This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chapter 5 The Kepler Problem This problem is deﬁned by the equation of motion dp oar
a? : "71:3— 5 where p is the linear momentum of a particle of mass m, 7' is its position vector,
and a is a positive constant. Eq. (5.1) represents a set of three second order
differential equations, whose solutions give the trajectory r(t) of the particle.
We will not solve the equation explicitly in this chapter. Instead, the geometrical
nature of the trajectory Will be determined by the constants of motion of this
problem. The force given by (5.1) is a special example of a central force (one which is
along the radial direction). Consequently, the torque (with respect to the origin
of the coordinate system) vanishes, and the angular momentum is conserved: dL_ :1? _ 0, (5.2) that is, L is a constant of motion. This fact can also be demonstrated explicitly
as follows. From L = 7' X p we have dL_clr d _p a: _
E_Exp+rxa_mxp+( ——)r><r—. (5.3) Deﬁne the RungeLenz vector JVI as follows: (5.4) 26 CHAPTER 5. THE KEPLER PROBLEM We will show that M is also a constant of motion. Indeed, dM 1dp p dL d r
_:___ L __ __ _ _
dt mdt X +m X dt adt (r)
a cl 7'
WWW—dab)
oz r'r 1dr
=—7;3r><(r><p)—a *T—2+;E (5.5)
___0‘_ . _ 2 Di _O‘_p
— W301?" p) TP)+T2r W
——93 5i: 7' +aﬂ——a—Trr+ furl—O
— r3 clt rg— r3 ar2_ ’ where the fourth equality follows from the “bac  cab” identity (1.51) and the
next to last equality from (2.39), which implies that v  r 2 rr.
The constants of motion L and M satisfy the following conditions: LrzLMzO, (5.6) which imply that L is perpendicular to the plane of the orbit and M lies on the
plane of the orbit. Eq.(5.6) can be demonstrated easily: Lr:r(rxp)=p'(rxr):O, (5.7)
L 1
LMzL(px —9£>=—p(LXL)—gr~L=O. (5.8)
m r m r
To find the equation of the orbit we calculate M r:
 L L2
Mrzr—(P—Q—arzﬂ—ar. (5.9)
m m Choosing the :c—axis to be along the direction of M , we can write the above
equation in terms of polar coordinates (r, 0) as (Fig. 5.1) L2
Mr cosé’ : — — or, (5.10)
m
or
L2
r 2 ———~—]\—/—[— (5.11)
am (1 + 3 cos 0) This is the equation of a conic section, with the eccentricity equal to M / a
(Fig. 5.2). Fig. 5.1, Fig. 5.2 ...
View Full
Document
 Spring '08
 LAM
 mechanics

Click to edit the document details