PHY5200 F06

Chapter 8: Two-Body Central Force Problems

Reading

Taylor 8.6-8.8 (today) and dark matter on (Wednesday).

Recall:

r(φ) = (l²/μk) / (1 + εcosφ) = C / (1 + εcosφ)

The Bounded Orbits

When ε<1, the orbits are ellipses, that is, the expression for the orbit can be cast in the form

(x+d)²/a² + y²/b² = 1

This will be left as an exercise for you. There is a displacement, d, in the x direction because the ellipse is centered at (x,y) = (-d, 0) rather than at (0,0). But the origin is a focus of the ellipse.

The separation, r, oscillates in synch with cosφ, with a minimum value, rmin = c/(1 + ε) when φ = 0, and a maximum value rmax = c/(1 - ε) when φ = π. For orbits of planets, comets, or other bodies about the sun, the location of minimum separation is called the perihelion (peri-, Greek for around or near and -helion, Greek for sun) and the location of maximum separation is called the aphelion (ap- Greek for away). For orbits around the earth, these locations are called perigee and apigee (gee Greek for earth). If ε=0, then r = c is constant and the orbit is circular.

The Orbital period; Kepler's Third Law

Kepler's third law states that the square of the period is proportional to the cube of the semi-major axis, τ² ∝ a³. By comparing the period of planetary orbits measured in earth years with the semimajor axes of the orbits measured in astronomical units (AU), the size of the earth's semimajor axis, we can see that Kepler's third law works very well (to at least a part in a thousand).

Period Semimajor axis Eccentricity
Planet τ (yr) τ² (yr²) a (AU) a³ (AU³) ε
Mercury 0.241 0.0581 0.387 0.0580 0.206
Venus 0.615 0.378 0.723 0.378 0.007
Earth 1.000 1.000 1.000 1.000 0.017
Mars 1.881 3.538 1.524 3.540 0.093
Jupiter 11.86 140.7 5.203 140.8 0.048
Saturn 29.46 867.9 9.539 868.0 0.056
Uranus 84.01 7058. 19.18 7056 0.047
Neptune 164.8 27160 30.06 27160 0.009

Let's derive Kepler's third law from Newton's law. Start with the result we obtained for Kepler's second law, dA/dt = l/2μ. We can integrate this over one period to find

∫(dA/dt)dt = ∫ dA = l/2μ ∫ dt = τl/2μ

The total area of an ellipse with semimajor axis a and semiminor axis b is A = πab. So the integral over the area becomes

πab = τl/2μ

Now, square both sides, solve for τ², and express all lengths in terms of the semi-major axis, a. Use the relations b² = a²(1-ε²)

τ² = 4π²μ²a4(1-&epsilon²)/l²

Relation Between Energy and Eccentricity

The Unbounded Kepler Orbits

Summary of Kepler Orbits

Changes of Orbit

A Tangential Thrust at Perigee

Discovery of Neptune: 1845--1846

The orbit of known planets are perturbed by other planets, especially Jupiter and Saturn. These perturbations were known, and approximations were used to calculate them. The calculations were in good agreement with the observed orbits for all the planets except Uranus, the outermost of the known planets. A number of possible explanations were considered, for instance:

Both options were seriously considered. Two astronomers, Adams (in 1845) and Leverrier (in 1846) used Newtonian mechanics to predict the position of a planet that would account for the deviations of the orbit of Uranus. Both results agreed, and in 1846, astronomers at the Berlin observatory discovered the planet and named it Neptune. This discovery was regarded as a triumph of Newtonian mechanics and the law of Universal Gravitation.

Dark Matter


© Robert Harr 2006