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    The Variation of Acceleration Due to Gravity on Earth

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    A body weighs 98N on a spring balance at the North Pole. What will be its weight recorded on the same scale if it is shifted to the equator?

    Radius of earth is 6400km

    © BrainMass Inc. brainmass.com December 24, 2021, 4:43 pm ad1c9bdddf
    https://brainmass.com/physics/solar-system/variation-acceleration-gravity-earth-weight-4132

    SOLUTION This solution is FREE courtesy of BrainMass!

    At the poles, the apparent weight is the same as the true weight.

    Thus, 98N = mg = m (9.8 m/s^2)

    or m = 10kg. At the equator, the value of g will be

    g1 = g - w^2 R

    Where w is the angular speed of the earth.

    w = 2pi/(24 x 60 x 60) rad/s
    = 7.27x10^-5 rad/s

    Thus g1 = 9.8 - [7.27x10^-5 ]^2 x R

    Where R = 6400 x 10^3 m

    And the weight of the object at the equator will be,

    = m x g1
    = 10 x {9.8 - [7.27x10^-5 ]^2 x R }
    = 97.6 Newton

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 4:43 pm ad1c9bdddf>
    https://brainmass.com/physics/solar-system/variation-acceleration-gravity-earth-weight-4132

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