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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)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

• The general result for the Orbit θ(r) was:

θ(r) = ∫(/r2)(2μ)-½[E - U(r) - {2(2μr2)}]-½dr• Put U(r) = -kr-1 into 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 + {2E2(μk2)}]½

• Choosing the integration constant so that r = rmin when θ = 0

gives the Orbit for the Inverse Square Law Force: cosθ = [(α/r) - 1]/ε (1)

where α [2(μk)] & ε [ 1 + {2E2(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

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 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 + {2E2(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!

• 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 = rmin at θ = 0

rmin The Pericenter

rmax The Apocenter

Any radial turning point An Apside

• For orbits about the sun:

rmin The Perihelion rmax The Aphelion

• For orbits about the earth:

rmin The Perigee rmax The Apogee

• Conic Section: (α/r) = 1 + ε cosθ

ε = [ 1 + {2E2(μk2)}]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. 9.10.

0 < ε < 1 Vmin < 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- ε2]-1 = (½)(k)|E|-1

b α[1- ε2]-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 rmin & 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 + {2E2(μk2)}]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)()-1dA• The Period τ = the time to sweep out the

ellipse area:τ = ∫dt = (2μ)()-1∫dA = (2μ)()-1A

• 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 & 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)()-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

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 let μ 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.

• 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

• 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!

Kepler’s Laws