A body is weighed by a spring balance to be $1.000~kg$ at the north pole. How much will it approxima

Question

A body is weighed by a spring balance to be $1.000~kg$ at the north pole. How much will it approximately weigh at the equator? (Account for the earth's rotation only, assume $R = 6400~km$, $g = 9.8~m~s^{-2}$)

Professor PC Thomas Classes · Answer

At the pole, $g_p = g$. At the equator, $g_{eq} = g - R\omega^2$. Change in $g = 6.4 	imes 10^6 	imes (\frac{2\pi}{24 	imes 3600})^2 \approx 0.034~m~s^{-2}$. The new weight is $1.000 	imes \frac{9.8 - 0.034}{9.8} \approx 0.997~kg$.