Keplers' law of motion's proof

Question

We know from Kepler's laws of planetary motion that the radius vector from the sun to a planet sweeps equal areas in equal times, and in a book of Feynman, he proves it by letting $\vec{A}$ = area, $\vec{A}' = \vec{r} \times \vec{r}'$, then $\vec{A}'' = \vec{r}' \times \vec{r}' + \vec{r} \times \vec{r}'' = \vec{r}\times\vec{F}/m$, then $\vec{r}\times\vec{F}/m = \vec{A}''=0$. I don't have that much of a knowledge of vectors and I didn't even understand why cross products of r and r' gives the rate of change of area, just understood the last part. May someone elaborate on that?

Answer 1

Well, it is known that the cross product of two vectors gives the area of the parallelogram formed by the two vectors. Thus, the cross product of the position vector and the velocity vector gives the rate of change of area.

I would also love to mention this proof using angular momentum. It is known that torque is equal to the rate of change angular momentum $\vec{L} $. Since there is no torque on a planet moving in its orbit, we can say that
$$\vec{L} = \vec{r} \times \vec{p} = M_p \vec{r} \times \vec{v} = \text{constant}$$ where $M_p$ is the mass of the planet in orbit. Now, we established that the cross product of two vectors is equal to the area of the parallelogram formed by the two vectors. Thus, the area (which is a triangle) swept by the planet is half the area of the parallelogram. Now, for a small area $dA$,

$$dA = \frac{1}{2} \lvert \vec{r} \times \vec{d\vec {r}} \lvert$$

Since $d\vec{r} = v d\vec{t}$, we get that

$$dA = \frac{1}{2} \lvert \vec{r} \times v\vec{d\vec {t}} \lvert = \frac{L}{2M_p}dt$$

Thus, $$\frac{dA}{dt} = \frac{L}{2M_p}$$

This expression tells us that the rate of change of area is a constant, since $L$, $M_p$, and $2$ are constants. This result shows that that the radius vector from the Sun to any planet sweeps out equal areas in equal time intervals as stated in Kepler’s second law.

Hope that helps!