7.2.3 Earth’s Gravitational Field

From outer space, the Earth looks round. On Earth, however, the Earth looks flat. Similarly, on a cosmic scale, the Earth’s gravitational field is radial and non-uniform in strength. But on a terrestrial scale, it is vertically downward and uniform in strength.

Modeling the Earth as a uniform sphere of radius  {{R}_{E}}=6370\text{ km} and mass M=5.97\times {{10}^{{24}}}\text{ kg}, we can calculate the theoretical value of g on the surface of the Earth to be

\displaystyle g=\frac{{GM}}{{{{R}_{E}}^{2}}}=\frac{{(6.67\times {{{10}}^{{-11}}})(5.97\times {{{10}}^{{24}}})}}{{{{{(6370\times {{{10}}^{3}})}}^{2}}}}=9.81\text{ N k}{{\text{g}}^{{-1}}}.

9.81 m s-2 is of course the published value for the acceleration of free fall. This is no coincidence because gravitational field strength is also the gravitational acceleration. An object free-falling on Earth’s surface must thus, in theory, accelerate at g=9.81\text{ m }{{\text{s}}^{{-2}}}. In practice, however, 9.81 m s-2 is only the average value of g. The main reasons for g to vary are:

1) Non-uniform density

The Earth is not uniform in composition. An oil field would cause the local g to be lower, but a mineral ore would cause the local g to be higher.

2) Non-uniform radius

The Earth is not exactly a sphere. The radius of Earth is 6,378 kilometers at the equator and 6,356 km at the poles – a difference of 22 km. The bulge at the waistline results in a general trend of g decreasing from the poles to the equator.

3) Earth’s rotation

The Earth’s rotation actually causes the acceleration of free fall to be less than the gravitational acceleration, further contributing to the trend of g decreasing from the poles to the equator. The reason for this will be discussed in detail in the next section.

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