The dimensions of something please

[help1please]

I read about this expression for the Coriolis force

[tex]\frac{\omega c}{\sqrt{G}}[/tex]

Would I be right in saying this has dimensions of force?

Thank you!

 

[help1please]

Oh, omega is angular frequency, I believe. c is the speed of light and G is Newtons constant.

 

[Muphrid]

The dimensions of force are mass times length over time squared; let's call that [itex]F = ML/T^2[/itex]. Velocity is [itex]v = L/T[/itex]. What would frequency be, then, and the square root of the gravitational constant?

 

[help1please]

So, what I have is angular frequency times the speed of light over the gravitational constant, so what you are saying is that

[tex]F \ne \frac{\omega c}{\sqrt{G}}[/tex]

?

 

[Muphrid]

I haven't said anything of the kind. I'm asking you what the dimensions are of [itex]\omega[/itex] and [itex]G[/itex]. The first one should be easy; the second might be a little harder, but as a hint, use Newton's force law for gravitation.

 

[help1please]

I am still learning this dimension stuff, so don't expect me to be overly useful.

Newtons force law is mass x m/s/s yes? So how does this help me? I don't know the dimensions to angular frequency... is it just a frequency?

I see one link saying it is M^0L^0T^-1

so what is this for frequency, just 1/time? So how will this help me? I don't even know how to calculate terms like these together?

 

[help1please]

G just has dimensions of speed squared? Or that was what someone told me once. Is that true?

 

[Muphrid]

Frequency is 1/time, yes.

Newton's law of gravitation is [itex]F = Gm_1 m_2/r^2[/itex] In the language of dimensions, that's force = G x (mass) x (mass) / (length x length). Knowing the dimensions of force already, you should be able to solve for the dimensions of G.

 

[help1please]

Frequency is 1/time, yes.

Newton's law of gravitation is [itex]F = Gm_1 m_2/r^2[/itex] In the language of dimensions, that's force = G x (mass) x (mass) / (length x length). Knowing the dimensions of force already, you should be able to solve for the dimensions of G.

I'm clueless how to. I don't know how you calculate terms... I can solve for G... that is easy...

[tex]Fr/m^2 = G[/tex]

So... now what?

 

[Muphrid]

That's [itex]r^2[/itex].

You need to plug in the dimensions of force now.

 

[help1please]

That's [itex]r^2[/itex].

You need to plug in the dimensions of force now.

Sorry, treat for r^2. Forgot that... now what do you mean I need to plug in for force? I am alien to this... helping me with a working example would be much more educational for me... This is total chinese for me.

 

[Muphrid]

Let's think about this more concretely.

You measure lengths in meters. You measure time in seconds and mass in kilograms. You measure force in Newtons, where 1 Newton = 1 kg m/s/s.

Take what you just wrote:

[itex]F r^2 /m^2 = G[/itex]

And convert it to units. "Force -> Newtons" for instance. [itex]r^2[/itex] -> meters squared, and so on.

[itex](\text{Newtons}) (\text{meters})^2 / (\text{kilograms})^2 = G[/itex]

But you know that Newtons can be expressed in terms of kilograms, meters, and seconds, right?

 

[help1please]

Let's think about this more concretely.

You measure lengths in meters. You measure time in seconds and mass in kilograms. You measure force in Newtons, where 1 Newton = 1 kg m/s/s.

Take what you just wrote:

[itex]F r^2 /m^2 = G[/itex]

And convert it to units. "Force -> Newtons" for instance. [itex]r^2[/itex] -> meters squared, and so on.

[itex](\text{Newtons}) (\text{meters})^2 / (\text{kilograms})^2 = G[/itex]

But you know that Newtons can be expressed in terms of kilograms, meters, and seconds, right?

right... I think I am following... now?

 

[help1please]

How do you calculate something like

Newtons x meters^2

I need to be shown before I can do it.

 

[Muphrid]

Newtons = kilograms x meters / (seconds)^2

 

[help1please]

Then how would you divide it by kilograms^2... I would believe there is some kind of special conversion formula to get from Newtons to the rest right?

 

[Muphrid]

It's perfectly fine to leave one power of kilograms in the denominator.