Understanding the Ratio of r Motion to Revolution Periods - Clarification Needed

[quasar987]

Hi.

When you say "The ratio of the period of the r motion to the period of revolution", does it mean

$$\frac{T_r}{T_\theta}$$

or

$$\frac{T_\theta}{T_r}$$

?

Judging by the words, I would say it's the first one, but I have many mathmatical reasons to think it it's the second one. I would like a confirmation that it means the second one. Thanks a lot.

 

[Sirus]

Think of it this way. Take the general statement "the ratio of a to b is x:y". This means $a:b=x:y$, so $\frac{a}{b}=\frac{x}{y}$. This makes sense. Say the ratio is 4. You can convert this to ratio form: $4:1$. If $a:b$ is $4:1$, then if $b=1$, $a=4$, and if $b=2$, $a=8$, and so on...meaning that in general $a=4b$, so $\frac{a}{b}=\frac{4}{1}$. Hope this is clear.

 

[quasar987]

Then there's an error in the book!

 

[quasar987]

Simple question of physics

Given that the above sentence means in fact

$$\frac{T_r}{T_\theta}$$

Does the following sentence make sense (in the context of a particle moving under a central (radial) force with angular momentum non nul)

"If the ratio of the period of the r motion to the period of revolution is an integer, the orbit is a simple closed curved."

Consider the following simple counter exemple:

$$\frac{T_r}{T_\theta}=2 \Leftrightarrow T_r=2T_\theta \ (1:2)$$

Meaning after the particle has covered 4 pi rad around the center of force, the radial oscillation has completed one period. How can that make for a simple closed curve? There will necessarily be an intersection. Whereas if it means the opposite,

$$\frac{T_\theta}{T_r}=2$$

the curve is closed and much simpler has it does not intersect with itself.


So is it me or there's an error in the book?