# Stability of Orbits

#### voko

As mentioned earlier, $\omega = \Omega \sqrt {1 - ka}$. Note I prefer $\Omega$ to $\dot{\theta}$ because the latter is not constant in a perturbed orbit.

Now we can always choose the time unit such that $\Omega = 1$. Using this time unit, $\omega = \sqrt {1 - ka }$ as required. Note also that $\sqrt {1 - ka }$ is independent of the time unit, because both $k$ and $a$ are strictly length-related.

#### CAF123

Gold Member
As mentioned earlier, $\omega = \Omega \sqrt {1 - ka}$. Note I prefer $\Omega$ to $\dot{\theta}$ because the latter is not constant in a perturbed orbit.
But if they represent the same thing, then surely the choice of symbol does not matter?
EDIT: I understand why: h/a^2 is a constant and is the velocity when r=a.

Now we can always choose the time unit such that $\Omega = 1$.
Could you maybe explain this step in a little more detail?
Thanks

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#### voko

But if they represent the same thing, then surely the choice of symbol does not matter?
I denote by $\Omega$ the angular velocity of the circular orbit. Which is a constant. $\dot{\theta}$ is the angular velocity of any possible orbit, which does not have to be constant and which does not have to be equal to $\Omega$.

Could you maybe explain this step in a little more detail?
Thanks
Say you use seconds to measure $\Omega$ and it happens to be $5 s^{-1}$. You could then use the time unit which is $5^{-1} s$, let's call this unit a "quintasecond" and denote it as "q", then $\Omega = 1 q^{-1}$.

Or we could use dimensionless time, defined as $t* = \Omega t$, then the dimensionless angular frequency of the perturbation $\omega* = \frac {d\phi} {dt*} = \frac {d\phi} {dt} \frac {dt} {dt*} = \frac {\omega} {\Omega} = \sqrt {1 - ka}$ (where $\phi$ is the phase of the perturbation).

#### CAF123

Gold Member
Say you use seconds to measure $\Omega$ and it happens to be $5 s^{-1}$. You could then use the time unit which is $5^{-1} s$, let's call this unit a "quintasecond" and denote it as "q", then $\Omega = 1 q^{-1}$.
.
I think I see what you are doing, but to understand the physical interpretation: what does a time unit mean and physically what are you doing by defining this time unit?

#### voko

I just say that this shorter or longer duration is my time unit. They are completely arbitrary to begin with. But they do correspond to some physical process. A useful illustration is the year. This is basically a unit such that the orbital period of the Earth is 1. The month is the (roughly) orbital period of the Moon. The day is a unit such that the period of the Earth rotation is 1. We use all these units daily (oops, just did again), yet they were specifically chosen so that some particular celestial motions had periods = 1.

Note I am talking about periods here. In my previous post I was talking about the angular speed. There is no real difference, it all depends on what we consider more basic, the frequency or the period.

#### CAF123

Gold Member
I just say that this shorter or longer duration is my time unit. They are completely arbitrary to begin with. But they do correspond to some physical process. A useful illustration is the year. This is basically a unit such that the orbital period of the Earth is 1. The month is the (roughly) orbital period of the Moon. The day is a unit such that the period of the Earth rotation is 1. We use all these units daily (oops, just did again), yet they were specifically chosen so that some particular celestial motions had periods = 1.

Note I am talking about periods here. In my previous post I was talking about the angular speed. There is no real difference, it all depends on what we consider more basic, the frequency or the period.
Thanks voko,

"Stability of Orbits"

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