537] 
SOLUTIONS OF A SMITH’S PRIZE PAPER FOR 1871. 
513 
11. A particle is attracted to two centres of force, one of them at the origin, the 
other revolving about the origin in a circle in the plane of xy with a uniform angular 
velocity n': find the equations of motion; and writing v for the velocity of the particle 
and A for the resolved area (about the fixed centre) in the plane of xy, show that 
dA 
there is a first integral giving the value of v- — 4>n in terms of the coordinates of 
the particle and of the revolving centre. 
Take %, y, 0 for the coordinates of the moving centre, so that 
f = a cos n't, y = a sin n't; 
the equations of motion are 
d 2 x . x . x — P 
dt 2 r r p 
d 2 z , z , z 
= o ’ 
where 
We have 
But 
whence 
r- — a? + y 2 + z 2 , 
p> = (x- £) 2 + (y - yf + ¿ 2 . 
r dr = x dx -t-y dy + z dz, 
p dp = (x — |) (dx — d%) + (y — y) (dy — dy) + z dz. 
d% = — n'a sin n't dt — — n'y dt, 
dy = n'a cos n't dt = nfi dt, 
p dp = (x — %)dx + (y — y) dy + z dz 
+ n [y (x-Ç)-Ç (y - y)] dt 
= (x — %)dx + (y — y) dy + z dz 
- n' [x (y - y) - y (x - £)] dt. 
Hence from the equations of motion 
2 fdx d 2 x dy d 2 y dz d 2 z\ 
\dt dt 2 dt dt 2 dt dtV 
d?y d 2 x\ 
c , ( d 2 y d 2 x 
— 2 n x — y 
V dt 2 J dt 2 
fir „ f dx dy dz\ 
= -7 2 { x di + ydt +z dt< 
*{<—+ + * 
2 n' {®(y-rj)-y(x- f)} 
dz) 
dt) 
C. VIII. 
65