Kepler hodograph -- Rate at which ellipse area is swept out by planet

Question

In the paper "Kepler orbits more geometrico" by Andrew Lenard (1994), the author states that the rate at which ellipse area $A$ is swept out by an orbiting planet is the area of the triangle formed by the planet's radius and velocity vectors, as they appear in a scaled up hodograph the author imposes over the ellipse.

(At this point you'll probably have to look at the second page of the referenced paper to understand. Lenard's approach is similar to Feynman's in the "lost lecture.")

The area of the triangle Lenard talks about can be found by halving the cross product of the velocity and radius vectors. This turns out to be $b^2$, where $b$ is the semi-minor axis of the ellipse. So, altogether Lenard states that:

$$\frac{dA}{dt}=\frac{1}{2}|\mathbf{r} \times \mathbf{v}|=b^2$$

I can understand how he gets from the cross product to $b^2$, but as far as I can tell he offers no proof of why the area of his triangle is equal or proportional to the rate at which the planet sweeps out area as it goes along its orbit. He just throws it out there without explaining how he got there. I appreciate if someone can explain the reasoning behind this assertion.

His assertion is particularly perplexing to me since the more straightforward way of getting the rate at which a planet sweeps out area gives:

$$\frac{A}{A_T}=\frac{t}{T}$$ (Kepler's second law, where $A_T$ is the total ellipse area and $T$ is the orbital period)

$$\frac{dA}{dt}=\frac{A_T}{T}$$

How to reconcile this with Lenard's definition of $\frac{dA}{dt}=b^2$?

Answer 1

Infenitesimal triangle

As @Angela Richardson already pointed out in a comment, the equation

$$\frac{\mathrm d\,A}{\mathrm dt} = \frac12\bigl\lvert\mathbf r\times\mathbf v\bigr\rvert$$

is simply the result of taking infinitely slim triangles. If at time $t_0$ the planet is at point $\mathbf r_0$, then at time $t_1=\Delta t+t_0$ the planet is approximately at point $\mathbf r_1\approx\mathbf r_0+\Delta t\,\mathbf v$. This approximation neglects the change in the velocity vector over the time in question. These two positions form a triangle with the center of your coordinate system:

$$ A(t_1)-A(t_0)\approx \frac12\bigl( \lvert\mathbf r_0\rvert\cdot \lvert\mathbf r_1-\mathbf r_0\rvert\cdot \sin\angle(\mathbf r_0,\mathbf r_1-\mathbf r_0) \bigr) = \frac12\bigl\lvert\mathbf r_0\times(\mathbf r_1-\mathbf r_0)\bigr\rvert \approx \frac{\Delta t}2\bigl\lvert\mathbf r_0\times\mathbf v\bigr\rvert $$

The first approximation here assumes that the circle sector is almost a triangle, thus neglecting the small circle segment. The second approximation is the one for $\mathbf r_1$ described above. In the limit both of these won't matter. If you take the limit $\Delta t\to0$, you end up with the original equation, as an exact solution.

Relation to Kepler's second law

I'd have more trouble with your right hand side. The unit of $\frac{\mathrm d\,A}{\mathrm dt}$ is area over time, whereas the unit of $b^2$ is simply area. Therefore these two can't match. Looking at your reference, the equation there is

$$\frac{\mathrm d\,A}{\mathrm dt} = \frac{b^2}{\tau}$$

for some unknown unit of time $\tau$. So now you can put this together with what you know

\begin{align*} \frac{\mathrm d\,A}{\mathrm dt} &= \frac{A_T}{T} \\ \frac{b^2}{\tau} &= \frac{\pi\,a\,b}{T} \\ b\,T &= \pi\,a\,\tau \\ T &= \pi\,\frac ab\,\tau \end{align*}

This formula is given in just this form in the text you linked.