# Earth as an ellipsoid

Hi everybody.

I need help. I had model the planet Earth as a perfect sphere.

Now I want to model planet Earth as an ellipsoid. However I got stump by a mystery.

If Earth is an ellipsoid and I am standing at point P (see diagram), then where is my Zenith as far as Astronomy is concerned?

Answer1 is my feet points to (xc,yc) ie. the center of the Earth

Answer2 is my feet points to (xn,yn) ie the normal to the surface of the ellipsoid

I am really confused.

Regards

This is not really a Julia language problem. You need a little background in geodesy. Look up “earth ellipsoid” and “geodetic coordinates” in Wikipedia. You might also consult some books on astrodynamics which treat these ideas. Example: Fundamentals of Astrodynamics and Applications, by David A. Vallado.

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A zip with the corresponding PDF can be downloaded from the author’s public website.

According to the reference above, that is the difference between geodetic and geocentric latitude. In celestial navigation, the zenith distances and altitudes refer to the local plumb line, which aligns itself with gravity and points to the astronomical zenith. Assuming a homogeneous mass distribution, the plumb line coincides with the local normal to the ellipsoid which points to the geodetic zenith. Astronomical and geodetic zenith are then identical.

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might be better, because the zenith concept originates from the position of the Sun for an observer on the Earth. The Sun is at zenith when it is directly overhead. To maintain this interpretation, Answer2 i.e. normal to surface is preferable.

Wow, that’s interesting. I know it’s a quirk (or deep consequence) of the inverse square law that the gravitational attraction of a uniform sphere is the same as a point mass at the sphere’s center outside the sphere, and that at any point inside a spherical shell the gravity from opposing portions cut off by a cone cancels exactly. But I am a little surprised that the gravitational center of an ellipse is the ellipse center. Perhaps it follows simply from the same inverse-square oppositional principle,

Wow, that’s interesting. I know it’s a quirk (or deep consequence) of the inverse square law that the gravitational attraction of a uniform sphere is the same as a point mass at the sphere’s center outside the sphere, and that at any point inside a spherical shell the gravity from opposing portions cut off by a cone cancels exactly. But I am a little surprised that the gravitational center of an ellipse is the ellipse center. Perhaps it follows simply from the same inverse-square oppositional principle,

The inverse square law applies BUT you forgot that the earth is SPINNING.

To the total force (of person at point P) is Gravity_vector + Centripetal_vector

That is why the total force points to the (xn,yn) instead of the center of the Earth

Perhaps the best way to think about it is to IMAGINE if the earth is made out of water (ie. water world). So what would the shape of the earth/waterworld be? It would be an ellipsoid so the answer2 is CORRECT because otherwise the water at point P would be rising up against the Gravitional Potential on one side (and going down the Gravitional Potential on the OTHER side) and this is of course ridiculous.

Postscript: I would prefer the term “centre of the earth” as the term “center of the earth” makes me imagine there is a shopping center on earth somewhere.

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For a general ellipsoid of revolution not rotating, the gravity force at surface may point somewhere else. See this reference.

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