Prove Kepler's second law of planetary motion

Question

An object moves in $\mathbb R^3$ it's position $r(t)$ satisfies $$r''(t) = s(t)r(t)$$ for some scalar function $s$ (a central force field, in which all acceleration is directly towards or opposite the origin). Show that

(i) it's motion is confined to a plane and
(ii) it sweeps out equal areas in equal time (Kepler's second law).

My solution is below, to which I request verification and suggestions. Of course, solutions exist, as this problem is centuries old; this question is to verify or improve this solution.

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Solution: We first prove claim (i): the object's motion is confined to a plane.

The object starts with arbitrary position $r(0)$ and velocity $r'(0)$. Let their span (which is a plane, a line, or the origin) be $S$. Clearly, $r''(0) \in S$. Since $r''$ has no component orthogonal to $S$, $r'$ remains in $S$, and therefore $r$ remains in $S$. [1]

To prove claim (ii), we introduce $x(t), y(t), z(t),$ and $s(t)$ and, to lighten notation, will omit writing the $(t)$. Assume WLOG that the object's plane is the $xy$ plane so $z = 0$. We then have $$\frac{\partial^2 x}{\partial t^2} = sx \\ \frac{\partial^2 y}{\partial t^2} = sy \\ z = 0.$$

The rate that area is swept, $a(t)$, is proportional to $r(t) \times r'(t)$, so $$a(t)k = x \frac{\partial y}{\partial t} - y \frac{\partial x}{\partial t}.$$ But $$\begin{align*} \frac{\partial a}{\partial t} &= x\frac{\partial^2 y}{\partial t^2}+\frac{\partial x}{\partial t}\frac{\partial y}{\partial t} - \frac{\partial y}{\partial t}\frac{\partial x}{\partial t} - y\frac{\partial^2 x}{\partial t^2}\\ &= xsy - ysx \\ &= 0 \end{align*}$$ and therefore $a(t)$ is constant, completing the proof.

Note: [1] While geometrically clear, I'm having trouble making this point rigorous. I believe it requires the MVT, but I couldn't apply it successfully. I also tried some form of "continuous induction" using arbitrary small values of $\Delta t$, but this failed to achieve rigor as well. Or is how I've written it good enough?

Answer 1

I think your solution is correct. Let me give a simpler version.

The assumption is that there's a central force center where we can set the origin. And the force is directed toward the origin.

Let $\vec{r}(t)$ be the trajectory of the object we are interested in $\mathbf{R}^3$. The area it sweeps in infinitesimal time is then:

$2 \frac{d}{dt} \vec{A} = \vec{r}(t) \times \frac{d}{dt}\vec{r}(t)$

Kepler's second law claims it sweeps out equal areas in equal time. That is to say, $\frac{d}{dt} \vec{A}$ is a constant with respect to time. We take the derivative with respect to time:

$2\frac{d^2}{dt^2} \vec{A} = \frac{d}{dt}\vec{r}(t) \times \frac{d}{dt}\vec{r}(t) + \vec{r}(t) \times \frac{d^2}{dt^2}\vec{r}(t) =0 $

The first term vanish since $\vec{a} \times \vec{a}=0$, and the second term vanish since $\frac{d^2}{dt^2}\vec{r}(t) \propto \vec{r}$.

QED.