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Accepted value of G with longitude

  1. Sep 6, 2007 #1
    Hey I have a lab write up due, and on it I need the accepted value of g and to get it at my longitude, any idea what this means, and how I can figure it out
  2. jcsd
  3. Sep 7, 2007 #2
    Your longitude [by which, we'll pretend you actually said latitude]? Do you mean altitude, or do you have some experiment that can distinguish centripetal acceleration from gravitational acceleration (without being sensitive to geological variations specific to your location)?

    I suggest "10ms^-2".
    Last edited: Sep 7, 2007
  4. Sep 7, 2007 #3


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    Here is a link:

    http://geophysics.ou.edu/solid_earth/notes/potential/igf.htm [Broken]
    Last edited by a moderator: May 3, 2017
  5. Sep 7, 2007 #4
    thanks man so its 9.7803267714 right? and then the other equation next to it with lambda is to fix for longitude?
  6. Sep 7, 2007 #5


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    No lambda is latititude, there is no theoretical reason to fix for longitidue.
    Note that the equation is an avaerage over the earth, after about the 3rd decimal place the local geology has a larger effect, so you probably don't want to quote more sig figures.
  7. Sep 7, 2007 #6
    Can someone knock up a quick example to show the OP how ridiculous the number of SF he has used there is?

    What kind of mass corresponds to a 0.0000000001 m/s/s change in g? A pebble? A boulder?

    It's something i'd be interested to know. And I think that it's important that the OP has it hammered home how carefully you have to handle SF.
  8. Sep 8, 2007 #7
    Using Newton's equation of gravitation g = -(Gm) / r^2

    G = Gravitational Constant r = Distance from centre of body g = "Force" due to gravity m = mass

    Assuming the radius of the object is 1 meter

    0.0000000001 = (6.67300 × 10 ^ -11 x mass) / ( 1 ^ 2)

    Mass = 0.0000000001 / 6.67300 × 10 ^ -11

    Mass = 1.4985763 kg

    So yeah a large pebble....
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