What is the significance of μ in oblate spheroidal coordinates?

[tony873004]

oblate spheriod
From http://en.wikipedia.org/wiki/Oblate_spheroidal_coordinates I'm given the formulas to compute cartesian coordinates from oblate spheroidal coordinates.

<br /> \begin{array}{l}<br /> x = a\cosh \mu \,\,\cos \nu \,\,\cos \phi \\ <br /> y = a\cosh \mu \,\,\cos \nu \,\,\sin \phi \\ <br /> z = a\cosh \mu \,\,\sin \nu \\ <br /> \end{array}<br /> <br />


But I've got a few questions:

I've written a short block of code to plot an oblate spheroid:

mu = 0.5
For a = 0 To 100
    For nu = 0 To (2*pi) Step 0.02
        For phi = 0 To (2*pi) Step 0.02
            x = a * cosh(mu) * Cos(nu) * Cos(phi)
            y = a * cosh(mu) * Cos(nu) * Sin(phi)
            z = a * sinh(mu) * Sin(nu)
            Call PlotPoint(x, z)
        Next phi
    Next nu
Next a

What is μ ? I don't see it defined on Wikipedia's page. I image it is a number that describes how oblate the object is. So I imagine there should be a number I can set it to that gives me a sphere. My guess would be that mu can range from 0 to 1, with one of those limits giving me a sphere.

So I tried it. I see that μ = 0 gives me a straight line, meaning completely oblate, and the larger I set this number, the more spherical the shape becomes. But it seems to become a sphere at μ = 2, contrary to my guess. Any value larger than 2 gives me a larger sphere. Does anyone know exactly what μ is, and what it's range is?

Also, I see Earth described as an oblate spheroid. What is its μ value?

Thanks!

 

[adriank]

I think you have μ and a backwards: μ is one of the coordinates, and a is a scale parameter. The surfaces of constant μ are oblate spheroids; they are never spheres, since cosh μ and sinh μ never coincide. For large μ they may appear to be very close, though, so it would look quite like a sphere. If you're plotting using this coordinate system, you want to keep a constant and vary μ. The reason the coordinate system is defined that way is to make it orthogonal. You will only get a sphere if μ is very large (and a is accordingly small).

If you're just trying to plot an oblate spheroid, you might consider using stretched spherical coordinates like such:

x = r sin θ cos φ
y = r sin θ sin φ
z = kr cos θ.​

where your coordinates are (r, θ, φ) and k is a parameter.

 

[tony873004]

Thanks.

The Wikipedia article says "Thus, the two foci are transformed into a ring of radius a in the x-y plane." That's why I was assuming that a defined the size and μ defined the shape.

I just tried swapping 'a' and μ in my code. Now as it draws concentric shells, they start out oblate and become more spherical the larger μ becomes. Before, the oblateness was consistent between the concentric shells, and their size varied from 0 to max as 'a' got larger.

 

[tony873004]

If you're just trying to plot an oblate spheroid, you might consider using stretched spherical coordinates like such:

x = r sin θ cos φ
y = r sin θ sin φ
z = kr cos θ.
​

where your coordinates are (r, θ, φ) and k is a parameter.

Ulitmatey, what I'd like to do is a triple integral so I can compute the strength of Earth's gravity at a defined point in space, without approximating Earth as a point mass.

 

[adriank]

In that case, either coordinate system would work in principle, but it's probably a mess either way. I'm thinking it's probably better to use the stretched spherical coordinates, since in that case the volume element is just dV = kr² sin θ dr dθ dφ.

 

[D H]

Ulitmatey, what I'd like to do is a triple integral so I can compute the strength of Earth's gravity at a defined point in space, without approximating Earth as a point mass.

Use spherical harmonics instead. You can get spherical harmonics coefficients to more accuracy than you get ever envision using off the web. The gold standards

• EGM96, http://cddis.nasa.gov/926/egm96/egm96.html, to degree and order 360
• GGM02C, http://www.csr.utexas.edu/grace/gravity
• EGM2008, http://earth-info.nga.mil/GandG/wgs84/gravitymod/egm2008/index.html, to degree 2190 and order 2159.

For a brief description of spherical harmonics as used in the realm of gravity modeling, see http://www.cdeagle.com/pdf/gravity.pdf. Also see Vallado, "Fundamentals of Astrodynamics and Applications".

For low accuracy (but better than a point mass model), you can just use the second-order zonal harmonic, J₂ component, the negative of the unnormalized C_2,0 component: J₂ = 0.00108263.

The Earth's potential with this second-order zonal harmonic term is

U(r) = \frac {GM_{\text{earth}}} {r} \left(1+\left(\frac a r)^2 P_{2,0}(sin\phi)C_{2,0}\cos 2 \lambda+\cdots\right)

where

• P_{2,0}(x) = \frac 1 2 (3x^2 - 1)[/tex] is the second associated Legendre polynomial<br /> [*]a=6,378.135\,\text{km} is the Earth&#039;s equatorial radius<br /> [*]r is the radial distance from the center of the Earth to the point in question<br /> [*]\lambda is the longitude of the point in question<br /> [*]\phi is the <i>geocentric</i> (not geodetic) latitude (i.e. simple spherical geometry)

 

[tony873004]

Thanks Adriank and DH. How does this formula know how oblate Earth is? Is it in C 2,0? For C 2,0, do I use .00108263, or the negative of that? Is this the number that describes how oblate Earth is?

Private Sub Command1_Click()
    Dim U As Double, G As Double, M As Double, a As Double, r As Double, C20 As Double
    Dim U2 As Double
    G = 6.6725985E-11: M = 5.97369125232006E+24: a = 6378135: r = 100000000#: lambda = 0.1: phi = 0.1: C20 = 0.00108263
    
    U = (G * M / r ^ 2) * (1 + (a / r) ^ 2 * P20(Sin(phi)) * C20 * Cos(2 * lambda))
    U2 = G * M / r ^ 2 'point mass
End Sub

Private Function P20(x As Double) As Double
    P20 = 0.5 * (3 * x ^ 2 - 1)
End Function

In the above code, I squared the first instance of r to get acceleration, rather than potential. Is that valid to do in this equation? It seems to behave like I want, with the answer approaching the point mass approximation as r becomes large.