
c ' = T7-|(^ 2 - t) ^+i O' 3 - V jfi + S) W +|(1 -f‘ 2 ) r2 §7? (“+ ms -^r )’ 
= IV 0 + s P 2 + A f* 4 ) _ T (^ ~ s') (* + 3 s ^ 2 ) + gT (f* 4 - J + 
_3\Pa 
35/ a b 
+ ^(1 
9 M « / 3 m 2 \ 
S f‘)3#V m + mS -"2“J- 
2 
Equate the coefficients of (m 2 and those of (jA, and we have two equations for deter- 
mining s and A, thus showing that equilibrium is possible. These are, 
and 
whence 
and 
22. The resultant attraction in the direction of r, is obtained by differentiating 
/#> as found in (21.), and changing its sign. This produces 
\( A ~T)-F r +-§{(* ~W + r ( l ~ ft) \ m + m '* 
Ma 
( 
m 
3 
m £ + 
3 
Ni 
a ( 1 
e \ 
2 Pa 
3 a 
V s - 
*2 
2 
4 
ft) - 
a 3 
; Vt 
5 / 
21 ~aF 
Ma . 
0 
Na 
3 £ 
1 
Pa 
YY ( A + m i 
a 3 
T 
+ y 
— = 0; 
a b 
Ma 
/ 
m 
. 6 A 
+ 
3 0 
+ 
m £ 
11 o\ 
_ Na 
3 a 
V s 
~ 2~ 
+ T 
V 
4 
~T ‘ 
- y £“) - 
“ 5 a 3 ’ 
Ma | 
5 m 
s\ 
1 Pa 
3a ‘ 
V A - 3 
£ 2 
_ 
-) 
= — 
~9 ~aF ’ 
4tt^ 
Ma 
- 3 f 
3 r 2 
5 \ 
2 7 3 a 
i r » 
since the terms arising from differentiating the expression for V in (20.) with respect 
to a, cancel each other. 
For r write its value, and arrange the result according to powers of (m, and this 
formula becomes 
4 7Tg> - 
M a 
3 
~a (l — m + m 2 — yr + ft (2 s + rti + 2 m 2 — yp-) + ft (2 A + 2 2 — rm)) j 
! 
N a / 1 . ,/3 4 x ,12 \ . Pfl/1 2 10 , , 5\ 
“ V“ 5 + ^ ( 5 - 5 V + ^ 5“ £ j + ^ V21 “ ^ 21 + ^ 9 ) 
. Ma C t 3 m 27 , 9 9 2 _ , 3 . 
= 45 r ?3?(l+ s -T~r4 m 2tqw 2 -f 2 + yA 
( 5 7 ^ 15 99 19 \ / 15 \ 7 
Y m - 2 + Y m 2 - T m 2 - y 2 2 + y A) + (4 22 - 3 A - ~ 2 ms )j> 
