Calculating Earth's Shape as an Ellipsoid

In summary, an engineering student's professor gave a homework to calculate the real shape of the Earth (as an ellipsoid), considering centrifugal force. They used the equation ∇p + ρg + ρa = 0 and solved for Earth as a sphere. However, the professor said that there are different types of ellipsoids, so the student attempted to find an equation for an ellipsoid that produces an approximate ellipse in the x-z plane.
  • #1
norsktramp
4
0

Homework Statement



I'm an engineering student, and my professor of the mechanics course gave a homework to my class last week, we were intended to calculate the real shape of the Earth (as an ellipsoid) by taking the centrifugal force in account, using the equation [itex]a' = a -wX(wXr)[/itex]. For that, we also used Euler's equation [itex]∇p + ρg + ρa = 0 .[/itex] My professor then solved the problem considering Earth as a sphere, as I'm going to show it below. So my question is, how can I calculate the shape of the Earth (as an ellipsoid).

Variables and meaning : ∇p = gradient of pressure, a' = aceleration in a non inercial reference, w = angular velocity, g = gravity. (all of them are vectors)


Homework Equations



He told us that in a non inercial reference, the imaginary acelerations could be considered as a field aceleration, so he combined Euler's formula with the centrifugal force, by substituting it in the place of gravity. so we got:[tex]∇p -ρ[wX(wXr)] + ρa = 0.[/tex]

a = GM/r2 runitary = GM/r3 rvector,[tex]r^2 = x^2 + z^2[/tex].

considering w in the z axis we will obtain a relation for ∂p/∂x and ∂p/∂z, then we will solve the equation obtaining: [tex]p(x,z) = -ρGM/(√x^2 + z^2) -ρw^2*x^2/2 + C[/tex] C being a constant.

considering p as: [itex]p(0,R) = po = -ρGM/R + C[/itex], we obtain [itex]C = po + ρGM/R[/itex]
so [tex] p(x,z) = -ρGM/(√x^2 + z^2) -ρw^2*x^2/2 + po + ρGM/R[/tex]
To find the formula of the surface of the Earth we got to make p = po.
Now, making an assumption that w → 0 (not true for the earth), we will obtain [itex]ρGM/(√x^2 + z^2) = ρGM/R[/itex] [tex]x^2+z^2 = R^2[/tex] Obtaining a sphere equation.


The Attempt at a Solution


I want to know how to calculate the Earth's shape not considering w → 0, in a way I will obtain an ellipsoid. I also know a lot of things are mispelled and with a dificult to read format, but I ask for your patience and help. If you need any extra information please ask for it.
 
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  • #2
So basically, you need to follow his method and obtain the equation for an ellipsoid.

But there are different types of ellipsoids, so which is it? I would believe that Earth looks mostly like an oblate spheroid, and since you wrote real shape, would that be it? Or is it just a normal ellipsoid?

Anyhow, how much of his example, do you think, you can use? I mean could you approximate the radius of the spherical Earth to be equal to the length, c, as in this picture:

http://en.wikipedia.org/wiki/File:Ellipsoid_revolution_prolate_and_oblate_aac.svg

I am still not entirely sure, how to solve it though, but got to start with getting the facts straight.
 
  • #3
I followed my professor's solution method, and calculated C = po + ρGM/R1, R1 being the small radius of the Earth's ellipsoid, then, I made p(x,z) = po, which led me to [tex]GM/R1 = w^2x^2/2 + GM/√(x^2 + z^2) [/tex]
I was hopping that the ellipse equation would emerge naturally (as the circle equation did previously), but I can't find a way to get it. I hope it clarify things up.
 
  • #4
I'm not sure whether this theory is supposed to produce an exact ellipsoid (with two equal axes), or if that's just an approximation. If you substitute x = r cos(θ), z = s sin(θ) in your equation, where s is not much less than r, you can take a first-order approximation and get an equation which, with the right value of s, gives an ellipse in the x-z plane.
 
  • #5
I'm not sure whether this theory is supposed to produce an exact ellipsoid (with two equal axes), or if that's just an approximation. If you substitute x = r cos(θ), z = s sin(θ) in your equation, where s is not much less than r, you can take a first-order approximation and get an equation which, with the right value of s, gives an ellipse in the x-z plane.

First of all, I would like the deduction of the problem to be made with no aproximations, and second, I really can't see how this is going to help me, if you could write the equation for your suggestion i would apreciate.
 
  • #6
norsktramp said:
First of all, I would like the deduction of the problem to be made with no aproximations, and second, I really can't see how this is going to help me, if you could write the equation for your suggestion i would apreciate.
You can make the substitution and see how far you get. Use the binomial theorem/Taylor expansion to turn (1+<something small>)-1/2 into (1-<something small>/2)
 
  • #7
I posted it in the introductory physics section because its really about a simple subject, but this problem is showing itself to be extremely hard to solve (at least for me). I'm really worried because this question will have some influence on my score at the college, and no one in my classroom could solve it, so, if anyone got any other suggestions, or even better, if someone could solve this problem I would be extrelly gratefull for that.
 
  • #8
norsktramp said:
I posted it in the introductory physics section because its really about a simple subject, but this problem is showing itself to be extremely hard to solve (at least for me). I'm really worried because this question will have some influence on my score at the college, and no one in my classroom could solve it, so, if anyone got any other suggestions, or even better, if someone could solve this problem I would be extrelly gratefull for that.
How far can you get with what I suggested? What part do I need to explain better?
 

Related to Calculating Earth's Shape as an Ellipsoid

1. How is the Earth's shape determined to be an ellipsoid?

The Earth's shape is determined to be an ellipsoid through various methods such as satellite measurements, geodetic surveys, and mathematical calculations. These methods take into account the Earth's rotation, gravitational forces, and surface features to accurately determine its shape.

2. Why is the Earth's shape not perfectly spherical?

The Earth's shape is not perfectly spherical due to its rotation. The centrifugal force caused by the Earth's rotation causes it to bulge out at the equator, making it slightly wider than it is at the poles. This bulging effect gives the Earth an ellipsoidal shape.

3. How does calculating the Earth's shape as an ellipsoid affect map projections?

Calculating the Earth's shape as an ellipsoid is crucial in creating accurate map projections. Since the Earth is not a perfect sphere, cartographers use ellipsoidal models to create maps that accurately depict the Earth's surface features and distances.

4. How does the Earth's shape as an ellipsoid affect navigation?

Knowing the Earth's shape as an ellipsoid is important for navigation, particularly for calculating distances and directions. Navigational tools such as GPS systems use ellipsoidal models to accurately determine locations and routes on the Earth's surface.

5. Does the Earth's shape as an ellipsoid change over time?

Yes, the Earth's shape as an ellipsoid is constantly changing due to various natural processes such as plate tectonics, sea level changes, and variations in the Earth's rotation. However, these changes are minimal and do not significantly affect the overall ellipsoidal shape of the Earth.

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