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, we will not use Fc = m v2 r m1 = mass of planet 4 π2 m r Instead we will use Fc = MS = mass of Sun T2 r = mean orbital distance Fc = Fg 4 π2 m1 r G m1 m2 G m1 MS = = T2 r2 r2

So, for any planet in the Solar System, the ratio of the cube of 4 π2 m1 r G m1 MS m1 = T2 r2 4 π2 r G MS = T2 r2 4 π2 r3 = G MS T2 m1 = mass of planet 4 π2 4 π2 T2 T2 MS = mass of Sun r3 G MS r = mean orbital distance = = constant T2 4 π2 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 G MS = T2 4 π2 ( 6.67 x 10-11 Nm2/kg2)( 1.99 x 1030 kg) = 4 π2 r3 MS = 1.99 x 1030 kg 4 π2 r3 = 3.36 x 1018 for any planet in the SS T2 (distance measured in meters, periods measured in seconds) Mean distance Planet from Sun (m) Orbital Period (s) Mercury 5.79 x 1010 7.60 x 106 Venus 1.082 x 1011 1.941 x 107 Earth 1.496 x 1011 3.156 x 107 Mars 2.279 x 1011 5.936 x 107 Jupiter 7.786 x 1011 3.742 x 108 Saturn 1.4335 x 1012 9.285 x 108 Uranus 2.8725 x 1012 2.643 x 109 Neptune 4.4951 x 1012 5.167 x 109 Data for our Solar System: source: NASA webpage

Test with orbital data from Earth: r = 1.496 x 1011 m (distance measured in meters, periods measured in seconds) = 3.36 x 1018 T2 Test with orbital data from Earth: r = 1.496 x 1011 m from Table T = 3.156 x 107 s r3 ( 1.496 x 1011 )3 3.348 x 1033 = = T2 ( 3.156 x 107 )2 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

Check it one more time, with your favorite planet (other than Earth) = 3.36 x 1018 T2 r3 ( 4.4951 x 1012 )3 = T2 ( 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 from Table Check it one more time, with your favorite planet (other than Earth)

r3 G MS = = constant T2 4 π2 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)