1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Centripetal acceleration at equator
Loading...