502 
Mr. Airy on the 
(19.) Let m be the ratio of the centrifugal force at the 
equator to gravity at the equator. The centrifugal force 
there = ^ a ( 1 + e ) = -y- • pfr 2 a (1 + e) : the gra- 
vity there = ~~ (1 — e) — . 3 a j : therefore 
a(2 + 2e) = JLQ( m — me) a. 3 m ; whence 
3 n a ft (a) to — to e ft (a) / to g 3 m x \ 
2 T z a 2, 2 + ze f 3»i ' a 2 l 2 4 j 
Making this substitution, gravity = {(1 ■— e — m 
+ e, + 7 ” e + {»’ + 7 A )+ (-^ — e + 2 e—^me 
~ J m ‘+ y me + 3A). A*. fUjq 
= — e ~- i f- + eS +-?7 ffle + fw 9 + -iA) 
x i 1 +(“ — e + e 9 — ^-me + j- A) . X 9 — (•£ me 
— “ -{- 3 A) . A 2 .i — x 3 | . If then the equatorial gravity 
be represented by G, the gravity in latitude l will be 
G l 1 + ('T i “" e + e2 “"lT me + 7 A ) sin . 2 1 
■ — m e — -f 3 A) sin 2 l . cos 2 1 j 
Whatever therefore be the law of density in the interior of 
the earth, the gravity at any point of the surface can be ex- 
pressed from a knowledge of the form of that surface only. 
This is an extension of Clairaut’s theorem. 
(20). We shall now proceed to find an expression for the 
length of an arc of the meridian included between two given 
latitudes. If 6 be the corrected latitude, and u = the radius 
f 1 + _L . 
of curvature is — d \ d ^ Here u^~(i — e.T— f* 2 + 
