Question
A body is weighed by a spring balance to be 1.000kg at the north pole. How much will it weigh at the equator? Account for the earth's rotation only.
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Answer
Let $g^{\prime}$ be the acceleration due to gravity at equation $\&$ that of pole $=g g^{\prime}=g-\omega^2 R$ $=9.81-\left(7.3 \times 10^{-5}\right)^2 \times 6400 \times 10^3=9.81-0.034=9.776 \mathrm{~m} / \mathrm{s}^2 \mathrm{mg}^{\prime}=1 \mathrm{~kg} \times 9.776 \mathrm{~m} / \mathrm{s}^2=9.776 \mathrm{~N}$ or $0.997 \mathrm{~kg}$ The body will weigh $0.997 \mathrm{~kg}$ at equator.
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