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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?

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

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

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