Typical interaction between the press and a scientist?!

1. Typical interaction between the press and a scientist?!

2. Planetary Motion:The Kepler / Newton Problem:Section 8.7 • Lets look in detail at the orbits for the Inverse Square Law force: F(r) = -kr-2 ; U(r) = -kr-1 • The most important special case of Central Force motion! • Special case of interest: Motion of the planets (& other objects) about the sun. The force is then given by Newton’s Universal Law of Gravitation  k = GmM; m = planet mass, M = sun mass.

3. The relative coordinate used in the 2 body problem has been solved in terms of the reduced massμ: μ-1 m-1 + M-1 = (m-1)[1+ mM-1] (1) • Table 8-1 (p. 304) shows that for all planets m < < M  μ-1 m-1orμ m (2) • To get corrections to this, write: μ = (m)[1+ mM-1]-1 &expand in a Taylor’s series for small mM-1 = (m/M)  μ m[1 - (m/M) + (m/M)2 - ... ] (3) From Table 8-1, even for the largest planet, Jupiter, (m/M)  10-3, so (2) is ALWAYS a good approximation! Note:(2)  In the following, we have (k/μ)  GM

4. The general result for the Orbit θ(r) was: θ(r) = ∫(/r2)(2μ)-½[E - U(r) - {2(2μr2)}]-½dr • Put U(r) = -kr-1into this: θ(r) = ∫(/r2)(2μ)-½[E + kr-1 - {2(2μr2)}]-½dr • Integrate by first changing variables: Let u  r-1 θ(u) = (2μ)-½∫du[E + k u - {2(2μ)}u2]-½ • This integral is tabulated. The result is easily found to be: (after letting r = u-1): θ(r) = cos-1{[(α/r) -1]/ε}+ constant where α [2(μk)] &ε [ 1 + {2E2(μk2)}]½

5. Choosing the integration constant so that r = rminwhen θ = 0 gives theOrbit for the Inverse Square Law Force:cosθ = [(α/r) - 1]/ε (1) where α [2(μk)] & ε [ 1 + {2E2(μk2)}]½ • Rewrite (1) in standard form as: (α/r) = 1 + ε cosθ (2) • (2) is the equation of a CONIC SECTION (from analytic geometry!) • Orbit properties: ε Eccentricity 2α Latus Rectum

6. Pure Math Review: Conic Sections • The plane polar coordinate form for a Conic Section: (α/r) = 1 + ε cosθ (2) ε Eccentricity 2α Latus Rectum • They could (of course) also be written in the (perhaps more familiar) x & y coordinates as (there are several equivalent forms!): α = [x2 + y2]½ + ε x (2´) • Conic Sections are curves formed by the intersection of a plane and a cone. • A Conic Sectionis a curve formed by the loci of points (formed in a plane) where the ratio of the distance from a fixed point (the focus) to a fixed line (the directorix) is a constant.

7. Conic Section: (α/r) = 1 + ε cosθ • The type of curve depends on the eccentricityε. See the figure 

8. Conic Section Orbits • Conic Section:(α/r) = 1 + ε cosθ • The shape of curve (in our case, the orbit!) clearly depends on the eccentricity: ε [ 1 + {2E2(μk2)}]½ This clearly depends on the energy E & the angular momentum ! ε> 1  E > 0  A Hyperbola ε= 1  E = 0  A Parabola 0 < ε< 1  Vmin < E < 0  An Ellipse ε= 0  E = Vmin  A Circle ε< 0  E < Vmin  Not Allowed!

9. Conic Section:(α/r) = 1 + ε cosθ • Conic section orbit terminology: Recall the discussion of radial turning points & that we chose the integration constant for θ(r) so that r = rminat θ = 0 rmin The Pericenter rmax The Apocenter Any radial turning point  An Apside • For orbits about the sun: rmin The Perihelionrmax The Aphelion • For orbits about the earth: rmin The Perigee rmax The Apogee

10. Conic Section:(α/r) = 1 + ε cosθ ε= [ 1 + {2E2(μk2)}]½α= [2(μk)] • The Physicsthat goes with different orbit shapes: ε> 1  E > 0  Hyperbola This occurs for the repulsive Coulomb force. For example, 2 particles of like charge scattering off each other; (+,+) or (-,-). See our text, Sect. 9.10. 0 < ε< 1  Vmin < E < 0  Ellipse This occurs for the attractive Coulomb force ANDfor the Gravitational Force The orbits of all of the planets(& several other solar system objects!)are ellipseswith the Sun at a focus.

11. Data on the Orbits of the Planets& Other Solar System Objects Mass

12. Planetary Orbits • Simple algebra relates the geometry (the semi-major & semi-minor axes, a & b) of the elliptical planetary orbits to the physical properties of the orbit (energyE & angular momentum). Also usek = GmM. See figure. Results:a  α[1- ε2]-1 = (½)(k)|E|-1 b  α[1- ε2]-½ = (½)()(μ|E|)-½  Thesemi-major axis depends only on energy E. The semi-minor axis depends on both energy E & angular momentum 

13. Similarly, algebra gives the apsidal distancesrmin& rmax in terms of the eccentricity & the other parameters: rmin = a(1- ε) = α(1 + ε)-1 rmax = a(1+ ε) = α(1 - ε)-1 Recall: ε= [1 + {2E2(μk2)}]½α= [2(μk)]

14. Planetary Orbit Periods • For general central force, we had a constant areal velocity: (dA/dt) = ()(2μ)-1 = const • Use this to compute the period of the orbit!  dt = (2μ)()-1dA • The Periodτ =the time to sweep out the ellipse area: τ = ∫dt = (2μ)()-1∫dA = (2μ)()-1A

15. The elliptical orbit period is thus: τ = (2μ)()-1A (1) where A = the ellipse area. • From analytic geometry the area of an ellipse is: A  πab (2) where a & b are the semi-major & semi-minor axes. • From our discussion, a & bin terms of k, E &  are: a = (½)(k)|E|-1b = (½)()(μ|E|)-½(3) (1), (2) & (3) together τ = πk[(½)μ]½|E|-(3/2)

16. The elliptical orbit period is thus: τ = πk[(½)μ]½|E|-(3/2)(4) • An alternative, more common form for the period can be obtained using the fact that: b = (αa)½with α  [2(μk)] (5) along with (1) & (2), which, when combined give: τ = (2μ)()-1A = (2μ)()-1 πab (6) (5) & (6) combined give: τ2 = [(4π2μ)(k)–1]a3 (7) In words, (7) says that the square of the period is proportional to the cube of the semi-major axis of the elliptic orbitKepler’s Third Law

17. Kepler’s Laws • Kepler’s Third Law is:τ2 = [(4π2μ)(k)–1]a3 (7) The square of the period is proportional to the cube of the semi-major axis of the elliptic orbit. • Note: The reduced mass μ enters the proportionality constant! As derived empirically by Kepler, his 3rd Law states that (7) is true with same proportionality constant for all planets. This ignores the difference between reduced the mass μ & the planet mass m. We had: μ-1 m-1 + M-1 = (m-1)[1+ mM-1] so that μ = (m)[1+ mM-1]-1 m[1 - (m/M) + (m/M)2 - ... ] Note that k = GmM, so that if we letμ m (m << M) we have μ(k)–1 (GM)–1In his approximation, (7) becomes:τ2 = [(4π2)(GM)–1]a3(m << M) So Kepler was only approximately correct! He was only correct if we ignore the difference between μ & m.

18. Kepler’s First Law:The planets move in elliptic orbits about the Sun with the Sun at one focus. • Kepler proved this empirically. Newton proved this from the Universal Law of Gravitation & calculus. Now, we’ve done so also! • Kepler’s Second Law:The area per unit time swept out by a radius vector from the Sun to a planet is constant. (Constant areal velocity): (dA/dt) = ()(2μ)-1 = constant • Kepler proved this empirically. We’ve proven it in general for any central force. • Kepler’s Third Law:The square of a planet’s period is proportional to the cube of the semi-major axis of the planet’s orbit: τ2 = [(4π2μ)(k)–1]a3

19. Kepler’s Laws • Halley’s Comet, which passed around the sun early in 1986, moves in a highly elliptical orbit with eccentricity ε= 0.967 & period τ= 76 years. Calculate its minimum & maximum distances from the sun. • Use the formulas we just derived & find: rmin = 8.8  1010 m Inside Venus’s orbit & almost to Mercury’s orbit! rmax = 5.27  1012 m Outside of Neptune’s orbit & near to Pluto’s orbit!