PHY5200 F07

Chapter 8: Two-Body Central Force Problems

Reading

Taylor 8.5-8.6 (today) 8.7-8.8 and dark matter on (Friday).

Recall:

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

The Kepler Orbits

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

Derivation of Kepler's Third Law

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

We still haven't exposed all the a dependence. The square of the angular momentum depends on a by l² = ckμ = a(1-\epsilon²)kμ yielding

τ² = 4π²μa³/k

Finally, because Kepler's third law holds that the same constant of proportionality works for all planets, we must eliminate the dependence on the mass of the planet. This occurs in the factors μ and k, but k = GMμ ≈ GMsμ, where we leave the factor of μ since it will cancel out, and we approximate M = Ms + m by just Ms, finally yielding

τ² = (4π²/GMs)a³

This form clearly has all the a dependence exposed, and the constant is the same for all planets. This is Kepler's third law. The additional advantage of this derivation is that we know have the constant of proportionality.

Example: Period of Low-Orbit Earth Satellite

Find the period of a satellite in a circular orbit 300km above the surface of the earth.

The perod is given by τ = (2π/√GMe)a3/2 where Me is the mass of the earth, a is the radius of the circular orbit, a = Re + 300km, and Re is the radius of the earth. Recall that g = GMe/Re², and can be used to write the period as τ = (2π/√g)(Re + h)3/2/Re = (2π/√9.8 m/s²)(6.38×106m + 3.0×105m)3/2/6.38×106m = 5187s = 86.5min.

The well-known approximation is that low-orbit satellites orbit in about 90min.


© 2007 Robert Harr