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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.

The relative coordinate used in the 2 body problem has been solved in terms of the reduced mass μ: 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 -1 or μ  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) ] (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

The general result for the Orbit θ(r) was: θ(r) = ∫( /r 2 )(2μ) -½ [E - U(r) - { 2  (2μr 2 )}] -½ dr Put U(r) = -kr -1 into this: θ(r) = ∫( /r 2 )(2μ) -½ [E + kr -1 - { 2  (2μr 2 )}] -½ dr Integrate by first changing variables: Let u  r -1 θ(u) = (2μ) -½ ∫du[E + k u - { 2  (2μ)}u 2 ] -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  (μk 2 )}] ½

Choosing the integration constant so that r = r min when θ = 0 gives the Orbit for the Inverse Square Law Force: cosθ = [(α/r) - 1]/ε (1) where α  [ 2  (μk)] & ε  [ 1 + {2E 2  (μk 2 )}] 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

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 = [x 2 + y 2 ] ½ + ε x (2´) Conic Sections are curves formed by the intersection of a plane and a cone. A Conic Section is 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.

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

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  (μk 2 )}] 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  V min < E < 0  An Ellipse ε = 0  E = V min  A Circle ε < 0  E < V min  Not Allowed!

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 = r min at θ = 0 r min  The Pericenter r max  The Apocenter Any radial turning point  An Apside For orbits about the sun: r min  The Perihelion r max  The Aphelion For orbits about the earth: r min  The Perigee r max  The Apogee

Conic Section: (α/r) = 1 + ε cosθ ε = [ 1 + {2E 2  (μk 2 )}] ½ α = [ 2  (μk)] The Physics that 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 < ε < 1  V min < E < 0  Ellipse This occurs for the attractive Coulomb force AND for the Gravitational Force The orbits of all of the planets (& several other solar system objects!) are ellipses with the Sun at a focus.

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

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 (energy E & angular momentum ). Also use k = GmM. See figure. Results: a  α[1- ε 2 ] -1 = (½)(k)|E| -1 b  α[1- ε 2 ] -½ = (½)( )(μ|E|) -½  The semi-major axis depends only on energy E. The semi- minor axis depends on both energy E & angular momentum Planetary Orbits

Similarly, algebra gives the apsidal distances r min & r max in terms of the eccentricity & the other parameters: r min = a(1- ε) = α (1 + ε) -1 r max = a(1+ ε) = α (1 - ε) -1 Recall: ε = [1 + {2E 2  (μk 2 )}] ½ α = [ 2  (μk)]

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)( ) -1 dA The Period τ = the time to sweep out the ellipse area: τ = ∫dt = (2μ)( ) -1 ∫dA = (2μ)( ) -1 A

The elliptical orbit period is thus: τ = (2μ)( ) -1 A (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 & b in terms of k, E & are: a = (½)(k)|E| -1 b = (½)( )(μ|E|) -½ (3) (1), (2) & (3) together  τ = πk[(½)μ] ½ |E| -(3/2)

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) ½ with α  [ 2  (μk)] (5) along with (1) & (2), which, when combined give: τ = (2μ)( ) -1 A = (2μ)( ) -1 πab (6) (5) & (6) combined give: τ 2 = [(4π 2 μ)(k) –1 ]a 3 (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

Kepler’s Laws Kepler’s Third Law is: τ 2 = [(4π 2 μ)(k) –1 ]a 3 (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 3 rd 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) ] Note that k = GmM, so that if we let μ  m (m << M) we have μ(k) –1  (GM) –1  In his approximation, (7) becomes: τ 2 = [(4π 2 )(GM) –1 ]a 3 (m << M) So Kepler was only approximately correct! He was only correct if we ignore the difference between μ & m.

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:τ 2 = [(4π 2 μ)(k) –1 ]a 3

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