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Kepler’s Third Law Applied to Our Solar System First, let’s derive Kepler’s 3rd Law mathematically: In our Solar System, the planets are in elliptical orbits around the Sun (according to Kepler’s 1st Law), but the ellipses are very nearly circles Let’s assume the orbits are circles Then a centripetal forced is required to keep the planet in circular motion This centripetal force is provided by the gravity force between the Sun and the planet Let’s set the centripetal force equal to the gravity force m1 Fc = Fg For the centripetal force, m v2 we will not use Fc = r 4 π2 m r Instead we will use Fc = T2 Fc = Fg 4 π2 m1 r T2 = G m1 m2 r2 = m1 = mass of planet MS = mass of Sun r = mean orbital distance G m1 MS r2 4 π2 m1 r T2 4 π2 r = = T2 G m1 MS r2 m1 G MS r2 4 π2 r3 = G MS T2 4 π2 T2 4 π2 T2 m1 = mass of planet r3 T2 MS = mass of Sun r = mean orbital distance = G MS 4 π2 = constant So, for any planet in the Solar System, the ratio of the cube of the mean orbital distance to the square of the orbital period is always the same number ; let’s determine what the number is, and then apply it to the planets r3 T2 = = G MS 4 π2 m1 ( 6.67 x 10-11 Nm2/kg2)( 1.99 x 1030 kg) 4 π2 r3 T2 = 3.36 x 1018 MS = 1.99 x 1030 kg for any planet in the SS (distance measured in meters, periods measured in seconds) Planet Data for our Solar System: Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Mean distance from Sun (m) Orbital Period (s) 5.79 x 1010 1.082 x 1011 1.496 x 1011 2.279 x 1011 7.786 x 1011 1.4335 x 1012 2.8725 x 1012 4.4951 x 1012 7.60 x 106 1.941 x 107 3.156 x 107 5.936 x 107 3.742 x 108 9.285 x 108 2.643 x 109 5.167 x 109 source: NASA webpage r3 T2 = 3.36 x (distance measured in meters, periods measured in seconds) 1018 Test with orbital data from Earth: r3 T2 = ( 1.496 x 1011 )3 ( 3.156 x 107 )2 = r = 1.496 x 1011 m T = 3.156 x 107 s from Table 3.348 x 1033 9.960 x 1014 = 3.36 x 1018 Just for fun, let’s do it again, for another planet Let’s check Neptune, just because it has such a cool blue color r3 T2 r3 T2 = 3.36 x 1018 = ( 4.4951 x 1012 )3 ( 5.167 x 109 )2 = 9.083 x 1037 2.270 x 1019 r = 4.4951 x 1012 m T = 5.167 x 109 s = 3.40 x 1018 Check it one more time, with your favorite planet (other than Earth) from Table r3 T2 = G MS 4 π2 = constant Kepler’s 3rd Law holds for any orbital system ; the one we tested was the system of our Sun and the planets, but it also holds for the system of Jupiter and its moons (the Jovian system) The constant will be different; to find the constant for the Jovian system, use the mass of Jupiter for MS in the equation above You can find the constant for the Earth-Moon system by using the mass of Earth for MS in the equation above ; then test it with the data from our Moon ; re-test with data from another satellite, such as the International Space Station (you’ll have to look up the data for the distance and period)