Transcript Sect. 8.4
Ch. 8: Summary So Far
• We’re doing the “2 body”, conservative central force problem!
• 2 bodies (m1 & m2) with a central force directed along the line
between them F(r) = -U(r) = -(dU/dr)r
• Started with a 6d, 2 body problem. By transforming to CM &
relative coordinates (R,r) reduced it to 2, 3d 1 body
problems. One (the CM motion, R) is trivial & free particlelike. The 2nd is for a “particle” with mass μ (m1m2)(m1+m2)
= reduced mass interacting with F(r).
• Lagrangian for relative motion: L (½)μ|v|2 - U(r)
• Angular momentum conservation:
– First, constant DIRECTION, effectively reduced the 3d relative
coordinate problem (relative motion) from 3d to 2d, motion in a plane
( L)! Then, constant MAGNITUDE ( μr2θ = constant), reduced
the 2d problem of motion in a plane from problem with 2 degrees of
freedom to a problem with one generalized coordinate = r .
• The Lagrangian is: L = (½)μr2 + [2(2μr2)] - U(r)
• We also used conservation of angular momentum to prove
Kepler’s 2nd Law (areal velocity is constant): The radius
vector from the force center sweeps out equal areas in equal
times. Valid for any central force, not just gravitation!
• The total mechanical energy is conserved since the
central force is conservative: E = T + U = constant
E = (½)μ(r2 + r2θ2) + U(r) = const
– Using μr2θ = constant:
E = (½)μr2 + [2(2μr2)] + U(r) = const
Equations of Motion Section 8.4
E = (½)μr2 + [2(2μr2)] + U(r) = const (1)
• Conservation of energy allows us to get solutions to
the equations of motion in terms of both r(t) and r(θ)
or θ(r) The orbit of the particle(s)!
– Equation of motion to get r(t): One degree of freedom
Very similar to 1d problem discussed in earlier chapters!
• Solve for r = (dr/dt) : r = ({2/μ}[E - U(r)] - [2(μ2r2)])½
– Note: This gives r(r), the phase diagram for the relative
coordinate & velocity. We can qualitatively analyze the (r
part of the) motion using it, just as we did in 1d in Ch. 2.
• Solve for dt & integrate to get t(r). In principle, then
invert this to get r(t).
r = ({2/μ}[E - U(r)] - [2(μ2r2)])½
• Solve for dt & integrate to get t(r).
t(r) = ∫dr({2/μ}[E - U(r)] - [2(μ2r2)])-½
– Note the square root in the denominator!
• Often, we are much more interested in the
path in the r - θ plane:
r(θ) or θ(r) The Orbit.
• Note the chain rule: (dθ/dr) = (dθ/dt)(dt/dr) = (dθ/dt)(dr/dt)
Or: (dθ/dr) = (θ/r) Also, μr2θ = const θ = [/(μr2)]
(dθ/dr) = [/(μr2)] ({2/μ}[E - U(r)] - [2(μ2r2)])-½
Or: (dθ/dr) = (/r2)(2μ)-½[E - U(r) -{2(2μr2)}]-½
• Integrating this will (in principle) give the orbit θ(r) .
• Integrating (dθ/dr) formally gives
θ(r) = ∫(/r2)(2μ)-½[E - U(r) - {2(2μr2)}]-½dr
• Once the central force is specified, we know U(r) & can, in
principle, do the integral & get the orbit θ(r), or, (if
this can be inverted!) r(θ). This is quite remarkable!
Assuming only a central force law & nothing else:
We have reduced the original 6d problem of 2
particles to a 2d problem with only 1 variable
(r). The solution can (in principle) be obtained
simply by doing the above (1d) integral!
Orbits
• The general equation for the orbit (for any
Central Potential U(r)) is:
(2μ)½θ(r) = ∫ (r-2dr)[D(r)]-1 (1)
where D(r) [E - U(r) - {(2μ)-12r-2}]½
• In general, (1) must be evaluated numerically. This
is true even for most power law forces of the form:
F(r) = k rn ; U(r) = k rn+1
• For a few integer & fractional values of n, we can
express (1) in terms of certain elliptic integrals.
(2μ)½θ(r) = ∫ (r-2dr)[D(r)]-1 (1)
D(r) [E - U(r) - {(2μ)-12r-2}]½
• It can be proven (see the graduate text by Goldstein), that
only for n = 1, -2, and -3 can (1) be integrated to give
“simple” functions (like trigonometric functions).
n = 1, F(r) = kr. The isotropic, 3d simple
harmonic oscillator.
n = -2, F(r) = kr-2. Inverse square law force:
Gravitation, Coulomb. Treated in detail this chapter!
– Other cases: Homework Problems!
• Recap: We’ve solved the problem for r(t) & for the orbit
θ(r) or r(θ) using conservation laws exclusively:
– We combined conservation of angular momentum with
conservation of energy into a single result which gives the
orbit θ(r) in terms of a SINGLE INTEGRAL!
• Before doing this in detail for the r-2 force, it’s useful
to first take another (equivalent, of course!) approach
which will result in a differential equation for r(θ)
• Back to the Lagrangian for the relative coordinate
(before using conservation of angular momentum):
Or:
L (½) μ|v|2 - U(r)
L = (½) μ(r2 + r2θ2) - U(r)
L = (½) μ(r2 + r2θ2) - U(r)
• The Lagrange equation for r is:
(∂L/∂r) - (d/dt)(∂L/∂r) = 0
(∂L/∂r) = μrθ2 - (dU/dr)
(d/dt)(∂L/∂r) = μr;
- (dU/dr) = F(r)
Differential Equation of Motion for a
“particle”, mass μ subject to the Central Force F(r):
μ(r- rθ2) = F(r)
(2)
(2) is just Newton’s 2nd Law in plane polar coordinates!
• The Differential Equation of Motion:
μ(r- rθ2) = F(r)
(2)
• Its most convenient to solve (2) by first making a change of
variables: Let u (1/r) = r-1
(du/dr) = - (1/r2) = - r-2 ( -u2)
• We are more interested in the orbit θ(r) or r(θ) than in r(t).
Manipulation (with repeated use of the chain rule!):
(du/dθ) = (du/dr)(dr/dθ) = - r-2(dr/dθ) = - r-2(dr/dt)(dt/dθ)
= - r-2(dr/dt)(dθ/dt) = - r-2(r/θ)
(du/dθ) = - r-2(r/θ)
(3)
• Conservation of angular momentum gives: μr2θ = const
θ = [/(μr2)] (4)
• Combining (3) & (4): (du/dθ) = -(μ/)r
(5)
• Similar manipulation for:
(d2u/dθ2) = (d/dθ)[-(μ/)r] = -(μ/)(dt/dθ)(dr/dt) = -(μ/)(r/θ)
Again substituting θ = [/(μr2)]: (d2u/dθ2) = - (μ2/2)(r2 r) (6)
• Solving (6) for r (& using u2 = r-2) :
r = - (2/μ2) u2(d2u/dθ2)
(7)
μr2θ = const; θ = [/(μr2)]
(but u = r-1)
rθ2 = (2/μ2) u3
(8)
• The Equation of Motion is:
μ(r - rθ2) = F(r)
(2)
• Putting (7) & (8) into (2) gives (on simplifying!):
(d2u/dθ2) + u = - (μ/2)u-2F(u-1) (9)
• DIFFERENTIAL EQUATION for the ORBIT:
(d2u/dθ2) + u = - (μ/2)u-2F(u-1)
(9)
With u = r-1. Alternatively, we could write:
(d2[r-1]/dθ2) + r-1 = - (μ/2)r2F(r)
(9)
• Note! Because of the right hand side, (9) & (9) are nonlinear
differential equations in general! The EXCEPTION is when
F(r) r-2 (Inverse Square Law), where the right side = const!
• (9) or (9) could, in principle, be used to solve for the orbit
r(θ) or θ(r) given the force law F(r).
– The result, of course, will be the same as if the previous
integral version of θ(r) is evaluated. Can show that the
integral form for θ(r) is the solution to (9) & (9) .
• DIFFERENTIAL EQUATION for the ORBIT:
(d2u/dθ2) + u = - (μ/2)u-2F(u-1)
(9)
With u = r-1. Alternatively, we could write:
(d2[r-1]/dθ2) + r-1 = - (μ/2)r2F(r)
(9)
• Usually, rather than solve (9) or (9) for orbit, given
F(r), it’s usually more convenient to use the integral
formulation.
• However, where (9) or (9) is the most useful is for
THE INVERSE PROBLEM:
Given a known orbit r(θ) or θ(r),
determine the force law F(r) !!
Examples 8.1, 8.2, 8.3
• 8.1: Find the force law for a central force
field that allows a particle to move in a
logarithmic spiral orbit given by r = k eαθ,
where k and α are constants.
• 8.2: Find r(t) and θ(t) for the same case as
in Example 8.1
• 8.3: What is the total energy of the orbit of
the previous two examples?