sign in sign up
<<
Home , Central force, Constant of motion, Force, Johannes Kepler, Velocity

1

PHYS 705: Classical Central Problems I 2

Two-Body Problem

Historical Background: ’s Laws on celestial bodies (~1605)

- based his 3 laws on observational data from - Formulate his famous 3 laws: - of each is an with at one of its foci - Equal swept out in equal by an orbit - The ratio t 2 R 3 is the same for all , where t is the period and R is the semi-major axis -All these results were obtained through amazing sheer mathematical efforts.

Before we get back to this important celestial application, we will address the general problem… 3

Kepler’s Notes 4

Kepler’s Notes 5

Two-Body Central Force Problem

Set up of the general problem: 2 interact via directed along the line that connects them (central force): strong form of 3rd law

First Step: Central force problems can be reduced to an effective 1-body problem: Change to : CM R and relative position r,

r' r 2 ' r1 r2 CM R r1

' ' ' r r2  r 1 r1 Rr  1 r r2  r 1 ' r2 Rr  2

Instead of using r 1 & r 2 , use R & r (CM & relative position) 6

Two-Body Central Force Problem

From def. of CM: MR mmmi rrr i 1 1 2 2 i r ' ' r2 m1Rr  1  m 2 Rr  2  r ' ' 1 MR m1  m 2  R m1r 1  m 2 r 2 r' ' ' 2 r' m1r 1 m 2 r 2  0 CM 1 ' ' R Combining it with r  r 2  r 1 , we have

'm2 ' m 2 ' r1 r 2  rr  1  m1 m 1

m2  ' m 2 1 r1  r m1  m 1 m ' m2 ' ' 1 r r and rrr2 1  r 1 m m m1 m 2 1 2 7

Two-Body Central Force Problem

Now, form the Lagrangian: ' m2 r1  r m1 m 2 TTCM  T about CM ' m1 r2  r 2 2 M 2m1 ' m 2 ' m1 m 2 R  r1    r 2  2 2 2  M 1 mm2 mm 2  R 21 2  2 1  r 2 2 2 2 2  mm1 2   mm 1  2  

M 21 mm1 2  2 R    r (express in terms of R & r only) 2 2 m1 m 2 

We have a central force so that U(r) depends on the relative r only. 8

Two-Body Central Force Problem

So, M 21 mm1 2  2 LTUR   r  Ur( ) 2 2 m1 m 2 

Note that R does not appear in L (cyclic) so that P R is conserved, i.e., MR  const

CM is either stationary or moving uniformly. Pick an inertial ref. frame (CM frame) in which CM is not movingR  0 and we can ignore the 1st term in L.

The result is then:

1 2 mm1 2 Same as a single particle Lr Ur( ), where   2 m1 m 2 with  moving in U(r). 9

Reference Frames for Central Force Problems

Motion in the relative coordinate frame with  R (fixed), r  (3 dofs) :

 m2

' m2 r1  r m1 m 2 r ' r 2 ' m1 r2  r m1 m 2 m 1 CM r1 '

m1

Effective one body problem with of m 1 and m 2 in CM frame reduced mass  and relative

position r (dist from m 1 to m 2 ). 10

Reference Frames for Central Force Problems

Motion in original  r 1 , r 2  with 6 dofs (not all independent):

m1 and m 2 circle each other in space as their m2 r CM move with constant CM along a fixed R r direction. m1 2

r1 11

Two-Body Central Force Problem

The reduced problem is a much simpler problem with 3 dofs instead of 6.

Also, we often have m 1  m 2 (e.g., sun-)

Then,

mm1 2 mm 1 2     m2 mm1 2 m 1  the CM is also close to m and ' 1 r r2 12

Two-Body Central Force Problem

Further reduction of the two-body central force problem:

 Since U(r) is central, i.e., F always directs // toward CM along r,

So, we have, dL N0and  N 0 or L  const dt (Alternatively, as we will see later, q is cyclic !)  L will points in a constant direction fixed by initial condition. Then, L  rp  =const, and r & p must be  to L always, i.e., motion has to be planar.  If L  0 (trivial case), then the is a 1D motion through the origin. 13

Two-Body Central Force Problem

With all these considerations, it is natural to use spherical coordinates and to orient the polar axis with L.

Then, we can analyze the problem entirely in a 2D polar plane  L.

Lastly, using the fact that U(r)= U(r) depends only on the of the distance, we can show that the Lagrangian is effectively one-dimensional.

To see that, start with L in  r , q  : m L rr2  2q 2  Ur( ) (we will simply call m= from now on) 2   ˆ (Recall, vr   r rˆ  r q  θ in polar coord.) 14

Two-Body Central Force Problem

Since U(r) is central, q is cyclic (does not appear) in L.

And, the generalized with respect to q is conserved,

L p  mr2q  l const q q Replacing q  with the constant l in the Lagrangian, we then have an effectively 1D system in r only: q  l mr 2

2 m2 m 2 2 m2 l L r  rq  Ur( ) r 2  Ur( ) 2 2 2 2mr

Note: q is ignorable but it is NOT a constant. r and q change while l is fixed in time. Motion is still in 2D in the CM frame*. 15

Two-Body Central Force Problem

Now, we calculate the EOM for both r and q : q dL   L m 2 2 2     0 L rr q   Ur( ) dt q   q 2

L L  0  mr 2q q q (L is cyclic in q !)

This gives: mr2q  l  const

[One can recognize this as the constant ang momentum: l  I   ( mr 2 ) q  .] (This is the equivalent statement on the conservation of L as before.) 16

Two-Body Central Force Problem

Connecting back to Kepler’s 2nd law…

rdq Note that the swept out by r with an 1 r dq dq is dA r2 dq 2 dA1 dq 1 And, r2  r 2q dt2 dt 2

d 2 d 1 2  From EL equation, we have mr q   0 r q   0 (*) dt dt 2 

Thus, Eq (*) establishes Kepler’s 2nd Law! 17

Two-Body Central Force Problem

m Now, we are back to the EOM for r: L rr2  2q 2  Ur( ) 2  

r L L2 dU  mr mrq  r r dr

d L   L 2 dU     0 gives mr mrq   0 dt r   r dr

Combining with the q equation q   l mr 2 , we have,

2 l  dU mr mr 2   0 mr  dr l2 dU mr    mr3 dr 18

Two-Body Central Force Problem

We have seen that the q equation give one constant of motion l. We can get another constant of motion E since: L   0 so that h (Jacobi / function) is conserved. t

 U doesn’t depend on q  j and r ( q j ) does not depend on t explicitly so that h = E  total energy is conserved. L check: h q j  L j q j 1 rmr  qq mr2   mrr 2  2 q  2 U    2    1 1 1 l 2 mr2  mr 2q 2  U mr 2   UE 2 2 2  2mr 2 19

Two-Body Central Force Problem

Here is another way to argue that E is conserved. Let start with the r EOM, dU l2 d l 2  mr  3  U  2  dr mr dr2 mr  Multiplying r on both sides,

2 dm2  d l  dr mrr r   U  2  dt2  dr 2 mr  dt 2 dm2  d  l  r   U  2  (reverse chain rule) dt 2  dt  2 mr  dm l 2  So, 2 E is conserved! r 2  Ur( )   0 dt2 2 mr  E 20

m Two-Body Central Force Problem L rr2  2q 2  Ur( ) 2   Summary: We get 2 EOMs and 2 of motion (l, E) for this problem.

mr2q  l

2 2 dU l m2 l mr   3 Er   Ur( ) dr mr 2 2mr 2

st Note: The E equation effectively is the 1 integral of the r equation. We can rewrite it explicitly as,

2 l 2  r  EUr ( )  2  m2 mr  21

Two-Body Central Force Problem

Integrating once more time gives: r  dr ' t   With initial condition  2  2 l  r0 at t=0  E Ur( ')  2  r0 m2 mr '  Similar, we can integrate the q  equation and get, t  ldt With initial condition q(t )  2  q0  mr( t ) q0 at t=0 0 So, the problem is completely solved with t(r) and q (t). The problem has been reduced to a quadrature (doing the integrals). 22

Graphical Analysis of Central Force Problem

Using the concept of an effective , one can get a useful qualitative understanding of the problem without actually integrating!

Let consider the r equation:

dU l 2 The last two terms combined can be mr   3 dr mr considered as an effective force f'( r )

This looks like a 1D problem: a single particle moving in 1 under the influence of an effective force, dU l 2 f'( r )   dr mr 3 23

Graphical Analysis of Central Force Problem dU l 2 mr    dr mr 3 Case 1 l  0 :

No , then, it is really a 1D problem. (uninteresting case)

Case 2 l  0 :

There is now a “effective” extra force term that is not directly associated with the two objects:

2 2 2 2 2 2 l 2 mrq mrr q   r q  v2   m m q mr3 mr 3 mr 3 r r ˆˆ ˆ where vq is the θ component of velocity vrr rθ  q So, we recognize this as the centrifugal acc. term in a co-rotating ref. frame. 24

Graphical Analysis of Central Force Problem

For the l  0 case, we can combine the RHS into an effective potential term:

dU l2 dl2  d Ur'( ) mr 3 U  2   dr mr dr2mr  dr l 2 where Ur'() Ur ()  2mr 2 We will consider three examples:

aU.  kr  f  kr 2 (gravitational, EM) bU. ar3  f  3 ar 4 cUkr.2 / 2  f  kr (Hooke’s law: HMO)

Note: zero point for U(r) is different for c) as compared with a) & b). 25

Central Force Problem: Inverse Square Force k l 2 Example a: inverse-square force U'( r )   r2 mr 2

For a fixed l, we can plot: Note the asymptotic: U, U ' l 2 2mr 2 1.U , U ' 0 as r  2.U , U ' as r  0 U'( r ) 3. 1r dominates 1 rasr2  4. 1r2 dominates 1 rasr  0 r This gives,

k U'  U k r as r   2 2 r U' l 2 mr as r  0  a well in the mi ddle 26

Central Force Problem: Inverse Square Force

Now, for a fixed value of total E > 0 & consider what kind of orbit is possible? 1 1 k l 2 Recall E mr2  Ur'( )  mr 2  2 2  r 2 mr 2 Notice that

1 2  mr E  Ur'( )  0 since the radial KE can’t be negative. 2 

 rmins.t. EUr  '( min ), r  0 r  0 here r The orbit is  to the line from the origin to the orbit: v

 r can’t go beyond rmin, the turning point [where E  Ur '( min ) ]

2 since r 0 and r    27

Central Force Problem: Inverse Square Force E > 0 (unbounded orbit) U'( r ) r 1 mr2  E  Ur'( ) 2 

E as r  r U ' 0 E U'  E ( const )

rmin ( turning pt ) r  const

(close by) (far away) At turning pt, there is no The system becomes less instantaneous radial motion. influence by U(r) and more All motion is angular. like free particles. 28

Central Force Problem: Inverse Square Force

2 q U ' E U'  mr 2 (radial KE) U' U  mr 2q 2 2 (angular KE) E E

2 r  r

U '  r 2q 2 U

U

(close by) r rmin (far away) r  2 2 rq gets big  q gets big rq  0 andq  0 and r  const  All motion is angular here  All motion is radial here 29

Central Force Problem: Inverse Square Force

In the relative position space r, the orbit will look like:

(far away) q 0 and r  const as if it were a free particle  origin

rmin

here is the point of closest approach (turning point)

r  0 q is large 30

Central Force Problem: Inverse Square Force

- As we have seen, if E > 0, for r large, the particles act like free particles with T = E.

- Now, if E = 0, for r large, the particles will be stopped with T = E = 0.

- What happens to the orbit when E < 0?  orbits will be bounded

-Orbit is bounded between r & r U ' turning pts again r  0 1 2 -These are called “apsides” or “apsidal distances” -Similar to previous turning r1 r2 r points, are to E circles: r = r1 & r = r2 -q is largest at inner (U’-U U is largest at r1) 31

Central Force Problem: Inverse Square Force

So, for bounded orbits (E < 0), the orbits in r space look like:

(for the inverse-square force)

 We will talk more about bounded r r 2 1 orbits later.  In particular, cond for closed orbits

If E is at the minimum (r= r0) , the bounded orbit will

r0 have a constant radius r0 and it will be a circular orbit. r

E 32

Central Force Problem: Inverse Square Force

U ' turning pts (apsides) r  0 E E2  0( unbounded ) rmin

E2

r1 r2 r

E1 origin

E0

r E E  0( bounded ) min U 1

r2 r1

E E0 ( circular ) 33

Central Force Problem: Stronger Attractive U

Example b: Stronger Attractive Potential U ar3  f 3 ar 4 a l 2 This gives, U'( r )   r32 mr 2 l 2 2mr 2 1 dominates r 2 E1 1. E=E1: orbit is unbounded

2. E=E2: orbit is bounded away E2 r3 r1 r2 from r2 or bounded within r1  r particles can collide 3. E=E : orbit is bounded E3 3

1 within r3 and it can go thru  3 dominates U ' U r origin  particles can collide 34

Central Force Problem: Stronger Attractive U

What is the condition for the particle to go thru the origin?

From the total energy equation, we have, 1 l 2 mr2  EUr'()  EUr ()  2 2mr 2  Let, U( r )  r n

For the origin to be accessible, r   0 near the origin. Thus, we must have,  l 2  l 2 E   0 or Er 2   rn 2 mr 2 rn2 2 m 35

Central Force Problem: Stronger Attractive U

 l 2 Er 2   rn2 2 m As the orbit gets closer to the origin r  0 , LHS 0

For this inequality to be satisfied in the limit r  0 , RHS must be negative for r small. This can certainly be true for n  2 , attractive force stronger than inverse-square type. l 2 If n = 2, then Er 2    must hold at the limit r  0 2m Then, this is true if l2 l 2 0,ie . .,   2m 2 m 36

Central Force Problem: Hooke’s Law

Example c: Hooke’s Law/SHO Ukr2 2  f  kr (Homework problem) kr2 l 2 U'( r )   2 2mr 2 l0: UU '   U’ is the green line 

U ' Motion is 1D on a line and is simple harmonic U E l  0: Motion is on a 2D plane but l 2 orbit will be bounded and 2mr 2 r typically elliptic