It was noted earlier that : d r2 dt
:
::
= 2rr + r2 = r 2r + r
::
= 0:
:
(303) :
Thus as r 6= 0, then r2 is constant, and so the angular momentum L = mr2 is also constant. Hence :
=
L mr2
(304)
and the radial equation can therefore be written in the form ::
m r
L2 m2 r3
=
K : r2
(305)
Instead of solving for r and as functions of time we will consider the shape of the orbit, i.e. determine r as a function of . To do this we de…ne u = 1=r. We then have r
=
r
=
dr d 1 1 du = = = dt dt u u2 dt L L du 1 du = : 2 2 u d mr md
1 du d ; u2 d dt
(306) (307)
Di¤erentiating again we get ::
r
::
r
= =
dr d L du = dt dt md L2 2 d2 u u : m2 d 2
=
d d
L du md
d = dt
L d2 u md 2
L mr2
(308) (309)
In terms of u the radial equation (305)is m Multiplying through by
L2 2 d2 u u m2 d 2
u3
L2 m2
= Ku2 :
(310)
m= L2 u2 gives d2 u +u= d 2
mK : L2
(311)
In terms of the variable y = u + mK=L2 this is the equation of simple harmonic motion, d2 y +y =0 d 2 with solution y = A cos (
0 ).
(312)
Thus the general solution of eq(311) is u = A cos (
0)
mK 1 = : L2 r
(313)
This is of the same form as the general equation of a conic section, (see eq(317)) u=
1 1 = (1 + e cos ) : r h
We can choose 0 = 0 as this de…nes the orientation of the trajectory. If K is negative, i.e. attractive force, the trajectory is of the form shown in …g 48. If K is positive, i.e. repulsive force, the trajectory is of this form in …g 49
40
(314)