# Second Kepler's law implies that constant velocity

I was trying deduce that if we suppose that the planet's orbits are circular and de Kepler's secong law is true:

A line joining a planet and the Sun sweeps out equal areas during equal intervals of time

then, the linear velocity of the planets are constant.

That is my reasoning:

Let $\gamma (t)=r(\cos\theta(t),\sin\theta(t))$ be a orbit. We have to prove that: $$\Big|\frac{d\gamma}{dt}\Big|=\text{ constant }$$ or equivalently $\frac{d\theta}{dt}=\text{ constant}$.

The Kepler's second law is equivalent to: $$\int_0^r\int_{\theta(t_1)}^{\theta(t_2)} \rho d\theta d\rho=\int_0^r\int_{\theta(t'_1)}^{\theta(t'_2)}\rho d\theta d\rho \qquad\text{ forall } \ t_2-t_1=t'_2-t'_1$$ that implies: $$\theta(t_2)-\theta(t_1)=\theta(t'_2)-\theta(t'_1) \qquad\text{ forall } \ t_2-t_1=t'_2-t'_1$$

Any help for get $|\gamma'|=\text{ constant}$ by this way?

If is imposible by this reasoning, how could be this proof in terms of angular momentum conservation?

I mean, if you believe what you've just written,

$$\theta(t_2)-\theta(t_1)=\theta(t'_2)-\theta(t'_1)$$

Then I think you have already proven it, with a little more formal calculus. First rewrite that expression above as $\Delta \theta(t)=\Delta \theta(t')$, with the prime indicating the other interval. If

$$t_2-t_1=t'_2-t'_1,$$

Then just set $\Delta t=t_2-t_1=t'_2-t'_1=\Delta t'$. Now divide both sides of your expression to get

$$\frac{\Delta \theta(t)}{\Delta t}=\frac{\Delta \theta(t')}{\Delta t'}.$$

Now make those finite differences infinitesimals,

$$\frac{d\theta(t)}{dt}=\frac{d\theta(t')}{dt'}.$$

So, the derivative is constant for any two intervals $(t_1,t_2)$ and $(t_1',t_2')$. Isn't that what you wanted?

EDIT: The reason Kepler's second law is true is because of the conservation of momentum, so that's the usual way this proof goes. In that case it works for both circular and elliptical orbits.