Question:
A body is weighed by a spring balance to be $1000 \mathrm{~kg}$ at the north pole. How much will it weigh at the equator? Account for the earth's rotation only.
Solution:
Let $g_{p}$ be the gravity at the poles , $g_{e}$ be the gravity at the equator
$g_{e}=g_{p}-\omega^{2} R$
$=9.81-\left(7.3 \times 10^{-5}\right)^{2} \times\left(7.3 \times 10^{-5}\right)^{2} \times 6400 \times 10^{3}$
$=9.766 \mathrm{~N} / \mathrm{m}^{2}$
Now, $\mathrm{mg}_{\mathrm{e}}=1 \mathrm{~kg} \times 9.766 \mathrm{~N} / \mathrm{m}^{2}$
$=9.766 \mathrm{~N}$
Thus the body weighs $9.766 \mathrm{~N}$ at the equator.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.