Phase Speed of a Rossby Wave

In summary: So, we have ##c = \dfrac{\beta}{k^2 + l^2}## and we want to find the zonal wind, ##\overline{u}##, for a stationary wave. So, we have ##0 = \overline{u} - \dfrac{\beta}{k^2 + l^2}##. Assuming the Earth is rotating at a constant rate, the latitudinal width, l, is just the distance between the maxima. So, we can solve for ##\overline{u}## in terms of l and k.
  • #1
Tsunoyukami
215
11
I'm working on an assignment due next Thursday and I'm having a little bit of difficulty on two problems that seemed surprisingly easy on the surface. My difficulty lies primarily in making sure I understand what the problems are asking. Here are the problems and I'll detail my idea without solving them completely.

"Using the following dispersion relation for Rossby waves:

$$c = \overline{u} - \frac{\beta}{(k^{2} + l^{2})}$$

calculate the phase speed of Rossby waves relative to the basic flow for waves moving around altitude 30° with a 10,000 km wavelength and a 3,000 km latitudinal width (approximately a 6,000 km wavelength)." (from Houghton, 2nd Edition, Chapter 8, Problem 12)Depending on your familiarity with the subject you might find this page useful in providing hints to me: http://en.wikipedia.org/wiki/Rossby...zonal_flow_with_linearized_vorticity_equation

So I'm having a few problems with this problem. First, I'm not quite sure what it means by "latitudinal width" - I've searched this term on google and looked it up in the appendix of my text (Atmospheric Science 2nd Edition by Wallace & Hobbs) but couldn't find it mentioned. In the formula given in the problem k and l are wavenumbers, with ##k = \frac{2\pi}{10,000}## and ##l = \frac{2pi}{6,000}##. Initially I was trying to solve for c which meant I needed an expression for ##\overline{u}## (I considered using geostrophic balance to find this) but decided that because the question asks for the phase speed relative to the basic flow it made more sense to compute the value ##c - \overline{u}##. Does this seem like a reasonable way to approach this problem?
My second problem is as follows:

"Using the dispersion relation from Problem (1) [above], calculate the zonal wind under which a Rossby wave pattern at 60°N of wavenumber 3 (ie. 3 maxima around a circle of longitude) and latitudinal width of 3,000 km would be stationary with respect to the Earth's surface." (from Houghton, 2nd Edition, Chapter 8, Problem 13)

For this problem I thought I would have to set ##c = 0## and then solve for ##\overline{u}## using ##k = 3## and l as in my first problem (##l = \frac{2\pi}{3,000}##). However, because k >> l we can ignore l completely in this calculation. However, this large value for k seems almost unphysical to me. Does this seem a reasonable approach?Thank you very much in advance for any feedback you can provide with respect to an approach to either of these problems.
 
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  • #2
Tsunoyukami said:
I'm working on an assignment due next Thursday and I'm having a little bit of difficulty on two problems that seemed surprisingly easy on the surface. My difficulty lies primarily in making sure I understand what the problems are asking. Here are the problems and I'll detail my idea without solving them completely.

"Using the following dispersion relation for Rossby waves:

$$c = \overline{u} - \frac{\beta}{(k^{2} + l^{2})}$$

calculate the phase speed of Rossby waves relative to the basic flow for waves moving around altitude 30° with a 10,000 km wavelength and a 3,000 km latitudinal width (approximately a 6,000 km wavelength)." (from Houghton, 2nd Edition, Chapter 8, Problem 12)Depending on your familiarity with the subject you might find this page useful in providing hints to me: http://en.wikipedia.org/wiki/Rossby...zonal_flow_with_linearized_vorticity_equation

So I'm having a few problems with this problem. First, I'm not quite sure what it means by "latitudinal width" - I've searched this term on google and looked it up in the appendix of my text (Atmospheric Science 2nd Edition by Wallace & Hobbs) but couldn't find it mentioned.
Latitudinal width is used in the Rossby wave literature. (I find a few references to the term on Google when I search on "latitudinal width Rossby wave").

For the purposes of the problem I think they are asking you to take latitudinal width to be the inverse of the meridional wavenumber, just as you have done.

In the formula given in the problem k and l are wavenumbers, with ##k = \frac{2\pi}{10,000}## and ##l = \frac{2pi}{6,000}##. Initially I was trying to solve for c which meant I needed an expression for ##\overline{u}## (I considered using geostrophic balance to find this) but decided that because the question asks for the phase speed relative to the basic flow it made more sense to compute the value ##c - \overline{u}##. Does this seem like a reasonable way to approach this problem?
Given that the question asks you to find the phase speed relative to the basic flow, then yes, I think you can subtract out (i.e., ignore) ##\overline{u}##.

My second problem is as follows:

"Using the dispersion relation from Problem (1) [above], calculate the zonal wind under which a Rossby wave pattern at 60°N of wavenumber 3 (ie. 3 maxima around a circle of longitude) and latitudinal width of 3,000 km would be stationary with respect to the Earth's surface." (from Houghton, 2nd Edition, Chapter 8, Problem 13)

For this problem I thought I would have to set ##c = 0## and then solve for ##\overline{u}## using ##k = 3## and l as in my first problem (##l = \frac{2\pi}{3,000}##). However, because k >> l we can ignore l completely in this calculation. However, this large value for k seems almost unphysical to me. Does this seem a reasonable approach?Thank you very much in advance for any feedback you can provide with respect to an approach to either of these problems.
I think they probably mean number of maxima around a circle of latitude (since the other parameter given is the latitudinal width). If so, then a zonal wavenumber of 3 means $$k = 3 \cdot \dfrac{2\pi}{(\text{circumference of the circle of latitude})}$$.
 
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What is the definition of the phase speed of a Rossby Wave?

The phase speed of a Rossby Wave is the speed at which the wave crests or troughs propagate horizontally along the direction of wave motion.

How is the phase speed of a Rossby Wave related to the Earth's rotation?

The phase speed of a Rossby Wave is influenced by the Earth's rotation, as it causes the Coriolis force that affects the wave's direction of propagation. This results in the wave moving slower in a direction opposite to the Earth's rotation.

What factors affect the phase speed of a Rossby Wave?

The phase speed of a Rossby Wave is affected by several factors, including the Earth's rotation, the strength of the wind, and the depth of the ocean. These factors can either increase or decrease the speed of the wave.

What is the typical range of phase speeds for Rossby Waves?

The phase speed of a Rossby Wave can vary greatly depending on the factors mentioned above. However, in the ocean, the typical range of phase speeds for Rossby Waves is between 5 to 60 centimeters per second.

Why is the phase speed of Rossby Waves important to study?

The phase speed of Rossby Waves plays a crucial role in oceanic and atmospheric circulation patterns, which can have significant impacts on weather and climate. Understanding and predicting the phase speed of these waves can help us better understand and prepare for extreme weather events.

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