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Variation of gravitation field strength?

  1. Jul 27, 2010 #1
    Hi all, this will be my first physics qns:D more to come...

    I just learnt that value of g at the equator is not exactly equal to the gravitational field strength. Can anyone explain with workings? i don't really understand my teacher workings.
    Also can i clarify if

    -We're assuming earth is perfectly sphere hence Radius (r) is same throughout and Angular speed is the same throughout the earth?

    *i suspect there is a printing error on my notes which makes cause my confusion over the understanding of this part hence getting some other people's workings ...
  2. jcsd
  3. Jul 27, 2010 #2
    IF the Earth were a perfect sphere, then g would have the same value at any point on its surface. In fact, though, the Earth is an oblate spheroid. In practical terms, this means there is less mass between your feet and the center of the Earth when you stand at the North Pole than when you stand at the Equator. This need not work out to a higher g on the equator--it's a question of distance as well as mass, so density figures in--but in our case, it does.

    The Earth's angular momentum will have a marginal effect on your scale if you are weighing something at the Equator, but it is important to note that this dynamic factor is NOT considered when calculating g.
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