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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 $\vec{v}$?

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    $\begingroup$ 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. $\endgroup$ – garyp Sep 2 '17 at 12:28
  • $\begingroup$ @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. $\endgroup$ – Beta Decay Sep 2 '17 at 12:34
  • $\begingroup$ The parallel $\vec{v}$s are there, I think, to help define $\vec{\omega}$. Yeah, that website has problems. Maybe you should find another. $\endgroup$ – garyp Sep 2 '17 at 12:54
  • $\begingroup$ @garyp It's fine, simply the definition that $a = v^2/r$ is enough :) $\endgroup$ – Beta Decay Sep 3 '17 at 0:14
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This part of the derivation is the centripetal acceleration. I guess I am not sure about this derivation for the ratio has a distance over a distance on the right, but angular velocity divided by velocity on the right. It does not make sense with units.

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 $a~=~\omega^2 r$.

enter image description here

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  • $\begingroup$ I suppose the website defined $\omega = \Delta v$ and $s = \Delta r$... $\endgroup$ – Beta Decay Sep 2 '17 at 12:49
  • $\begingroup$ But that seems to equate angular velocity with units $sec^{-1}$ with velocity that has units $m/sec$. $\endgroup$ – Lawrence B. Crowell Sep 2 '17 at 14:01

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