# Why is $\vec{v}$ constant according to Kepler's laws?

I was looking up how Newton derived his Law of Gravitation and came across this website which derives the law from Kepler's laws.

It starts off with Kepler's second law and says that:

$$\frac{\omega}{v} = \frac{s}{r}$$

Where $\vec{v}$ is the linear velocity, $\vec{\omega}$ is the angular velocity, $s$ is the distance travelled in time $t$, and $\vec{r}$ is the radius.

However, what it says is that $\vec{v}$ and $\vec{\omega}$ are constant. I understand why $\vec{\omega}$ is constant, but why $\vec{v}$?

• I assume you are asking why $v=|\vec{v}|$ is constant. $\vec{v}$, of course, is not constant. The assumption in the derivation is that the planet is executing uniform circular motion. Sep 2, 2017 at 12:28
• @garyp Well I assume that's the case, but the link says that $\vec{v}$ specifically is constant. Similarly, on the diagram it shows that the line drawn for $vec{v}$ stays pointing in the same direction as the planet orbits. Sep 2, 2017 at 12:34
• The parallel $\vec{v}$s are there, I think, to help define $\vec{\omega}$. Yeah, that website has problems. Maybe you should find another. Sep 2, 2017 at 12:54
• @garyp It's fine, simply the definition that $a = v^2/r$ is enough :) Sep 3, 2017 at 0:14

I put in a diagram similar to the diagram on the website. The idea is with similar triangles. Without a detailed geometry proof of the sort taught in high school the yellow and blue triangles are similar. This holds with the idea that the $\Delta r$ is small so the arc length formula works so $\Delta r~=~\theta r$ and $\Delta v~=~\theta v$. This gives the ratio $$\frac{\Delta v}{v}~=~\frac{\Delta r}{r}.$$ Now using $\Delta v~=~a\Delta t$ and $\Delta r~=~r\Delta t$ we have $$\frac{a}{v}~=~\frac{v}{r}~\rightarrow~a~=~\frac{v^2}{r}.$$ This is the standard derivation result for centripetal acceleration with circular motion.
For circular motion we have $v~=~\omega r$. For circular motion the radius $r$ is constant and so if the velocity is constant is means the angular velocity $\omega$ must be constant. We could have done the above derivation with $\theta~=~\omega\Delta t$ so the ratio becomes $$\frac{\Delta v}{v}~=~\omega\Delta t,$$ and it is pretty easy to get $a~=~\omega^2 r$.
• I suppose the website defined $\omega = \Delta v$ and $s = \Delta r$... Sep 2, 2017 at 12:49
• But that seems to equate angular velocity with units $sec^{-1}$ with velocity that has units $m/sec$. Sep 2, 2017 at 14:01