: Be hs Nika 
W. A. Norton on Comets. 91 
the particle will recede in the direction NZ, upon a different 
and extreme hypothesis; viz., that it is projected from a point 
at a distance 2r from the centre of the nucleus, with a velocity 
equal to that acquired in passing from the distance r to the 
distance 27, and that the repulsion of the nucleus acts in the di- 
rection NX, which, as we have seen, is inclined to NZ under an 
angle equal to the complement of a. The differential equation 
for finding the velocity in the direction NZ, will be 
(= sin a—& sin a) dz=vdv, 
22 
2 Bia, 
Integrating, —* sina—ksina.2--C="=. 
Putting V= velocity at distance 2r, 
V8 rk, : 
Cc =a t open a-+k sin a.2r 
Whence, L ewhath a (2r—z) +-pr? sin a Fs *) +— .. (9) 
aie or. 8 2 
’ 2 
To find V2, we = v’ ave dz, or = EC, 
% ‘2 salt | a ee ee ; 
0=F ; thus = pr? (5-3): Which gives — V2=pr, and 
v= V3. . 
Substituting in equ. 9, we have 
2 2r—z). 
=k sin a (2 pa 1) Oem sin a= 
Putting second member =0, we obtain 
me A Aw, 
(Rigoret vane 3" sin¢) (nemrly) (9 
The second term is small in comparison with the first, except 
for the smaller angles of emission. Taking a=45°, 
Se a eS 5 
ba sin a Od ksina’ 
: : | ie. ae 
We accordingly have, in general, Z= ee (nearly). This is 
the same result that was obtained by the first investigation. 
‘t it can differ but little from the truth may be seen by com- 
> Panng V= v pr, With the greatest possible velocity that the re- 
‘Pulsive energy of the nucleus could impart to @ particle, if 
by any force, and acting through an infinitely great 
