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?
Best Answer
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.