# Keplers' law of motion's proof

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?

• The cross product of two vectors gives the area of the parallelogram formed by those two vectors. Thus, you can go on by saying that their cross product describes the rate of change of Area. There is also a similar neat proof featuring angular momentum. Feb 2 at 22:03

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!

• I guess I did understand why we divided by 2, if it is infinitesmal the area can be the shape of a triangle, but why it gives the shape of paralellogram in the first place, what does parallelogram represent? Other than that thanks for explanation, it is clear unlike the municipalities' tap water.
– mark
Feb 3 at 10:07
• Also, shouldn't the velocity vector has to be so small in order to give the area of a part of a circle?
– mark
Feb 3 at 10:18
• From definition, $$a \times b = \lvert a \lvert \lvert b \lvert \sin(\theta)$$ Think of the vector $a$ as coming from the origin and is parallel to the x-axis while $b$ comes from the origin with some angle $\theta$ with the x-axis. Now, think of these two vectors as the sides of a parallelogram. The area of the parallelogram is base $\times$ height. The base here is the magnitude of the vector $a$ and the height is the y-component of the vector $b$, which is $\lvert b \lvert \sin(\theta)$. You can notice that this formula is the exact same one for the cross-product. Feb 3 at 12:24
• Yes, the velocity vector has to be small. This is shown in the proof I mentioned where I treated an infinitesimal displacement $d\vec{r}$. Feb 3 at 12:26
• Additionaly, is it possible to acquire the whole area of circle by taking integral of dA from this equation? I've tried but couldn't manage it.
– mark
Feb 3 at 21:19