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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

By Newtons law of gravitation, an object of mass *m* at distance *R* from the centre of the Earth experiences an attractive gravitational force *F* given by

where *G* is the gravitational constant and *M* is the Earths mass. So if *g* is the acceleration of free fall,

i.e.

Thats the formula relating *g* and *R*. Now, since the Earth is not a perfect sphere, *R* varies with location on Earth; hence *g* also varies from place to place. At the Equator (where

Lets try and calculate *g* at different places by assuming that the Earth is an oblate ellipsoid (i.e. the cross section of the Earth in a plane passing through a longitude is an ellipse that is flatter horizontally than vertically). In particular, we want to determine the acceleration of free of fall *g*[sub]θ[/sub] at latitude θ (N or S).

Let *E* be the Earths radius at the Equator and *P* the Earths ratius at the Poles. Then the equation of the cross-sectional ellipse is

At latitude θ, if the distance to the centre of the Earth is *R*[sub]θ[/sub], we have

Substituting this into the equation of the ellipse gives

Hence

i.e.

Neat formula, eh? For example, the value of *g* in London (latitude 51.5°N) is 9.841 m s[sup]−2[/sup], while the value of *g* in Melbourne (latitude 37.8°S) is 9.801 m s[sup]−2[/sup].

Furthermore, differentiating *g*[sub]θ[/sub] with respect to θ gives

This is positive for θ between 0° and 90°; hence *g*[sub]θ[/sub] is a strictly increasing function of θ between the Equator and the Poles proving that the further you live from the Equator, the more overweight you apparently become!

The above is a summary of the discussion in this thread: http://z8.invisionfree.com/DYK/index.php?showtopic=426

*Last edited by JaneFairfax (2008-03-03 10:32:27)*

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