76 KINETICS OF A PARTICLE. [144. 



144. Kepler's equation (30) gives the time as a function of <j> ; by 

 means of (28), it establishes the relation between t and r\ by means 

 of (29), it connects / with 0. It is, however, a transcendental equation 

 and cannot be solved for < in a finite form. 



For orbits with a small eccentricity e, an approximate solution can 

 be obtained by writing the equation in the form 



< = nt-\- e sin<, 

 and substituting under the sine for < its approximate value nt : 



<f> = nt + e sin nt. (3 1 ) 



This amounts to neglecting terms containing powers of e above the 

 first power. 



Substituting this value of < in (28), we have with the same approxi- 

 mation 



r= a(i ecosnt). (32) 



To find in terms of /, we have, from the equation of the ellipse, 

 r a ( i e 2 ) ( i -f- e cos 6) ~ l = a ( i e cos 0) , neglecting again terms in 

 & ; hence, r z = 2 (i 2e cos 0) . Substituting this value in the equation 

 of areas, r z dO = cdt V/xtf (i e*)dt, we find 



(i - 2e cos 0)dO = Y^<# = ndt; 

 whence, by integration, since & o for / = o, 



2e sin = nt, 

 or finally, 6 = nt + 2e sin nt. . (33) 



Thus we have in (31), (32), (33) approximate expressions for <f>, 

 r, and 6 directly in terms of the time. The quantity 2<? sin /, by which 

 the true anomaly exceeds the mean anomaly nt, is called the equation 

 of the centre. 



145, Exercises. 



(1) A particle describes an ellipse under the action of a central 

 force. Determine the law of force by means of formula (n), Art. 

 116: (a) when the centre of force is at the centre of the ellipse; 

 (b) when it is at a focus. 



(2) A particle is attracted by a fixed centre according to Newton's 

 law. What must be the initial velocity if the orbit is to be circular ? 



