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Centripetal acceleration at equator

  1. Jan 25, 2010 #1
    The centripetal acceleration of a person standing at the equator is about 0.03 m s–2.

    (Radius of the Earth = 6.4 × 106 m.)

    The size of the force R provides a measure of the apparent strength of the gravitational field. Show that the apparent field strength g at the equator differs from that at the poles by about 0.3%.

    The answer is:

    (0.034 ÷ 9.81) × 100% = 0.35%

    I dont understand this. How do we know that the acc at the pole is 9.81, and that 0.034 is the difference in the 2 accerelations?
  2. jcsd
  3. Jan 25, 2010 #2


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    The acceleration due to gravity 'g' you can measure.
    You can also work it out from GM/r^2 ( M is the mass of the Earth and r is the radius)

    ps. 9.81m/s^2 - is an average for the whole planet,
    g is around 9.79 m/s^2 at the equator and 9.83 m/s^2 at the poles - it also varies with local rock types
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