PHYS 705: Classical MechanicsCentral Force Problems I
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Two-Body Central Force Problem
- Based his 3 laws on observational data from Tycho Brahe
- Formulate his famous 3 laws:
- Orbit of each planet is an ellipse with sun at one of its foci
- Equal areas swept out in equal time by an orbit
- The ratio is the same for all planets, where is the period and R
is the semi-major axis
-All these results were obtained through amazing sheer mathematical efforts.
Historical Background: Kepler’s Laws on celestial bodies (~1605)
2 3R
Before we get back to this important celestial application, we will address
the general problem…
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Kepler’s Notes
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Two-Body Central Force Problem
Set up of the general problem: 2 masses interact via forces 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:
1r2r
r
2 1 r r r' '2 1 r r r
R
'1r
'2r
CM
'1 1
'2 2
r R r
r R r
Instead of using , use (CM & relative position)1 2&r r &R r
Change to generalized coordinates: CM position R and relative position r,
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Two-Body Central Force Problem
From def. of CM:
1 1 2 2
' '1 1 2 2
i ii
M m m m
m m
M
R r r r
R r R r
R 1 2m m R ' '1 1 2 2m m r r 1r
2rr
R
'1r
'2r
CM' '
1 1 2 2 0m m r r
Combining it with , we have' '2 1 r r r
' ' '2 21 2 1
1 1
'2 21
1 1
' 21
1 2
1
m mm m
m mm m
mm m
r r r r
r r
r r ' ' 12 1
1 2
mm m
r r r rand
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Two-Body Central Force Problem
Now, form the Lagrangian:
(express in terms of only)
2 22 ' '1 21 22 2 2
CM aboutCM
T T T
m mM
R r r
We have a central force so that U(r) depends on the relative distance r only.
&R r
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2 22 21 2 2 1
2 21 2 1 2
2 21 2
1 2
12 2
12 2
m m m mMm m m m
m mMm m
R r
R r
' 21
1 2
mm m
r r
' 12
1 2
mm m
r r
Two-Body Central Force Problem
So,2 21 2
1 2
1 ( )2 2
m mML T U U rm m
R r
Note that R does not appear in L (cyclic) so that EL equation for R will only give:
M constR
CM is either stationary or moving uniformly.
Pick an inertial ref. frame (CM frame) in which CM is not
moving and we can ignore the 1st term in L.
The result is then:
2 1 2
1 2
1 ( ), where 2
m mL U rm m
r Same as a single particle
with mass moving in U(r).
0R
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Motion of and in CM frame
Reference Frames for Central Force Problems
Motion in the relative coordinate frame with (3 dofs) :
2m
(fixed),R r
Effective one body problem with reduced mass and relative position (dist from to ).2m1m
r
1m
r
' 12
1 2
mm m
r r
' 21
1 2
mm m
r r
1m
2m
2 'r
CM1 'r
1m
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Reference Frames for Central Force Problems
Motion in original space with 6 dofs (not all independent):
R
r
1m
2m
CM
2r
1r
1 2,r r
and circle each
other in space as their
CM move with constant
velocity along a fixed
direction.
2m1m
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Two-Body Central Force Problem
The reduced problem is a much simpler problem with 3 dofs instead of 6.
Also, we often have (e.g., sun-earth)1 2m m
the CM is so close to and
Then,1 2 1 2
21 2 1
m m m m mm m m
1m '2r r
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Two-Body Central Force Problem
Further reduction of the two-body central force problem:
So, we have,
0 0danddt
LN N
Since U(r) is central, i.e., F always directs toward CM along r,
constLor
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L will points in a constant direction fixed by initial condition.
Then, =const, and r & p must be to L always, i.e., motion
has to be planar.
If (trivial case), then the trajectory is a 1D motion through
the origin.
L r p
0L
/ /
(Alternatively, as we will see later, is cyclic !)
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 .
(Recall, in polar coord.) ˆˆr r v r θ
2 2 2 ( )2mL r r U r (we will simply call m= from now on)
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Lastly, using the fact that U(r)= U(r) depends only on the magnitude of the
distance, we can show that the Lagrangian is effectively one-dimensional.
To see that, start with L in :
L
,r
Two-Body Central Force Problem
Since U(r) is central, is cyclic (does not appear) in L.
Replacing with the constant in the Lagrangian, we then have an effectively
1D system in r only:
2Lp mr l const
2l mr
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22
2 ( )2 2m lr U r
mr
And, the generalized momentum with respect to is conserved,
l
2 2 2 ( )2 2m mL r r U r
Note: is ignorable but it is NOT a constant. r and change while l is fixed
in time. Motion is still in 2D in the CM frame*.
Two-Body Central Force Problem
[One can recognize this as the constant ang momentum: .]
Now, we calculate the EOM for both r and :
This gives:
0d L Ldt
(L is cyclic in )
2 2 2 ( )2mL r r U r
0L
2L mr
2mr l const
2( )l I mr
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(This is the equivalent statement about the conservation of L as before.)
Two-Body Central Force Problem
Note that area swept out by r with an
infinitesimal rotation d is
Connecting back to Kepler’s 2nd law…
212
dA r d
Thus, Eq (*) establishes Kepler’s 2nd Law!
r drd
And, 2 21 12 2
dA dr rdt dt
From EL equation, we have 2 0d mrdt
21 0 (*)2
d rdt
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Combining with the equation , we have,
Two-Body Central Force Problem
Now, we are back to the EOM for r:
r
gives0d L Ldt r r
2L dUmrr dr
L mr
r
2l mr
2 0dUmr mrdr
2
2
2
3
0l dUmr mrmr dr
l dUmrmr dr
2 2 2 ( )2mL r r U r
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U doesn’t depend on and does not depend on t explicitly
so that h (Jacobi Integral/energy function) is conserved.
Two-Body Central Force Problem
We have seen that the equation give one constant of motion l.
0Lt
jq
2 2 2 2
22 2 2 2
2
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1 1 12 2 2 2
jj j
Lh q Lq
r mr mr m r r U
lmr mr U mr U Emr
We can get another constant of motion E since:
( )jqr
so that h = E total energy is conserved.
check:
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Two-Body Central Force Problem
Here is another way to argue that E is conserved. Let start with the r EOM,
2 2
3 22dU l d lmr Udr mr dr mr
Multiplying on both sides,r2
222 2
d m d lmrr r drUdt dr mr dt
So,2
22 ( ) 0
2 2d m lr U rdt mr
E
E is conserved!
(reverse chain rule)
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22
22 2dd m lr U
t dtd mr
Two-Body Central Force Problem
Summary: We get 2 EOMs and 2 integrals of motion (l, E) for this problem.
2
3
dU lmrdr mr
2
22 ( )
2 2m lE r U r
mr
Note: The E equation effectively is the 1st integral of the equation. We can
rewrite it explicitly as,
2mr l
r
2
2
2 ( )2
lr E U rm mr
2 2 2 ( )2mL r r U r
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Two-Body Central Force Problem
Integrating once more time gives:
0
2
2
'
2 ( ')2 '
r
r
drtlE U r
m mr
With initial condition
r0 at t=0
Similar, we can integrate the equation and get,
02
0
( )( )
t
ldttmr t
With initial condition
0 at t=0
So, the problem is completely solved with t(r) and (t). The problem has been
reduced to a quadrature (doing the integrals).
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Graphical Analysis of Central Force Problem
Using the concept of an effective potential, one can get an useful
qualitative understanding of the problem without actually integrating!
Let consider the r equation:
The last two terms combined can be
considered as an effective force
This looks like a 1D problem: a single particle moving in 1 dimension under the
influence of an effective force,
2
3
dU lmrdr mr
2
3'( ) dU lf rdr mr
'( )f r
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There is now a “effective” extra force term that is not directly associated with
the two objects:
Graphical Analysis of Central Force Problem
No angular momentum, then, it is really a 1D problem. (uninteresting case)
Case 1 :
2
3
dU lmrdr mr
ˆ ˆˆwhere is the component of velocity v v r r θ r θ
0l
Case 2 :0l
2 2 22 2 2 22
3 3 3
mr m r r r vl m mmr mr mr r r
So, we recognize this as the centrifugal acc. term in a co-rotating ref. frame.
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We will consider three examples:
Graphical Analysis of Central Force Problem
For the case, we can combine the RHS into an effective potential term:
3
2
2
2 '( )2
dU l d dmrdr mr
l U rUdr dmr r
where
0l
2
3 4
2
.
. 3
. / 2
a U k r f k rb U a r f a rc U kr f kr
Note: zero point for U(r) is different in these three cases.
2
2'( ) ( )2
lU r U rmr
(gravitational, EM)
(Hooke’s law: HMO)
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Central Force Problem: Inverse Square Force
Example a: inverse-square force 2
2'( )2
k lU rr mr
For a fixed l, we can plot:
2
22lmr
kr
'( )U r
r
, 'U U
2
2
1. , ' 02. , ' 0
3. 1 dominates 1
4. 1 dominates 1 0
U U as rU U as r
r r as r
r r as r
Note the asymptotic:
2 2
'' 2a well in the m ddle
0i
U U k r as rU l mr as r
This gives,
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Central Force Problem: Inverse Square Force
Now, for a fixed value of total E > 0 & consider what kind of orbit is possible?
22 2
2
1 1'( )2 2 2
k lE mr U r mrr mr
Recall
Notice that
21 '( ) 02
mr E U r since the radial KE can’t be negative.
min mins.t. '( ), 0r E U r r
r can’t go beyond rmin, the turning point [where ]min'( )E U r
The orbit is to the line from the origin to the orbit:r
v
0 herer
2since 0r and r
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Central Force Problem: Inverse Square Force
'( )U r
r
E
min ( )r turning pt
The system becomes less
influence by U(r) and more
like free particles.
(far away)
' 0' ( )
as rUE U E const
r const
rE > 0 (unbounded orbit)
(close by)At turning pt, there is no
instantaneous radial motion.
All motion is angular.
21 '( )2
mr E U r
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Central Force Problem: Inverse Square Force
'U
r
E
(close by)
gets big gets big
All motion is angular here
U
minr r
2r
(far away)
and and
r
20r 0
All motion is radial here
2
2 2
' 2
' 2
E U mr
U U mr
(radial KE)
(angular KE)
E
'UU
2r
2 2r
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r const
Central Force Problem: Inverse Square Force
In the relative position space r, the orbit will look like:
0 and r const (far away)
0r
here is the point of closest approach (turning point)
origin
minr
is large
as if it were a free particle
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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.
- What happens to the orbit when E < 0? orbits will be bounded
- Now, if E = 0, for r large, the particles will be stopped with T = E = 0.
'U
rE
1r 2r
turning pts again 0r -Orbit is bounded between r1 & r2
-These distances are called
“apsides” or “apsidal distances”
-Similar to previous turning
points, orbits are tangent to
circles: r = r1 & r = r2
- is largest at inner radius (U’-U
is largest at r1)
U
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r
E
0r
Central Force Problem: Inverse Square Force
So, for bounded orbits (E < 0), the orbits in r space look like:
1r 2r
(for the inverse-square force)
We will talk more about bounded
orbits later.
In particular, cond for closed orbits(fig. 3.7 is wrong, curvature must be concave for
attractive forces)
If E is at the minimum (r= r0) , the bounded orbit will
have a constant radius r0 and it will be a circular orbit.
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Central Force Problem: Inverse Square Force
'U
r
turning pts (apsides) 0r
U
2 0 ( )E E unbounded
origin
minr
minr2E
1E
1r 2r
1 0 ( )E E bounded
1r 2r
0E
0 ( )E E circular
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Central Force Problem: Stronger Attractive U
Example b: Stronger Attractive Potential2
3 2'( )2
a lU rr mr
3 43U a r f a r
This gives,
'U
r
1E
1r
U
2
22lmr
2E
3E
2r
3
1 dominatesr
2
1 dominatesr
1. E=E1: orbit is unbounded
2. E=E2: orbit is bounded away
from r2 or bounded within r1
particles can collide
3. E=E3: orbit is bounded
within r3 and it can go thru
origin particles can collide
3r
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Central Force Problem: Stronger Attractive U
What is the condition for the particle to go thru the origin?
22
2
1 '( ) ( )2 2
lmr E U r E U rmr
From the total energy equation, we have,
Let, ( ) nU rr
For the origin to be accessible, near the origin. Thus, we must have,
2
2 02n
lEr mr
0r 2
22 2n
lErr m or
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Central Force Problem: Stronger Attractive U
As the orbit gets closer to the origin , LHS 0
Then, this is true if2 2
0, . .,2 2l li em m
For this inequality to be satisfied in the limit , RHS must be negative for
r small.
0r
22
2 2n
lErr m
0r
This can certainly be true for , attractive force
stronger than inverse-square type.
2n
If n = 2, then must hold at the limit 2
2
2lErm
0r
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'U
r
EU
2
22lmr
Central Force Problem: Hooke’s Law
Example c: Hooke’s Law/SHO
2 2
2'( )2 2
kr lU rmr
2 2U kr f kr
U’ is the green line
Motion is 1D on a line and
is simple harmonic
0 : 'l U U
0 :l Motion is on a 2D plane but
orbit will be bounded and
typically elliptic
(Homework problem)
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