2. Kepler’s 1 st Law of Planetary Motion. 1. The orbits of the planets are ellipses, with the Sun at one focus of the ellipse. The Sun is not at the center of the ellipse, but is instead at one focus (generally there is nothing at the other focus of the ellipse). The planet then follows the ellipse in its orbit, which means that the Earth-Sun distance is constantly changing as the planet goes around its orbit.

3. The line joining the Sun and planet sweeps out equal areas in equal times, so the planet moves faster when it is nearer the Sun. Thus, a planet executes elliptical motion with constantly changing angular speed as it moves about its orbit. The point of nearest approach of the planet to the Sun is termed perihelion ; the point of greatest separation is termed aphelion. A planet moves fastest when it is near perihelion and slowest when it is near aphelion. 2. The line joining the planet to the Sun sweeps out equal areas in equal times as the planet travels around the ellipse.

4. Kepler's Third Law In this equation “P” represents the period of revolution for a planet and “R” represents the length of its semimajor axis. Kepler's Third Law implies that the period for a planet to orbit the Sun (P) increases rapidly with the radius (R) of its orbit. So, we can now use Kepler’s 3 rd Law to discover that Mercury, the innermost planet, takes only 88 days to orbit the Sun but the outermost planet (Pluto) takes 248 years to do the same. 3. The ratio of the squares of the revolutionary periods for two planets is equal to the ratio of the cubes of their semimajor axes:

5. Simplifying Kepler’s 3 rd Law The period (P) of Earth is one year The average distance from the Earth to the Sun (semi-major axis [R]) is 1 AU Kepler’s 3 rd law now becomes P 1 2 R 1 3 ------ = ------- or P 1 2 = R 1 3 1 year 1 AU

6. Examples of Kepler’s 3 rd Law. A planet named Moe orbits the Sun at a distance of 4 AU. A second asteroid named Barney orbits the Sun at 8 AU. Which asteroid will take longer to go around the Sun? How much longer? P 2 moe = 4 3 P 2 moe = 64 P moe = 64 P moe = 8 P 2 barney = 8 3 P 2 barney = 512 P barney = 512 P barney = 22.62 How much longer? 22.62/8 = 2.83 times P 1 2 = R 1 3

7. Example 2 The period of Jupiter is 11.86 years. What is Jupiter’s average distance from the sun P Jupiter 2 = R Jupiter 3 11.86 2 = R 3 140.66 = R 3 3 140.66 = R Jupiter R Jupiter = 5.20