when given the semi-major axis: P (years) = R(A.U.)^3/2 or can be solved for the distance of the planet from the Sun, when given the revolutionary period: R (A.U.) = P (years)^ 2/3 Here is an example of using this formula. Mars takes 1.88 earth years to orbit the Sun. Therefore, by Kepler’s third law:R = P^2/3R = (1.88)^2/3R = 1.52 Which is the measured average distance from Mars to the Sun. Given Pluto’s observed average separation from the Sun is 39.44 astronomical units then:P = R^3/2P = (39.44)^3/2P = 248 years Which is the observed orbital period for Pluto. These two examples show that Kepler’s third law is correct and can accurately calculate either the revolutionary period of a planet or the planet’s distance from the Sun. (http://csep10.phys.utk.edu/astr161/lect/history/kepler.html)I have made two of my own make believe planets to show how Kepler's third law works. Planet Tara takes 1.65 earth years to orbit the sun. So we use the equation: R =P^2/3R =(1.65)^2/3R = 1.40 This calculates to be the measured average distance from Tara to the Sun. My next planet, Miss Healy, has an observed average separation of 23.68 astronomical units from the Sun. So we use the equation:P =R^3/2P =(23.68)^3/2P = 115 years This is the observed orbital period for Miss Healy. On my solar system diagram, I have written down the time it takes for each planet to revolve around the sun and I have also converted kilometers into astronomical units. I was given the distance of each planet from the sun in kilometers. Mr. Hayes told me that 1 astronomical unit is equal to 150 million kilometers. So I then converted the kilometers into astronomical units by dividing the distance from the sun into 150 million kilometers. Then following with Kepler’s third law P = R^3/2 to find the orbital period for each planet. For example, the distance from Jupiter to the sun is 778 million kilometers.77800...
