Kepler explained the regularity of planet motions in terms of elliptical orbits. He discovered a relationship that relates the period (or orbital time) of a comet to the semi-major axis (or half the long axis of the orbit). If the period P is in years and the semi-major axis a is in astronomical units, we have:
(PYears)2 = (aA.U.)3
Obviously this is correct for Earth, since the equation would give 13 = 12, which is true. (As an exercise you can verify it for the other planets as well.)
Kepler made his discovery of the pattern of orbits long before Halley recognized that comets were cyclic phenomena. But the relationship above is true of all motion around the Sun, including the orbits of comets. For example, Chiron (originally classed as an asteroid but now known to be a comet) moves on an elliptical orbit between Saturn and Uranus, with a semi-major axis of 13.7 A.U. The cube of this number is 2571. Thus, if the period squared is 2571, the period must be & sqrt;2571 = 51 years. That is how long it takes Chiron to go around the Sun.
You can use Kepler's third law to predict distances from periods. For example, we know from historical records that Halley's Comet returns about every 75 years. What is its average distance from the Sun? The square of the period is 5625. Therefore, the cube of the semi major axis in A.U. is also 5776. Taking the cube root on a calculator, we find that the semi-major axis is about 18 A.U., or nearly as far as Uranus.
What is meant by the average distance of an orbit? Planets do not have extreme elliptical orbits; they look like slightly squashed circles. In this case, the foci of the ellipse are close together, the eccentricity is close to zero, and the semi-major axis is not much different from the radius of a circular orbit. Comet orbits are entirely different. Their ellipses are extremely elongated and the eccentricity is close to one. The foci of the ellipse are very far apart. In fact, the distance of closest approach to the Sun may be much smaller than the semi-major axis. It is therefore a good approximation to say that the maximum distance from the Sun is twice the semi-major axis. At its farthest point, Halley's Comet reaches about 36 A.U. from the Sun, between the orbits of Neptune and Pluto.
Comets in the Kuiper Belt and Oort Cloud are at distances of 100 and 100,000 A.U. from the Sun. What would be the period of such comets? Remember that the maximum distance is twice the semi-major axis. If you insert the semi-major axis values of 50 and 50,000 in the equation for Kepler's third law, you deduce periods of about 350 years and 11 million years, respectively!
The enormous period and elongation of orbits in the Oort cloud has an important implication. Kepler's second law states that an orbiting body will sweep out equal areas in equal amounts of time. With a highly elliptical orbit, this means a comet must travel much slower when it is far from the Sun than when it is close to the Sun. How much slower? Newton's gravity law can be used to show that the velocity in a circular orbit is:
v = &sqrt;(GM/r)
In this equation, G is the gravitational constant and M is the mass of the Sun. We can approximate the answer just by scaling from this relation where v ~ 1/&sqrt;r. Taking ratios so that G and M cancel out, we can say that:
vEarth / vcomet = (rcomet / rEarth)1/2
At 1 A.U., the Earth has an orbital velocity of about 30 kilometers per second. So at 50,000 A.U., a comet will have an orbital velocity of 30 (30 / 50,000)1/2 = 0.73 kilometers per second. The result is that a comet spends the vast majority of its orbit crawling along at an enormous distance from the Sun. A comet with a period of a million years will only spend one or two of those years within the planet orbits!