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#1 2008-03-03 04:51:07

JaneFairfax
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Variation of acceleration of free fall with latitude

By Newton’s 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 Earth’s mass. So if g is the acceleration of free fall,



i.e.



That’s 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
),
while at the Poles (
),
.

Let’s 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 gθ at latitude θ (N or S).

Let E be the Earth’s radius at the Equator and P the Earth’s 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 Rθ, we have



Substituting this into the equation of the ellipse gives



Hence



i.e.

                         


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

Furthermore, differentiating gθ with respect to θ gives



This is positive for θ between 0° and 90°; hence gθ 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! tongue

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

Last edited by JaneFairfax (2008-03-04 09:32:27)


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