Taylor 8.5-8.6 (today) 8.7-8.8 and dark matter on (Friday).
When ε<1, the orbits are ellipses, that is, the expression for the orbit can be cast in the form
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.
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
The total area of an ellipse with semimajor axis a and semiminor axis b is A = πab. So the integral over the area becomes
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)
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
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
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.
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.