1. Not finding help here? Sign up for a free 30min 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!

Rotation of the Earth and Apparent Weight?

  1. Oct 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Because of Earth’s rotation about its axis, a point on the Equator experiences a centripetal acceleration of 0.034 m/s2, while a point at the poles experiences no centripetal acceleration.
    What is the apparent weight at the equator of a person having a mass of 118.1 kg? The
    acceleration of gravity is 9.8 m/s^2 .

    Answer in units of N.

    2. Relevant equations


    3. The attempt at a solution

    So, at first this problem greatly confused me. Now I think I have an idea of how to approach it.


    This equation describes the net force toward the center of the Earth. Gravity acts toward the center of the Earth, while the normal force will resist it. If a person were to be standing on a scale, it would be the normal force that would produce the "apparent" reading. So then:



    F[tex]_{normal}[/tex]=1153.3646 N

    Is this line of thinking correct? Thanks in advance for any help provided.
  2. jcsd
  3. Oct 22, 2010 #2
    Yes that is correct.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Rotation of the Earth and Apparent Weight?
  1. Apparent Weight (Replies: 1)

  2. Apparent weight (Replies: 2)