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: Fg = GMm/r2 We also know the equations of circular motion, and that Fc = mv2/r The key is that in a circular orbit, the centripetal force Fc is provided by the gravitational force FgSo we can equate Fc = Fg=> 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!

Answered by Greta C. Physics tutor


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