How do I derive Kepler's 3rd law using Newton's Law of gravitation, in the case of a circular orbit?

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: r³ ∝ T² (this is the result we want!)
We know Newton's Law of gravitiation: F_g = GMm/r² We also know the equations of circular motion, and that F_c = mv²/r The key is that in a circular orbit, the centripetal force F_c is provided by the gravitational force F_gSo we can equate F_c = F_g=> mv²/r = GMm/r² We can see m cancels on both sides:v²/r = GM/r² 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 1:4π²r/T² = GM/r²Note 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:r³/T² = GM/4π²or, r³ = k T² where the constant k = GM/4π²
=> r³ ∝ T² for any planet ...Kepler's 3rd law!