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Homework Help: Weight at the North Pole vs. Equator Problem

  1. Dec 7, 2011 #1
    1. Suppose Earth is a perfect sphere of radius 6,370km. If a person weighs exactly 600N at the north pole, how much will the person weigh at the equator?


    r = 6,370km = 6,370,000m
    Weight at north pole = mg = 600N

    2. Relevant equations

    ω[itex]_{av}[/itex] = θ/t

    ƩF[itex]_{net}[/itex] = ma

    a[itex]_{c}[/itex] = v[itex]^{2}[/itex]/r

    3. The attempt at a solution

    Okay, so the first thing I did was to draw a picture. Then I noticed that ω[itex]_{av}[/itex] would be 0 at the North Pole because you don't have any rotational velocity! Then I set up a Free Body Diagram for a location at the equator and see that the normal force (N) is acting up perpendicular to the ground, and gravity is acting downward towards the center of the Earths gravity.

    I get

    ƩF[itex]_{y}[/itex] = ma[itex]_{y}[/itex]

    where a[itex]_{y}[/itex] is the centripetal acceleration a[itex]_{c}[/itex]


    ƩF[itex]_{y}[/itex] = mv[itex]^{2}[/itex]/r = mrω[itex]^{2}[/itex]

    and substituting in N and mg for ƩF[itex]_{y}[/itex]

    N-mg = mrω[itex]^{2}[/itex]

    since we are looking for N

    N = mrω[itex]^{2}[/itex]+mg

    and so when I substitute the values of m, r, ω, and g I get

    N = 61.16kg(6.37 x 10[itex]^{6}[/itex]m)(7.27 x 10[itex]^{-5}[/itex]rad/s)[itex]^{2}[/itex] + 61.16kg(9.81m/s[itex]^{2}[/itex])

    solving for N I get 602.06N

    But I thought that you weighed less at the equator? I believe I missed something fundamental, can someone please help me out here? Thank you for your time.
  2. jcsd
  3. Dec 7, 2011 #2


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    Science Advisor
    Gold Member

    It should be mg-N=mrω^2, since the centripetal acceleration points inwards towards the center, and so does mg, but not N.
  4. Dec 8, 2011 #3
    Oh! So at that point I should take downward as positive and solve for N = mg-mrω^2? That makes more sense I don't know why I didn't remember that the centripetal acceleration would be pointed towards the center of gravity as well, thank you.
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