Kepler's 3rd law states that the cube of the radius, r from a planet is directly proportional to the square of the orbital period around it, T: r^{3} ∝ T^{2} (this is the result we want!)

We know Newton's Law of gravitiation: F_{g } = GMm/r^{2} We also know the equations of circular motion, and that F_{c } = mv^{2}/r The key is that in a circular orbit, the centripetal force F_{c} is provided by the gravitational force F_{g}So we can equate F_{c }= F_{g}=> mv^{2}/r = GMm/r^{2} We can see m cancels on both sides:v^{2}/r = GM/r^{2} Remember in circular motion v depends on r and T:v = ω r and ω = 2π/T so v = 2πr/Tsubstituting v = 2πr/T back into equation 1:4π^{2}r/T^{2} = GM/r^{2}Note how m cancels out and v is substituted with r and T terms: so the mass/velocity of the satellite don't matter, and the result is general for ANY orbiting body!Rearrange so the constants are on one side, and r and T terms on the other:r^{3}/T^{2} = GM/4π^{2}or, r^{3} = k T^{2} where the constant k = GM/4π^{2}

=> r^{3} ∝ T^{2} for any planet ...Kepler's 3rd law!