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You are not logged in. #1 20080303 04:51:07
Variation of acceleration of free fall with latitudeBy 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 g_{θ} 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 crosssectional ellipse is At latitude θ, if the distance to the centre of the Earth is R_{θ}, 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^{−2}, while the value of g in Melbourne (latitude 37.8°S) is 9.801 m s^{−2}. Furthermore, differentiating g_{θ} with respect to θ gives This is positive for θ between 0° and 90°; hence g_{θ} 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 (20080304 09:32:27) 