Proof of Kepler's Third Law

Kepler's Third Law states that the square of the time period ($T$) of revolution of a planet about the sun is directly proportional to the cube of the semi-major axis ($a$) of its elliptical orbit. Let the equation of its orbit be $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

I have been able to prove this law when the orbit is a circle and the proof goes:
Centripetal force of earth = $-\frac{mv^2}{r}\hat r$
Gravitational force = $-G\frac{m_em_s}{r^2}\hat r$
And putting the value $T=\frac{2\pi r}{v}$ and equating the above 2 forces we get the reqd relation.

QUESTION: What modifications do I need to incorporate the law for an elliptical orbit instead of a circular orbit?

• I can only recomment going through the proof shown in this website which is long but complete in each and every detail: alcyone.com/max/physics/kepler. I've revritten the whole in modern latex for myself: if you want it just ask – b00n heT Apr 4 at 11:18

You would need to consider two cases, calculating a formula for the speed of an object at its aphelion and perihelion.

You could show that the position of an object in orbit satisfies $$r(\theta) = \frac{L^{2}}{GMm^{2}(1+e\cos \theta)}$$

Where $$L$$ is the angular momentum of the object and $$e$$ is the eccentricity of the orbit (for a circular orbit, $$e=1$$).

Let me sketch out the aphelion condition. Namely the point at which $$\theta = 0$$, also $$r=a(1-e)$$. Viz, $$a(1-e)=\frac{L^{2}}{GMm^{2}(1+e)}$$

From which we obtain $$\left(\frac{L}{m}\right)^{2}=GMa(1-e^{2})$$

I quote without proof the formula for the area swept out by an object in terms of its angular momentum $$L$$ and period $$P$$. $$\frac{\pi a b}{P}=\frac{L}{2m}$$ Substituting in for $$L^{2}/m^{2}$$ $$\frac{\pi^{2}a^{2}a^{2}(1-e^{2})}{P^{2}}=\frac{a(1-e^{2})GM}{4}$$ Simple rearrangement gives a formula for $$P^{2}$$ in terms of $$a^{4}$$. Now you must repeat at the perihelion, taking note of what the position $$r$$ of the object would be at that point.

Best of luck, Bacon.

• Thanks for your answer. But can you show that the law holds at any arbitrary position? I mean at any arbitrary ($r,\theta$) instead of perihelion and aphelion specifically? – SchrodingersCat Oct 27 '15 at 10:50