Central-Force Motion Chapter 8 Prof. Claude A Pruneau Physics and Astronomy Department Wayne State University

8.1 Introduction Consider motion of two particles affected by a force connecting the center of the two bodies. One of few problems that can be solved completely. Historically important e.g motion of planets, alpha-particle scattering on nuclei

8.2 Reduced Mass Description of a two-particle system Discussion restricted to frictionless (conservative) systems.

8.2 Reduced Mass (cont’d) Arbitrary Frame CM Frame Assume a force is acting between the two particles along a line joining them. Particle positions described in some arbitrary reference frame as positions r1 and r2, or in terms of the CM frame position, R, and the relative position vector r = r1 - r2 . m1 m1 CM CM m2 m2 Arbitrary Frame CM Frame

Lagrangian for a two-body system Assume the potential energy is only a function of the distance between the particles, U=U(r) Lagrangian may be written: Translational motion of the system uninteresting; Use R=0.

Two-body CM Coordinates We have Solving for r1 and r2 : Substitute in the Lagrangian r1 and r2 : 2 to 1 reduction

8.3 Conservation Theorems Particle of mass in a central force field described by the potential function U(r). Symmetry implies conservation of angular momentum. Radius vector and momentum lie in a plane normal to the angular momentum vector L. The problem reduced to 2 dimensions: i.e. along “r” and “q”.

Lagrangian cyclic in q implies: Angular momentum, pq, conjugate to q, is a conserved quantity. First integral of motion

The system’s symmetry permits the integration of one equation of motion. pq is the first integral of the motion. Denote it Note l can be negative or positive

Interpretation of l as Areal velocity The radius vector sweeps out an area dA in a time interval dt. The areal velocity is thus

Kepler’s 2nd law of planetary motion No particular assumption made about the form of U(r) implies: This result is NOT limited to an inverse-square law force but is valid for all central forces.

Since the motion of the CM is not interesting, only one degree of freedom remains to be considered. Linear momentum conservation adds nothing new here… Energy conservation provides the only remaining equation of motion. Total Energy

8.4 Equations of Motion Assume U(r) is specified. Solve for dr/dt: Solving for dt, and integrate to get a solution t = t(r). Invert it to get r = r(t) Alternatively, obtain q= q(r), starting with

Inversion of the result (if possible) yields the standard form (general) solution r = r(t). Because l is constant, dq/dt is a monotonic function of time. The above integral is in practice possible only for a limited number of cases…

Remarks with F(r) = rn, solutions may be expressed in terms of elliptic integrals for certain integers and fractional values of “n”. Solution may be expressed in terms of circular functions for n=1, -2, and –3. Case n = 1 is the harmonic oscillator. Case n = -2 is the inverse square law.

Solution using Lagrange equations Lagrange equation for “r” Use variable change Remember

Compute…

Solving Substitute back into Which is useful if one wishes to find the force law that produces a particular orbit r=r(q).

Example 8.1 – Log-spiral Find the force law for a central-force field that allows a particle to move in a logarithmic spiral orbit given by, , where k and a are constants. Now use To find: Force is Attractive and Inverse cube!

Example 8.2 – r(t), q(t) Answer: Determine the functions r(t) and q(t) for the problem in Ex 8.1. Solution: Start with: Rearrange, integrate: Answer:

Similarly for r(t), remember And write Answer (2): Where l and C are determined by the initial conditions

Example 8.3 – Total Energy What is the total energy of the orbit of the previous two examples? Solution: Need U…

Given the reference

8.5 Orbits in a central field Radial velocity of a particle in central field Vanishes at the roots of the radical

Vanishing of dr/dt implies turning points Two roots in general: rmin and rmax. Motion confined to an annular region between rmin and rmax. Certain combinations of E and l may lead to a single root: one then has a circular motion, and dr/dt = 0 at all times.

Periodic motion in U(r) implies the orbit is closed; I. e Periodic motion in U(r) implies the orbit is closed; I.e. loops on itself after a certain number of excursions about the center of force. The change in  while going from rmin to rmax is a function of the potential and need not be 180o. It can be calculated! Because the motion is symmetric in time:

Path closed only if Dq is a rational fraction of 2p. Dq= 2p(a/b) where a and b are integers. In this case, after b periods the particle will have completed a revolutions and returned to its original position. For a closed noncircular path exists only for n = -2 or +1.

8.6 Centrifugal Energy and Effective Potential In dr/dt, dq/dt, …, we have Where each term has the dimension of energy. Remember that Write

Interpret as a “potential energy” The associated force is: Traditionally called a centrifugal force. Although it is, STRICTLY SPEAKING, NOT A FORCE but rather a pseudo-force. We continue to use the term nonetheless…

The term can then be interpreted as the centrifugal potential energy, and included with U(r) to define an effective potential energy. V(r) is a fictitious potential that combines the real or actual potential U(r) with the energy term associated with the angular motion about the center of force. For an inverse-square law central-force motion, one gets:

E1 r E2 E3 Energy Turning point(s) (apsidal distances) r3 r4 r1 r2 unbound E1 r3 r4 r r1 r2 E2 bound E3

Values of E less than do not result in physically real motion; given velocity is imaginary. Techniques illustrated here are used in modern atomic, molecular and nuclear physics (but in the context of QM).

8.7 Planetary Motion – Kepler’s Problem Consider the specific case of an inverse-square force law. Integral soluble with variable substitution u=1/r. Define the origin of q so r is a minimum.

Define constants Then one can re-write: To get the equation of a conic section with one focus at the origin

The quantity, e, is called eccentricity, and 2a is termed the latus rectum of the orbit. Conic sections are formed by the intersection of a plane and a cone. More specifically … by the loci of points (formed by a plane) where the ratio of the distance from a fixed point (the focus) to a fixed line (called the directrix) is a constant.

Hyperbola, e>1 Parabola, e=1 Ellipse, 0<e<1 Directrix For parabola Circle, e=0

q=0 corresponds to a pericenter, i. e q=0 corresponds to a pericenter, i.e. rmin, whereas rmax corresponds to the apocenter. The general term for turning points is apsides. Planetary Motion: Major axis Minor axis

a b ae P P a

Kepler’s Third Law Period of elliptic motion: The area of an ellipse is: The period is then…. Noting One also finds: Kepler’s Third Law

Given the gravitational force: The square of the period: Where the last approx is realized for m1 << m2. Kepler’s statement is correct only if the mass m1 of a planet can be neglected with respect to the mass m2 of the sun. Correction needed for Jupiter given that it is 1/1000 of the mass of the Sun.

Kepler’s Laws Planets move in elliptical orbits about the sun with the sun at one focus. The area per unit time swept out by a radius vector from the sun to a planet is constant. The square of a planet’s period is proportional to the cube of the major axis of the planet’s orbit.

Example 8.4 Halley’s comet, which passed around the sun early in 1986, moves in a highly elliptical orbit with an eccentricity of 0.967 and a period of 76 years. Calculate its minimum and maximum distances from the sun. Solution: We thus find:

8.8 Orbital Dynamics