For a central force the position and the force are anti-parallel, so r F = 0.

For a central force the position and the force are anti-parallel, so rF=0. r F N is torque Newton II, angular So, angular momentum, L, is constant

Since the Angular Momentum, L, is constant: • Its magnitude is fixed • Its direction is fixed. L r p • By the definition of the cross product, both the position and the momentum are perpendicular to the angular momentum. • The angular momentum is constant. • Therefore, the position and momentum are restricted to a plane. The motion is restricted to a plane.

Show that the conservation of angular momentum implies that equal areas are swept out in equal times: r+dr r dr The green shaded area is dA The parallelogram formed by r and dr is twice the area of dA. But, by definition, the magnitude of rdr is the area of the parallelogram. r dr dr sin  (height)

Proof of Kepler’s Second Law The area swept out in time dt But dr is just v dt Divide by dt, and change v to (1/m)p The angular momentum is constant, so the rate at which area is swept out is also constant.

Central Forces in Polar Coordinates Let’s use the Lagrangian: Don’t confuse the Lagrangian, “L”, with the angular momentum, “L”. Motion is restricted to a plane, and the potential is that for a central force.

The Radial Equation

The Angular Equation But the term in parenthesis is just the angular momentum of a point particle, so conservation of angular momentum falls out of the constraints on the Lagrangian!

From the Lagrangian, we found Newton’s Laws in Polar Coordinates: Radial eqn.: Theta eqn.:

Obtain solutions for SHAPE, don’t need to solve for t Put it in terms of u and theta: Solve for r and its derivatives in terms of u and theta. Let

Define the Differential Equation for the Orbit Shape To reiterate: Replace r’s and thetas Now we have a differential eqn. of u and theta

Now, we plug our force law into the differential equation we have derived. The force is: This equation is similar to a simple harmonic oscillator with a constant force offset, except the independent variable is theta not time.

Manipulate the Shape Equation a Bit More: Choose theta nought so that theta equal to 0 yields the distance of closest approach “Look Ma, an ellipse!” “But Johnny, It doesn’t look like an ellipse???” “Oh Ma… don’t you know nothin’?”

We want to show that this equation is an ellipse

Review of Some Basic Ellipse Properties The ellipse below has the equation: Semiminor Axis b Semimajor axis: a

To solve for the eccentricity of an ellipse, use the defining relationship for the ellipse and solve equations for two special cases: (1) Pythagorean Triangle When touching the right edge (2) b r r Plug (2) into (1) a a

Put our Equation for the Ellipse in the Form of Our Shape Equation r r  a a Defining equation for an ellipse r'

Put our Equation for the Ellipse in the Form of Our Shape Equation From the last slide From the defining relationship

Define the Latus Rectum,  r The latus rectum is the distance to a focus from a point on the ellipse perpendicular to the major axis  2a Defining relationship for the ellipse Solve it for r Define a Pythagorean relationship Solve the resulting relationship for alpha

Finally, Show that our Central Force Yielded an Elliptical Orbit! So, our general equation for an ellipse is And the solution to our shape equation was They have the same form! We’ve derived Kepler’s Second Law. The trajectories of planets for a central force are described by ellipses.

Relations Between Orbit Parameters

Kepler’s Third Law Integration of the area dA/dt over one period gives A (1) Kepler’s Second Law Use Kepler II to relate integral and l. (2) Set (1) and (2) equal

Kepler’s Third Law Derivation Continued

Look How Well The Solar System Fits Kepler’s Third Law! Earth presumably fits this rule… Using Earth years for time, and Astronomical Units (1 AU = 1 Earth-Sun distance) for distance renders the constants equal to 1.

Universality of Gravitation: Dark Matter

Keplerian Motion vs. Constant Density Sphere Motion Keplerian Motion: All the mass of the galaxy would be assumed to be focused very close to the origin. Velocity curve, Assuming a Constant Density Sphere All the Way Out Only the mass interior to the star in question acts. Density, assuming a uniform density sphere. v r

Are Central Forces Conservative?

Central Forces Are Conservative!

Energy Equation of an Orbit in a Central Field

Orbital Energies in an inverse Square Law

Integrating Orbital Energy Equation

Integrating Orbital Energy Equation Continued

Total Energy of an Orbit To fit earlier form for an ellipse Earlier Relations Total energy of the orbit.

Ellipses, Parabolas, Hyperbolas E < 0 Ellipse or Circle e<1 E=0 Parabola e=1 E>0 Hyperbolic e>1

Maximum Velocity for An elliptical Orbit

Limits of Radial Motion E U is the effective potential

Limits Continued

Minimum Orbital Energy Radical ought to remain real for elliptical orbits. Value of E for which the radical is 0 Extremes of motion merge to one value.

Scattering and Bound States are All There Are! E < 0 E > 0

Energy Equation for a Central Force Again For attractive inverse square For repulsive inverse square

The Scattering Calculation Proceeds Exactly Like the Bound State Calculation But, we’re going to let k go to –Qq after we integrate

Integrate Both the Bound State And the Scattering Problem

Verify Integral in 6.10.4

Solve Both Bound State and Scattering Problem for u

Solve Both Problems for r Solution to the Scattering Problem Solution to the Bound State Problem

Scattering in an Inverse Square Field attractive Repulsive K goes to -Qq Solving the energy equation for r

A Drawing of the Scattering Problem Trajectory s o o b rmin Impact parameter b A small range of impact parameter scattering center

Asymptotes Trajectory s o o b rmin Impact parameter b A small range of impact parameter scattering center

Angles o 1 /2-o

For a central force the position and the force are anti-parallel, so r F = 0.