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Gravitational field, g

  1. Oct 11, 2005 #1
    Hi,
    This might be a dumb/stupid/ignorant/*insert name tag here* topic, but I was wondering-- what is a fairly accurate numerical representation of g?
    Most of my books use 9.8m/s^2, Fundamentals of Physics mentions 9.80665m/s^2 and I've heard some books use 9.82m/s^2...
    What is the most accurate accepted value? (Also, is there some power series to represent g?:rolleyes:)
    Thanks.
     
  2. jcsd
  3. Oct 11, 2005 #2

    Tom Mattson

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    You have to remember that [itex]g[/itex] is local (its value varies with your radial distance from the center of the Earth). It is given by:

    [tex]g=\frac{GM_{Earth}}{R_{Earth}^2}[/tex]

    So to find out how accurately [itex]g[/itex] is known, you need to know how accurately [itex]G[/itex], [itex]M_{Earth}[/itex], and [itex]R_{Earth}[/itex] are known.

    And no, you can't get [itex]g[/itex] from a power series.
     
  4. Oct 12, 2005 #3

    dextercioby

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    The value 9.80665 m/s^2 is the standard, exact (just like the speed of light) value of "g". Tom's formula assumes a spherical, nonrotating Earth.

    Daniel.
     
  5. Oct 12, 2005 #4

    David

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    As Tom points out, the value is local. But it depends on more than just your radius because there are local effects that distort the gravitational field. Indeed, detecting these distortions is one goal for people looking for deposits of minerals, ore, oil, etc.

    The varying can be quite significant. For example, in my undergrad university on teh second floor of hte physics building, the local value of g was 9.79616 m/s^2 (I think that was the number - after a multi-week-long Kater's pendulum experiment to measure it and make all sorts of corrections for systematics!) At my grad school, a thousand kilometres away, the value was closer to 9.81 m/s^2, I believe.
     
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