Math Is Fun Forum

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

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 g[sub]θ[/sub] 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 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)